Logic & Word Problemscomplementary-countingsum-constraint
Vasily has 20 balls. Each ball is either yellow, green, blue or black. Of the balls, exactly 17 are not green, exactly 15 are not black, and exactly 12 are not yellow. How many of his balls are blue?
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Answer: D — 4
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Hint 1 of 2
Turn each "not" statement into a count of that colour.
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Hint 2 of 2
Green = 20 − 17, black = 20 − 15, yellow = 20 − 12; the rest are blue.
Show solution
Approach: complementary counting
Not green = 17 means green = 3; not black = 15 means black = 5; not yellow = 12 means yellow = 8.
The picture shows the menu of a burger restaurant. The rain has washed away some of the numbers. The burgers are listed by price in increasing order, the cheapest being the “veggie” burger. What is the smallest possible price of the “deluxe” burger?
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Answer: B — 6.80
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Hint 1 of 2
Only the cents are readable; the whole-euro parts are hidden, but the prices increase down the list.
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Hint 2 of 2
Step down the menu choosing the smallest whole-euro part that keeps each price above the one before it.
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Approach: build the cheapest increasing price chain
Kangaroo Joey wants to jump through the maze. Joey starts in the square at the bottom left (see picture). The number of arrows in a square tells how long the jump must be: a square with three arrows means Joey jumps over two squares in the direction of the arrows and lands in the third square.
Where will Joey leave the maze?
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Answer: B — B
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Hint 1 of 2
Start in the bottom-left square and follow the arrows one jump at a time, counting squares carefully.
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Hint 2 of 2
Treat each square as an instruction: the arrow gives the direction and the number of arrows gives how many squares to advance; see which edge you eventually step off.
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Approach: trace the path square by square
Begin on the bottom-left square and read its arrows: move in the arrow's direction by as many squares as there are arrows (three arrows = land in the third square).
Repeat from each square you land on, always obeying the new square's arrows.
Following the chain of jumps, Joey is carried to the edge of the grid and exits through the side marked B.
The picture shows the footprints of the first few jumps of a kind of hopscotch. The prints show whether you may land on a square with the left foot, the right foot, or both feet. Mia starts with both feet on square number 1 and repeats the jumping pattern in the direction of the arrow.
On which of the following squares is Mia allowed to land with her right foot only?
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Answer: C — square no. 20
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Hint 1 of 2
Read off the four-square footprint pattern (both / right / both / left, say) and notice it repeats every four squares.
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Hint 2 of 2
Find which position in the repeating cycle is the 'right foot only' square, then check which listed number lands on that position.
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Approach: find the repeating cycle of footprints
The footprints in squares 1-4 set a pattern that then repeats every four squares in the arrow's direction.
Locate the position within that 4-square cycle that calls for the right foot only.
Each candidate number lands on a fixed spot in the cycle; only square no. 20 lines up with the right-foot-only position.
Mona wants to draw the figure shown without lifting her pen. The lengths of the segments are given. Mona can start anywhere and may go over segments more than once. What is the minimum distance, in centimetres, that Mona has to move her pen across the paper?
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Answer: B — 7 cm
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Hint 1 of 2
If you could draw the whole figure in one stroke without repeating, the pen distance would just be the total length of all the segments.
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Hint 2 of 2
Because some corners have an odd number of segments meeting, you must go over one short segment a second time; add that retraced length to the total.
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Approach: total segment length plus the unavoidable retraced piece
Add the lengths of all the segments in the figure: 3 + 2 + 1 = 6 cm if nothing is repeated.
A non-stop drawing can avoid repeats only when at most two corners have an odd number of segments; here too many corners are odd, so one segment must be traced twice.
The cheapest segment to repeat is the 1 cm piece, adding 1 cm: 6 + 1 = 7 cm.
Firefighter Fred wants to put out the fire. In the picture on the right, what is the smallest number of ladders he has to climb to reach the fire without jumping?
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Answer: C — 6
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Hint 1 of 2
Start at the firefighter and find every platform you can reach by climbing whole ladders, never jumping.
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Hint 2 of 2
Trace the route to the fire that uses the fewest ladders — each ladder climbed counts as one.
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Approach: trace the shortest ladder path from the firefighter up to the fire
Begin at the firefighter on the ground and only move along complete ladders between platforms.
Follow the chain of ladders upward that reaches the fire's platform.
Counting each ladder used, the shortest such route uses 6 ladders.
Five children each light a candle at the same time. Lisa blows out the candles at different times. Now they look as shown in the picture. Which candle did Lisa blow out first?
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Answer: D — D
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Hint 1 of 2
A candle that was blown out earlier has had more time to burn down, so it is shorter now.
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Hint 2 of 2
Find the shortest candle — that is the one extinguished first.
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Approach: compare burn lengths: longest burning means shortest now
All five candles were lit at the same moment.
The one blown out first kept burning the longest, so it is the shortest stub in the picture.
Matchsticks are arranged to form digits, as shown. For example, the number 15 needs 7 matchsticks, and so does the number 8. What is the biggest number you can build using exactly 7 matchsticks?
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Answer: D — 711
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Hint 1 of 2
Count the matchsticks each digit uses from the picture, then add up the sticks in each answer.
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Hint 2 of 2
Keep only the answers that use exactly 7 matchsticks, then take the largest of those.
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Approach: count sticks per digit, keep the total of 7, choose the biggest
From the picture, the digit 7 uses 3 sticks and the digit 1 uses 2 sticks.
So 711 uses 3 + 2 + 2 = 7 sticks; 31, 51 and 74 also each use 7 sticks, but 800 needs far more.
Among the numbers built from exactly 7 matchsticks, the largest is 711.
Maria switches the lights on and off according to the given plan (the bars show when each of Light 1, Light 2 and Light 3 is on, in minutes). For how many minutes in total are there exactly two lights on at the same time?
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Answer: C — 8
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Hint 1 of 2
Go minute by minute and count how many lights are on.
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Hint 2 of 2
Mark only the minutes where the count is exactly two.
Show solution
Approach: scan each minute and tally when exactly two bars overlap
Read each light's on-intervals from the chart and check every one-minute slot.
Counting the slots where precisely two of the three bars are present gives 8 such minutes.
Logic & Word ProblemsArithmetic & Operationssum-constraintcasework
Evita wants to write the numbers from 1 to 8, with one number in each field. The sum of the numbers in each row should be equal. The sum of the numbers in each of the four columns should also be the same. She has already written in the numbers 3, 4 and 8 (see diagram). Which number does she have to write in the dark field?
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Answer: E — 7
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Hint 1 of 2
The numbers 1..8 add to 36; use that to find each row sum and each column sum.
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Hint 2 of 2
Fit the remaining numbers around the given 3, 4 and 8 so every row and every column hits its target.
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Approach: use the fixed total to pin row/column sums, then place numbers
1 + 2 + ... + 8 = 36; with two equal rows each row sums to 18, and with four equal columns each column sums to 9.
Place the remaining numbers so each column totals 9 and each row totals 18, respecting the given 3, 4 and 8.
One of the five coins A, B, C, D or E should be moved to an empty square so that each row and each column ends up with exactly two coins. Which coin should be moved?
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Answer: C — C
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Hint 1 of 2
Count how many coins are in each row and in each column right now.
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Hint 2 of 2
Find the one row that has too many and the one column that has too many; the coin sitting where they cross is the one to move.
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Approach: balance rows and columns
Counting coins, one row has three coins (too many) and one row has only one (too few).
Likewise one column has three coins and another has only one.
The coin that sits in BOTH the overloaded row and the overloaded column is the one to move.
That coin is C; moving it to the empty cell of the short row and short column fixes every count to two.
In the four squares of a row there always have to be exactly two coins. In the four squares below each other there also always have to be exactly two coins. On which square does one more coin have to be placed?
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Answer: D — square D
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Hint 1 of 2
Count the coins in each row and in each column - which one still has only one?
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Hint 2 of 2
One row and one column are each short a coin; they meet at a single empty square.
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Approach: find the row and column that are short a coin
The third row down has only 1 coin, and the third column across also has only 1 coin.
Both still need one more coin to reach two.
The empty square where that row and column meet is square D, so the coin goes there.
Sofie wants to write the word KENGU by using letters from the boxes. She can only take one letter from each box. What letter must Sofie take from box 4?
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Answer: D — G
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Hint 1 of 2
Find the boxes that can only supply one needed letter and lock those in first.
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Hint 2 of 2
Box 3 forces the N, and the only K is in box 2 — then see what is left for box 4.
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Approach: fix the forced letters, then deduce box 4
Box 3 contains only N, so KENGU's N must come from box 3.
K appears only in box 2, so K comes from box 2.
That leaves E, G, U for boxes 1, 4, 5; box 1 gives E and box 5 gives U, so box 4 must give G.
Logic & Word Problemsbalance-reasoningwork-backward
Rosana has some balls of 3 different colours. Balls of the same colour have the same weight. What is the weight of each white ball?
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Answer: C — 5
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Hint 1 of 2
Look at the two scales that both have a black ball and a grey ball on them, and compare what is different.
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Hint 2 of 2
Find the grey ball's weight first, then the black ball's, and only then the white ball.
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Approach: compare the scales to find one colour at a time, then the white ball
One grey and one black ball together weigh 6, while one black and two grey balls together weigh 10.
The second scale is just the first scale with one extra grey ball, and it is 10 - 6 = 4 heavier, so one grey ball weighs 4 (and the black ball weighs 6 - 4 = 2).
Two white balls plus one grey ball weigh 14, so the two white balls weigh 14 - 4 = 10, which means each white ball weighs 5.
Byron is 5 cm taller than Aaron, but 10 cm shorter than Caron. Darren is 10 cm taller than Caron, but 5 cm shorter than Erin. Which of the following statements is true?
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Answer: E — Aaron is 30 cm shorter than Erin
Show hints
Hint 1 of 2
Anchor everyone to Aaron's height and step through the comparisons one at a time.
Counting & ProbabilityLogic & Word Problemscasework
Three soccer teams compete in a championship. Each team plays exactly once against each of the other teams. In each match the winning team earns 4 points and the losing team loses 1 point; in case of a tie, each team earns 2 points. Once the championship is over, what is the largest possible sum of the points obtained by the three teams?
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Answer: E — 12
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Hint 1 of 2
There are only three matches; figure out the point total a single match can produce.
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Hint 2 of 2
Compare a decisive match (4 and −1) with a tie (2 and 2) — which gives the bigger combined total?
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Approach: maximize the total per match
Three teams each playing each other once means 3 matches in total.
A decisive match adds 4 + (−1) = 3 points; a tie adds 2 + 2 = 4 points.
Ties give more, so make all three matches ties: 3 × 4 = 12 points.
Logic & Word ProblemsArithmetic & Operationsmagic-squaresum-constraintwork-backward
Juca wrote a whole number greater than zero in each box of the 3×3 board shown, so that the sums of the numbers in each row and in each column are equal. The only thing Juca remembers is that no number is used three times. What number is written in the center box?
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Answer: C — 4
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Hint 1 of 2
All three rows and columns share the same total; the top row 1 + 2 + 6 gives that total.
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Hint 2 of 2
Fill the grid from the common sum of 9, using that no value may appear three times.
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Approach: use the common row/column sum
Each row and column sums to 1 + 2 + 6 = 9.
The left column 1 + 3 + ? = 9 forces the bottom-left to be 5.
Filling in keeps the sums at 9; the center value can be 4 (center 5 would repeat a number three times, which is forbidden).
In a nursery group there are 14 girls and 12 boys. Half of the group go for a walk. What is the minimum number of girls that have to be amongst that group?
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Answer: E — 1
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Hint 1 of 3
First find how many children go for the walk: it is half of the whole group.
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Hint 2 of 3
To make girls as few as possible, imagine sending all the boys first.
Still stuck? Show hint 3 →
Hint 3 of 3
Once the boys are used up, see how many walking spots are still empty.
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Approach: fill the walking group with boys first
There are 14 + 12 = 26 children, so half of them, 13, go for the walk.
We want as few girls as possible, so picture sending boys first: there are only 12 boys, and they fill 12 of the 13 spots.
That leaves 13 − 12 = 1 spot, which must go to a girl, so the smallest number of girls is 1 (E).
There are 12 children in front of a zoo. Susi is the 7th from the front and Kim is the 2nd from the back. How many children are there between Susi and Kim?
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Answer: B — 3
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Hint 1 of 2
Find Susi's and Kim's spots in the line first, counting from the same end.
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Hint 2 of 2
Once you know both positions, count only the children strictly in between them.
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Approach: place both children by position, then count the gap
Counting from the front, Susi is in spot 7.
Kim is 2nd from the back of 12, so Kim is in spot 11.
The children between them sit in spots 8, 9 and 10.
There are two kinds of camels: bactrian camels that have 2 humps, and dromedaries that have 1 hump. Exactly 10 camels live in a certain zoo. Together they have 14 humps. How many bactrian camels are there in this zoo?
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Answer: D — 4
Show hints
Hint 1 of 2
If all 10 camels had just 1 hump, how many humps would that be?
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Hint 2 of 2
Every bactrian camel adds one extra hump beyond that; the extra humps tell you the count.
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Approach: start from all one-hump and add the extra humps
If all 10 camels had 1 hump each, that would be 10 humps.
There are 14 humps, so there are 4 extra humps.
Each bactrian camel has one extra hump, so there are 4 bactrian camels.
The diagram shows the calendar page of a certain month, but ink has run across parts of the page. Which day of the week does the 27th of that month fall on? (The weekday columns read Mon, Tue, Wed, Thu, Fri, Sat, Sun.)
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Answer: A — Monday
Show hints
Hint 1 of 2
Read the two visible dates to figure out which weekday column the month starts in.
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Hint 2 of 2
Once you know the weekday of the 1st, count forward 26 days using remainders mod 7.
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Approach: anchor the calendar, then step the weekday forward by a multiple of 7
The clear cells show Thursday is the 2nd and Friday the 3rd, so the 1st falls on Wednesday.
The 1st, 8th, 15th and 22nd are all Wednesdays.
From the 22nd, five more days (23,24,25,26,27) land on Monday.
Mike sets the table for 8 people: the fork has to lie to the left and the knife to the right of the plate. For how many people is the cutlery set correctly?
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Answer: A — 5
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Hint 1 of 2
Sit in each chair in turn; the setting is right only if the fork is on your left hand and the knife on your right hand.
Still stuck? Show hint 2 →
Hint 2 of 2
Remember that someone sitting on the far side of the table faces you, so their left and right are flipped from yours.
Show solution
Approach: stand in each person's place and check fork-on-left, knife-on-right from their seat
Picture yourself sitting in each of the 8 chairs, facing the plate.
From that seat the fork must be by your left hand and the knife by your right hand — tick the places that match.
Exactly five places pass the test, so the cutlery is set correctly for 5 people, answer A.
Ten islands are joined by 12 bridges (see the map). Every bridge is open. What is the least number of bridges that must be closed so that it is impossible to travel from island A to island B?
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Answer: B — 2
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Hint 1 of 2
You want to cut the fewest bridges so no path of open bridges remains from A to B.
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Hint 2 of 2
Look for a bottleneck: the smallest set of bridges whose removal disconnects A from B.
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Approach: find the minimum cut between A and B
Closing bridges blocks A from B only when every route between them is broken.
Search the map for the narrowest bottleneck — the fewest bridges lying on all A–B routes.
Removing those two bridges cuts every path between A and B, and one bridge is never enough.
Jane, Kate and Lynn go for a walk. Jane walks at the very front, Kate in the middle and Lynn at the very back. Jane weighs 500 kg more than Kate, and Kate weighs 1000 kg less than Lynn. Which of the pictures shows Jane, Kate and Lynn in the right order?
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Answer: A
Show hints
Hint 1 of 2
Turn the weight clues into an order: who is heaviest and who is lightest?
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Hint 2 of 2
Jane is heavier than Kate, and Lynn is heavier still, so rank all three, then match the picture's front-to-back sizes.
Show solution
Approach: order the three by weight, then match to the row
Jane = Kate + 500 and Lynn = Kate + 1000, so Lynn is heaviest, Jane is in the middle, Kate is lightest.
Walking order is Jane (front), Kate (middle), Lynn (back).
So the picture must show, front to back, a medium animal, then the smallest, then the largest.
Some girls are standing in a circle. The teacher makes them do a headcount: Bianca says one, her neighbour says two, and so on. If they count in a clockwise direction, Antonia says five. If they count in an anticlockwise direction, Antonia says eight. How many girls are forming the circle?
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Answer: C — 11
Show hints
Hint 1 of 2
Count the gaps between Bianca and Antonia going each way around the circle.
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Hint 2 of 2
The two arcs together make one full loop.
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Approach: add the two arc lengths
Clockwise, Antonia is the 5th, so 4 girls separate them that way.
Anticlockwise she is the 8th, so 7 girls separate them the other way.
Lisa has mounted 7 postcards on her fridge door using 8 strong magnets (the black dots). What is the maximum number of magnets she can remove without any postcards falling on the floor?
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Answer: C — 4
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Hint 1 of 2
A single magnet placed where two postcards overlap can hold both at once.
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Hint 2 of 2
Find the fewest magnets that still touch every postcard; the rest can be removed.
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Approach: keep the fewest magnets that still touch every postcard
Every one of the 7 postcards must keep at least one magnet on it, or it falls.
Where two postcards overlap, a single magnet can hold both at once, so the magnets that sit on overlaps do double duty.
Keeping just 4 well-placed magnets is enough to pin all 7 cards, so she can remove the other \(8 - 4 = 4\), choice (C).
Maria wants there to be a knife to the right of every plate and a fork to the left of it. To get the right order she always swaps one fork with one knife. What is the minimum number of swaps necessary?
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Answer: B — 2
Show hints
Hint 1 of 2
Mark every place where a fork is wrongly on the right or a knife wrongly on the left.
Still stuck? Show hint 2 →
Hint 2 of 2
One swap fixes one wrong fork together with one wrong knife at the same time.
Show solution
Approach: count the misplaced utensils, then pair them up
Check each place setting and mark every fork that is wrongly on the right and every knife wrongly on the left.
There are 2 such wrong utensils, and one swap trades a wrong fork for a wrong knife, fixing both at the same time.
Andrea was born sometime in the year 1997 and her sister Charlotte sometime in the year 2001. What is known for certain about the age difference of the two sisters? It is…
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Answer: E — not less than 3 years
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Hint 1 of 2
Push the two birth dates to their extremes to find the smallest and largest possible gap.
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Hint 2 of 2
Born in 1997 vs 2001 means the gap is bigger than 3 years but can fall short of 4.
Show solution
Approach: bound the age gap from both extremes
The earliest Andrea could be born is start of 1997; the latest Charlotte is end of 2001, giving an age gap just under 5 years.
The latest Andrea is end of 1997 and the earliest Charlotte is start of 2001, giving an age gap just over 3 years.
So the gap is always more than 3 years; nothing stronger is guaranteed, so it is not less than 3 years (E).
Sam paints the 9 small squares in the shape either white, grey or black. What is the minimum number he must paint over so that no two squares sharing a side have the same colour?
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Answer: A — 2
Show hints
Hint 1 of 2
Find every pair of side-touching squares that currently share a colour; those are the trouble spots.
Still stuck? Show hint 2 →
Hint 2 of 2
Try to fix several clashes at once by changing a single well-chosen square, and count the fewest squares you must repaint.
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Approach: spot the colour clashes and repaint the fewest squares to break them all
Scan the grid for neighbours that share a colour: the two grey squares touch, and the black squares touch.
Changing two carefully chosen squares is enough to separate every same-colour pair, while one change still leaves a clash somewhere.
The Mathematical Kangaroo takes place each year on the third Thursday of March. What is the latest possible date on which the competition could take place?
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Answer: D — 21 March
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Hint 1 of 2
The third Thursday is latest when the first Thursday falls as late in March as possible.
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Hint 2 of 2
The first Thursday is latest when March 1st is a Friday; then count two weeks forward.
Show solution
Approach: push the first Thursday as late as possible
The third Thursday is latest when the very first Thursday of March is as late as it can be.
If March 1st is a Friday, the first Thursday is the 7th.
Adding two more weeks gives the third Thursday on the 7th + 14 = 21st.
Anita has built fewer sandcastles than Hans but more than Stefan. Fabian has built more sandcastles than Anita and more than Hans. Bruno has built more sandcastles than Hans but fewer than Fabian. Who has built the most sandcastles?
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Answer: E — Fabian
Show hints
Hint 1 of 3
Picture the children standing in a line, tallest pile of sandcastles at the top.
Still stuck? Show hint 2 →
Hint 2 of 3
Each sentence tells you who stands above whom, so place them one at a time.
Still stuck? Show hint 3 →
Hint 3 of 3
Look for the one child who never has anybody above them.
Show solution
Approach: stand the children in order, most sandcastles at the top
Hans is above Anita, and Anita is above Stefan, so far Hans, then Anita, then Stefan.
Fabian is above both Anita and Hans, so Fabian goes near the top.
Bruno is above Hans but below Fabian, so the full line is Fabian, Bruno, Hans, Anita, Stefan.
Nobody is above Fabian, so the most sandcastles belong to Fabian.
Grey and white pearls are threaded on a string (see picture). Monika wants 5 grey pearls, but she can only pull pearls off from an end of the string, so she has to pull off some white pearls too. What is the smallest number of white pearls she has to pull off to get 5 grey pearls?
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Answer: B — 3
Show hints
Hint 1 of 2
She can only take pearls from one of the two ends, so compare the two ends.
Still stuck? Show hint 2 →
Hint 2 of 2
Pick the end where the fifth grey pearl is reached after passing the fewest white pearls, and count just those whites.
Show solution
Approach: scan inward from the better end and count the white pearls passed
Pearls come off only from an end, so check each end and stop once 5 grey pearls have come off.
Reading in from the end that reaches the fifth grey pearl soonest, only a few white pearls sit among those first five greys.
Counting just those white pearls gives a minimum of 3.
The clues describe what each child has or does (shown in the picture). Anna has one thing, Barbara gave Eva something, Josef has something, and Bob has something. Using the clues, which picture (A–E) shows Barbara?
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Answer: D
Show hints
Hint 1 of 3
Read the clues and cross off any picture that the clue gives to someone who is NOT Barbara.
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Hint 2 of 3
The clue about Barbara is the giving clue: she is the one who gave Eva her thing.
Still stuck? Show hint 3 →
Hint 3 of 3
Whoever is left after you label Anna, Josef and Bob, and who fits the giving clue, is Barbara.
Show solution
Approach: label the children you know, then the leftover picture matching the giving clue is Barbara
First label the pictures the clues hand directly to Anna, Josef and Bob.
The remaining clue says Barbara gave Eva her item, which points to one specific picture.
Counting & ProbabilityLogic & Word Problemscasework
A sack contains marbles in five different colours: 2 red, 3 blue, 10 white, 4 green and 3 black. You take marbles out of the bag without looking and without putting any back. What is the smallest number of marbles you must remove to be sure of having two of the same colour?
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Answer: E — 6
Show hints
Hint 1 of 2
Think about the worst luck possible while still avoiding a matching pair.
Still stuck? Show hint 2 →
Hint 2 of 2
This is a pigeonhole question: one of each colour first, then one more forces a repeat.
Show solution
Approach: pigeonhole / worst case
There are 5 different colours, and every colour has at least two marbles.
The worst case is drawing one marble of each colour: 5 marbles, all different.
The very next marble (the 6th) must repeat a colour.
Alex lights a candle every 10 minutes. Each candle burns for 40 minutes before going out. How many candles are burning 55 minutes after he lit the first candle?
Show answer
Answer: C — 4
Show hints
Hint 1 of 2
List when each candle is lit and how long it stays burning.
Still stuck? Show hint 2 →
Hint 2 of 2
At minute 55, check which lit candles have not yet burned out.
Show solution
Approach: track burning intervals
Candles are lit at minutes 0, 10, 20, 30, 40, 50, each burning for 40 minutes.
At minute 55: the ones lit at 0 and 10 have gone out (out by 40 and 50).
Each of the nine paths in a park is 100 m long. Anna wants to walk from A to B without using the same path twice. How long is the longest path she can choose?
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Answer: C — 700 m
Show hints
Hint 1 of 3
Try to walk along as many paths as you can without repeating one.
Still stuck? Show hint 2 →
Hint 2 of 3
Count how many paths meet at each junction — a route can pass straight through a junction only if an even number of paths meet there.
Still stuck? Show hint 3 →
Hint 3 of 3
Anna's start A and finish B are different corners, so she can't end where she began — check whether using all paths is even possible.
Show solution
Approach: count odd junctions (Euler trail)
Each junction needs paths to enter and leave in pairs; only the start and the finish are allowed to have an odd number of paths meeting them.
In this triforce network every one of the six junctions has an even number of paths (the three corners have 2 each, the three midpoints have 4 each), so a single route cannot start at corner A and end at the different corner B while using all 9 paths.
Leaving out the side that joins A to B (its two 100 m pieces) makes A and B the only odd junctions, and then all 7 remaining paths can be walked in one go.
In the picture on the right a number is written next to each marked point. Along every side of the large hexagon, the numbers at the points must add up to the same total. Two numbers are already filled in. Which number belongs at the point marked x?
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Answer: A — 1
Show hints
Hint 1 of 2
Every side of the hexagon must reach the same total, and two corner numbers are already fixed.
Still stuck? Show hint 2 →
Hint 2 of 2
Lock the common side-total from a side whose numbers you know, then carry it around to the side that contains x.
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Approach: use the equal side-sums to chase the value around to x
Every side of the large hexagon must carry the same point-total, and the sides through the corners 4 and 1 let you pin that common total.
Following the constraint from corner to corner around the hexagon forces each remaining value in turn.
Three racers take part in a Formula-1 race: Michael, Fernando and Sebastian. At the start Michael leads, ahead of Fernando, who is ahead of Sebastian. During the race Michael and Fernando overtake each other 9 times, Fernando and Sebastian 10 times, and Michael and Sebastian 11 times. In what order do the three finish?
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Answer: B — Fernando, Sebastian, Michael
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Hint 1 of 2
Two racers swap order exactly once each time they overtake one another, so only the parity of each count matters.
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Hint 2 of 2
Odd number of swaps reverses a pair's order; even keeps it the same.
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Approach: track each pair's order by the parity of its overtakes
Start: Michael, Fernando, Sebastian.
Michael–Fernando: 9 swaps (odd) → Fernando now ahead of Michael.
Fernando–Sebastian: 10 (even) → Fernando stays ahead of Sebastian.
Michael–Sebastian: 11 (odd) → Sebastian now ahead of Michael.
So Fernando > Sebastian > Michael: order Fernando, Sebastian, Michael.
Simon awoke one and a half hours ago. In three and a half hours he will catch a train to go to his grandma. How long before his train leaves did he wake up?
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Answer: E — Five hours
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Hint 1 of 2
He woke up before now; the train leaves after now.
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Hint 2 of 2
Add the time since he woke to the time still left before the train.
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Approach: add the two time gaps
From waking to now is 1½ hours; from now to the train is 3½ hours.
So from waking to the train is 1½ + 3½ = 5 hours, answer E.
Maria describes one of these five shapes in the following way: “It is not a square. It is grey. It is either round or three‑sided.” Which shape did she describe?
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Answer: B — B
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Hint 1 of 2
Use each clue to cross shapes off the list.
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Hint 2 of 2
Keep only the shape that is not a square, is grey, and is round or three-sided.
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Approach: eliminate by each clue
“Not a square” removes the two squares (A and C).
“It is grey” removes the white circle (D); “round or three-sided” removes the grey rectangle (E).
A staircase has 21 steps. Nick and Mike count the steps, one from bottom to top and the other from top to bottom. They meet at one step, which Nick counts as the 10th. Which number does Mike give this same step?
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Answer: C — the 11th
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Hint 1 of 2
They both count the very same step, just starting from opposite ends.
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Hint 2 of 2
On a 21-step staircase, think about how many steps sit below the meeting step and how many sit above it.
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Approach: count the same step from the other end
The two boys meet on one shared step on a staircase of 21 steps.
Mike counts that step from the top: the steps above it plus the meeting step itself make up his count, which lands on the 11th.
So Mike calls the meeting step the 11th — the answer is C.
Four friends each eat some ice cream. Mike eats more than Franz, Jaroslav eats more than Veit, and Jaroslav eats less than Franz. Put the friends in order by how much ice cream they ate, starting with the largest amount.
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Answer: C — Mike, Franz, Jaroslav, Veit
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Hint 1 of 2
Turn each clue into a simple "more than" comparison and chain them together.
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Hint 2 of 2
Jaroslav eats less than Franz but more than Veit, and Mike eats more than Franz.
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Approach: chain the inequalities into one order
Mike > Franz, Jaroslav > Veit, and Jaroslav < Franz.
Combine: Mike is biggest, then Franz, then Jaroslav, then Veit.
Largest first: Mike, Franz, Jaroslav, Veit — option C.
Brigitte goes on holiday to Verona and plans to cross all five of the famous old bridges over the Etsch (Adige) at least once. She starts at the train station and when she returns there she has crossed each of the five bridges but no others. During her walk she has crossed the river n times. What is a possible value for n?
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Answer: D — 6
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Hint 1 of 2
Each bridge she crosses takes her from one bank to the other, so think about which side she ends up on.
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Hint 2 of 2
Since she returns to the starting bank, the number of crossings must be even.
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Approach: parity of river crossings on a closed route
Every bridge she walks over flips her from one bank to the other, so the river crossing count must be even for her to come back to the station's bank.
That rules out the odd choices 3, 5 and 7, and 4 is too few to cover all five bridges.
Crossing the five bridges with one used twice gives 6 crossings, which works.
There are three boxes in front of me: one white, one red and one green. One box holds a chocolate bar, another holds an apple, and one box is empty. The chocolate bar is in either the white box or the red box. The apple is in neither the white box nor the green box. In which box is the chocolate bar?
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Answer: A — White
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Hint 1 of 2
Start with the apple: it is in neither the white nor the green box.
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Hint 2 of 2
Once you place the apple, the chocolate's two options shrink to one.
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Approach: elimination
The apple is in neither white nor green, so the apple is in the red box.
The chocolate is in the white or the red box, but red now holds the apple.
Therefore the chocolate is in the white box — answer A.
Every time a coin is put into the machine, the bottom ball falls out of one of the five tubes. How many coins does Barbara have to put in to be sure she gets at least one white ball?
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Answer: D — 11
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Hint 1 of 3
Imagine Barbara has the unluckiest day — every ball that is not white comes out first.
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Hint 2 of 3
Only the bottom ball of a tube can come out, so count the non-white balls that could fall before any white one.
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Hint 3 of 3
Add up all those non-white balls, then put in one more coin to be sure of a white ball.
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Approach: imagine the unluckiest order, then add one more coin
Each coin drops the bottom ball of some tube, so the worst luck is all the non-white balls coming out first.
Counting from the bottom of every tube up to its first white ball, 10 non-white balls could fall before any white one.
After those 10 unlucky balls, the next ball must be white, so she needs 10 + 1 = 11 coins. The answer is D.
Amir has stickers of ladybugs with 1, 2, 3 or 4 dots on their wings. He fills the grid so that every row and every column has a ladybug with 1, 2, 3 and 4 dots. What should the top row of the grid look like when he is finished?
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Answer: B
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Hint 1 of 3
In every row and every column you must use a 1-dot, 2-dot, 3-dot and 4-dot ladybug, each exactly once.
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Hint 2 of 3
Look down each column at the ladybugs already placed to see which dot-numbers are still missing.
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Hint 3 of 3
A top cell must hold the one dot-number its column does not have yet.
Show solution
Approach: fill the grid so every row and column has 1, 2, 3 and 4 once
Every row and every column must show a 1-dot, 2-dot, 3-dot and 4-dot ladybug, each just once.
Look down each column and find the one dot-number it is still missing — that number must go in the top cell.
Filling each top cell with its column's missing number gives the row shown in option B. The answer is B.
We consider a giant \(4 \times 4\) chessboard. A kangaroo is standing on each of the 16 squares. On each move, each kangaroo jumps to an adjacent square (up, down, left or right, but not diagonally). All kangaroos stay on the chessboard. Several kangaroos can be on one square at the same time. What is the maximum number of unoccupied squares that we can have after 100 moves?
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Answer: B — 14
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Hint 1 of 2
Colour the board like a checkerboard; what happens to a kangaroo's colour each jump?
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Hint 2 of 2
Every jump flips colour, so after an even number of moves the 8 dark-start and 8 light-start kangaroos stay split—each group can pile onto one square.
Show solution
Approach: checkerboard parity invariant
Each jump changes a kangaroo's square colour, so after 100 (even) moves 8 kangaroos sit on dark squares and 8 on light squares.
Each group can be gathered onto a single square, occupying just 2 squares total.
Hassan writes either the number 0 or the number 1 in each field of the table. The sum in each row, each column and each diagonal should be exactly 3. Hassan has entered 0 in one field and then fills out the table completely. What is the sum of the numbers in the fields with question marks?
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Answer: B — 2
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Hint 1 of 2
Each row and each column has 4 cells but must total 3 using only 0s and 1s — so what does that force about how many 0s are in each row and column?
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Hint 2 of 2
There is exactly one 0 in every row and exactly one 0 in every column; place the rest of the 0s so the two diagonals also each have a single 0.
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Approach: each row and column hides exactly one 0
A row of four 0s-and-1s that adds to 3 must be three 1s and a single 0, so every row has exactly one 0 — and by the same reasoning every column has exactly one 0, and each diagonal must also hold just one 0.
Starting from the given 0, the ‘one 0 per row, one per column, one per diagonal’ rule pins down where all four 0s go, so the whole grid is forced.
Looking at the four question-mark cells, two of them turn out to be 1 and two turn out to be 0, so their sum is 1 + 1 + 0 + 0 = 2, giving the answer (B) 2.
Six ladybirds have 1, 2, 3, 4, 5 and 6 spots. Marta takes four photos, each showing three different ladybirds, and each ladybird appears in the same number of photos. The first three photos are shown. How many spots do the three ladybirds in the fourth photo have in total?
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Answer: C — 12
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Hint 1 of 2
There are 4 photos with 3 ladybirds each, that is 12 ladybird-appearances shared equally among 6 ladybirds.
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Hint 2 of 2
So every ladybird appears in exactly 2 photos, which tells you the grand total of spots over all four photos.
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Approach: find the grand total of all four photos, then subtract the three shown
Four photos with three ladybirds each give 12 appearances; split equally among 6 ladybirds, so each appears in exactly 2 photos.
Then all four photos together show every spot-count twice: 2 × (1 + 2 + 3 + 4 + 5 + 6) = 2 × 21 = 42 spots.
The three shown photos hold 30 spots in total.
So the fourth photo has 42 − 30 = 12 spots, option C.
A witch has 10 apples, 9 bananas and 6 pears. One day she enchants all of her fruits into different types of fruit. For example, she turns each apple into either a banana or a pear. After that she has 15 apples, 7 bananas and 3 pears. How many apples did she turn into bananas?
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Answer: E — 7
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Hint 1 of 2
Every original fruit changes type, so all 10 apples leave and new apples arrive from other fruits.
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Hint 2 of 2
Set up the in/out counts for each fruit type and solve.
Show solution
Approach: track each fruit type in and out
All 9 bananas and all 6 pears must become apples or each other; the 15 final apples come only from old bananas and pears.
Since 9 + 6 = 15, every banana and every pear turned into an apple.
Then the 10 apples must supply all 7 final bananas and 3 final pears: 7 + 3 = 10.
Maria writes the numbers 1, 2, 3, 4, 5, 6 and 7, each exactly once, into the number wall. Each upper box equals the sum of the two boxes just below it. The bottom-left box already holds 6. Which number must she write in the box with the star?
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Answer: D — 4
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Hint 1 of 2
The box sitting on top of the 6 is 6 plus its neighbour, and it can be at most 7.
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Hint 2 of 2
That forces the neighbour, then keep building upward with 1 to 7 each used once.
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Approach: use that no box can exceed 7 to fix the numbers, then build upward
Every box is the sum of the two below it, and no number is bigger than 7.
The box above the 6 is 6 + (its right neighbour), so that neighbour must be 1, giving 6 + 1 = 7.
The leftover numbers 2, 3, 5 must fill the rest; placing the bottom row as 6, 1, 3, 2 gives the next row 7, 4, 5 - and that uses 1 to 7 exactly once.
The starred middle-top box is 1 + 3 = 4, option D.
Three square Martians and three round Jupiterians are sitting at a table as shown. One of the six has the key to the spaceship. Everyone from one planet always tells the truth, and everyone from the other planet always lies. When asked “Does any of your neighbours have the key?” all six answer as shown. Who has the key?
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Answer: B — B
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Hint 1 of 3
The key-holder's two neighbours truly have a neighbour with the key, while everyone non-adjacent to the key truly does not.
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Hint 2 of 3
Whoever holds the key answers about their own neighbours, who do NOT have it—so the holder's truthful answer would be "No."
Still stuck? Show hint 3 →
Hint 3 of 3
Test each candidate and keep the one that yields exactly three truth-tellers and three liars.
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Approach: assume each candidate holds the key and count consistent types
Around the table the neighbours are \(A\!-\!B\!-\!C\!-\!D\!-\!E\!-\!F\!-\!A\); the answers are A:Yes, B:Yes, C:No, D:No, E:No, F:Yes.
Suppose \(B\) has the key: then \(A\) (neighbour) truly says Yes; \(B\) sees no key neighbour yet says Yes (lie); \(C\) says No but a neighbour has it (lie); \(D,E\) truly say No; \(F\) says Yes but neither neighbour has it (lie).
That is truth-tellers \(A,D,E\) and liars \(B,C,F\)—exactly three each, the only candidate that works—so the key is with B, choice (B).
We want to place the numbers 1 through 8 in the eight squares of the figure shown in such a way that consecutive numbers are never in adjacent squares (not even diagonally adjacent). Which numbers can we write in the square marked with an X?
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Answer: B — 2 or 7
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Hint 1 of 3
Consecutive numbers may not touch even diagonally, so a number with many forbidden partners needs a roomy cell with few neighbours.
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Hint 2 of 3
Count how many cells each number must avoid: the ends 1 and 8 avoid just one each, while the most-crowded cell (touching the most others) needs a number that has very few neighbours to dodge.
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Hint 3 of 3
By the figure's symmetry, X and its mirror cell are the two most-connected cells, so list which values can sit in such a tight spot.
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Approach: match crowded cells to numbers with few forbidden neighbours
A value \(n\) (with \(1
Cell X is one of the two most-connected cells in the shape, so the number placed there must have few neighbours to conflict with; testing placements, only by putting a near-end value there can every other number be seated legally.
Working the arrangement out, the value in X must be 2 or 7 (the two cases are mirror images), answer B.
Annie, Bibi, Clara and Doris each live on a different floor of a four-storey building, and other people live there too. 25 people live above Annie, 5 people live below Bibi, 17 people live below Clara, and 22 people live above Doris. How many people live in the building in total?
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Answer: A — 27
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Hint 1 of 2
For any resident, (people above) + (themselves) + (people below) is the whole building, so each clue tells you both an 'above' and a 'below' count once you call the total \(N\).
Still stuck? Show hint 2 →
Hint 2 of 2
Annie has the most people above her, so she is the lowest of the four friends; rank the four and make their above/below counts line up.
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Approach: rank the four friends and reconcile the counts
Annie has 25 people above her — the most of the four — so she is the lowest friend, and the most people above means the fewest below; in fact only Clara (17 below) and the others can sit above her.
Ranking everyone in one line, Annie is 26th from the top, Doris 23rd, Bibi has 5 below and Clara 17 below; fitting these together with one person below Annie forces the total.
The counts only agree when the building holds 27 people, which is (A).
Anna, Bella, Che and Dimitry each have three shapes. Each child shares exactly one of their shapes with one other child. Anna has a triangle, a circle and a square; Bella has a heart, a square and a star; Che has a star, a triangle and a diamond. Which three shapes does Dimitry have?
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Answer: E
Show hints
Hint 1 of 3
Write out the shapes the other three children have and see which ones already come in matching pairs.
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Hint 2 of 3
Some shapes already appear for two children (a pair), but a few shapes appear for only one child so far.
Still stuck? Show hint 3 →
Hint 3 of 3
Dimitry must hold exactly the shapes that still need a partner, so that every shape ends up shared by two children.
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Approach: give Dimitry the shapes that still need a matching partner
Among Anna, Bella and Che, the square (Anna & Bella), the star (Bella & Che) and the triangle (Anna & Che) already come in pairs.
That leaves the circle (only Anna), the heart (only Bella) and the diamond (only Che) without a partner.
So Dimitry must have the circle, the heart and the diamond — this pairs every shape with exactly one other child, which is option E.
Zoran builds towers from three different building blocks (a triangle top, a rectangle, and an hourglass). The picture shows the heights of three towers. How high is the fourth tower?
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Answer: A — 12
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Hint 1 of 3
Each kind of block is always the same height, so the same block is worth the same number every time.
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Hint 2 of 3
Compare two towers that are almost the same — the difference in their heights tells you how tall the extra block is.
Still stuck? Show hint 3 →
Hint 3 of 3
Find the rectangle and the triangle heights, since the fourth tower is just those two stacked.
Show solution
Approach: compare towers to find each block's height
The tall tower (triangle + rectangle + hourglass) is 20, and the short one with just (triangle + hourglass) is 13; the only extra block is the rectangle, so the rectangle is 20 − 13 = 7.
The tower with (rectangle + hourglass) is 15, and the rectangle is 7, so the hourglass is 8; then in the tower of 13 the triangle is 13 − 8 = 5.
The fourth tower is just triangle + rectangle = 5 + 7 = 12.
Simon has four cups with matching saucers (see picture). He places a randomly chosen cup on each saucer. Which statement is definitely true?
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Answer: D — It is impossible that exactly 3 cups are on the matching saucers.
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Hint 1 of 2
Test each statement: can you picture an arrangement where it fails? If yes, that statement is not certain.
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Hint 2 of 2
Think about what happens to the last cup if exactly three are already on their matching saucers.
Show solution
Approach: rule out the impossible count of correct matches
If three cups sit on their matching saucers, then three cups and three saucers are correctly paired, leaving one cup and one saucer that must also match.
So having exactly three matches forces a fourth match — you cannot have exactly three.
The other options can each happen with some arrangement, so they are not certain.
The statement that is definitely true is D: it is impossible that exactly 3 cups are on the matching saucers.
Andrew throws arrows at a target. He starts with 10 arrows. Each time he hits the target, he gets 2 more arrows. In total Andrew throws 20 arrows, and then he has run out of arrows. How many times did Andrew hit the target?
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Answer: B — 5
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Hint 1 of 3
He only started with 10 arrows, but he threw 20, so where did the extra arrows come from?
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Hint 2 of 3
Every hit is like a little gift of 2 more arrows.
Still stuck? Show hint 3 →
Hint 3 of 3
Figure out how many extra arrows he earned, then see how many hits that took.
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Approach: count the extra arrows the hits gave him
He started with 10 arrows but threw 20 in total, so 10 extra arrows must have come from hitting the target.
Each hit gives 2 extra arrows, so we count by twos: 2, 4, 6, 8, 10 — that takes 5 hits to reach 10 extra arrows.
The two pictures show the same bridge at different times. All the cars are the same length. The numbers give the distances between the cars, and between a car and the end of the bridge. How long is each car?
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Answer: C — 5 metres
Show hints
Hint 1 of 3
It is the same bridge in both pictures, so the bridge is exactly the same total length both times.
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Hint 2 of 3
In each picture add up all the gap numbers, and count how many cars there are.
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Hint 3 of 3
The bottom picture has one fewer car but more gap — that extra gap must be exactly one car long.
Show solution
Approach: the same bridge length both times
Top picture: the gaps add to 1 + 2 + 1 + 2 = 6 metres, and there are 3 cars.
Bottom picture: the gaps add to 4 + 4 + 3 = 11 metres, and there are 2 cars; the bridge is the same length, so the bottom has 11 − 6 = 5 more metres of gap but exactly one fewer car.
That missing car is filling those extra 5 metres of gap, so each car is 5 metres long.
The map shows the seven subway lines of a city. The stations are shown by circles. Martin wants to colour in the subway lines on the plan. If two lines share a common station, they must have different colours. What is the smallest number of different colours he can use?
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Answer: A — 3
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Hint 1 of 3
Two lines need different colours only when they share a station, so first hunt for lines that all meet one another.
Still stuck? Show hint 2 →
Hint 2 of 3
If you can find three lines where every pair shares a station, those three already need three different colours — so you can never do it with just two.
Still stuck? Show hint 3 →
Hint 3 of 3
After that, try to colour the rest of the lines reusing only those three colours; if it works, three is the answer.
Show solution
Approach: find three lines that all meet (needs 3 colours), then colour everything with 3
Lines that cross at a shared station must get different colours.
On the map there are three lines that each share a station with the other two, so those three lines need three different colours — two colours can never be enough.
Going line by line, every remaining line shares stations with only lines you have already coloured, so it can always reuse one of the three colours.
Three colours are both needed and enough, so the smallest number is 3.
Daniel forms a 4×4-square out of the three pieces shown on the right and another piece. The sum of the four numbers in each row and in each column is the same. What does the fourth piece look like?
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Answer: A
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Hint 1 of 3
Place the three given pieces in the 4×4 grid and look at any row or column that is already complete to read off the common line sum.
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Hint 2 of 3
Once you know the target sum, each empty cell is forced: it must make its row and its column reach that total.
Still stuck? Show hint 3 →
Hint 3 of 3
The fourth piece must fit the leftover three cells with exactly the values those cells require.
Show solution
Approach: fit the three pieces, find the common line sum, then force the empty cells
Slot the three given strips into the 4×4 grid; a row or column that is already full reveals the common sum every line must reach.
Each remaining empty cell is then forced, since its value must complete both its row and its column to that sum.
The three forced values, in the shape of the leftover cells, read 1, 1, 3.
The picture on the right shows a honeycomb with 9 cells. Some cells contain honey. The number written in a cell tells how many of its neighbouring cells contain honey. How many cells are filled with honey?
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Answer: C — 6
Show hints
Hint 1 of 3
Each written number counts how many of that cell's touching neighbours hold honey — like a honey version of Minesweeper.
Still stuck? Show hint 2 →
Hint 2 of 3
Start at a cell with few neighbours: if a clue equals its number of neighbours, every one of them must be honey.
Still stuck? Show hint 3 →
Hint 3 of 3
Use each filled-in clue to force its neighbours, one cell at a time, until the whole comb is settled.
Show solution
Approach: use each clue to decide its neighbours, starting where a clue forces everything
Begin at an edge cell whose clue equals its number of touching neighbours — then all of those neighbours must hold honey.
Once those are fixed, neighbouring clues tell you which of their remaining cells are honey and which are empty.
Keep applying the clues, cell by cell, so that every number ends up matching the honey around it.
When the whole comb is consistent, six of the cells contain honey: 6 (C).
A tray of biscuits is in the kitchen (see picture). Three girls take biscuits from the tray, in some unknown order: Tina takes all the heart-shaped biscuits still on the tray, Emma takes all the white biscuits still on the tray, and Rosa takes all the large biscuits still on the tray. One girl ends up with 3 biscuits, one with 6, and one with 7. Which of the pictured groups of biscuits (A)–(E) is exactly what one of the girls took?
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Answer: E
Show hints
Hint 1 of 2
Whoever goes first takes a whole category from all 16 cookies; figure out an order giving takes of 3, 6, and 7.
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Hint 2 of 2
Rosa (large) going first takes 7, then Tina (hearts) takes 6, leaving Emma (white) just 3.
Show solution
Approach: find the order of takers that yields 7, 6, 3
There are 16 cookies, and the takes are 3, 6, 7, so the first taker removes 7.
If Rosa takes the 7 large cookies first, only small ones remain.
Tina then takes the 6 small hearts, leaving three small white round cookies.
Emma takes those three round white cookies — the set in option E.
Logic & Word Problemswork-backwardcareful-counting
Each of the children Ali, Lea, Josef, Vittorio and Sophie gets a birthday cake. The number on top of the cake shows how old the child is. Lea is two years older than Josef, but one year younger than Ali. Vittorio is the youngest. Which cake belongs to Sophie?
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Answer: C
Show hints
Hint 1 of 3
There are five cakes with five ages on them — match a child to each one.
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Hint 2 of 3
Use the clues to label the easy children first, and give the smallest cake to Vittorio.
Still stuck? Show hint 3 →
Hint 3 of 3
Sophie gets the one cake that is left over after everyone else is matched.
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Approach: match the easy children first; Sophie gets the leftover cake
Josef, then Lea (2 older than Josef), then Ali (1 older than Lea) line up in a row of ages.
Vittorio is the youngest, so he takes the smallest-numbered cake.
Four children are now placed, so the one cake left over is Sophie's: the cake showing 6, option C.
6 beavers and 2 kangaroos are standing on the fields 1 to 8 in a row. Of any three animals in a row there is always exactly one kangaroo. On which of these numbers stands a kangaroo?
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Answer: C — 3
Show hints
Hint 1 of 2
Split the eight fields into the blocks 1-2-3, 4-5-6 and note each must hold one kangaroo.
Still stuck? Show hint 2 →
Hint 2 of 2
Test where the two kangaroos can sit so every three-in-a-row has exactly one.
Show solution
Approach: place the two kangaroos so each consecutive triple has exactly one
Fields 1-2-3 must contain one kangaroo and fields 4-5-6 another, using both kangaroos.
Checking every window of three forces the kangaroos onto fields 3 and 6.
Among the offered choices, a kangaroo stands on field 3.
Hanni wants to colour in the circles in the diagram. When two circles are connected by a line they should have different colours. What is the minimum number of colours she needs?
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Answer: B — 3
Show hints
Hint 1 of 2
Two circles joined by a line must differ in colour.
Still stuck? Show hint 2 →
Hint 2 of 2
If three circles are all linked to each other, they need three different colours.
Show solution
Approach: find a triangle (forces 3) then show 3 colours suffice
The diagram contains three circles that are all connected to one another, so they need three distinct colours.
Trying to colour the whole diagram with just those three colours succeeds.
Logic & Word Problemswork-backwardcareful-counting
Three frogs live in a pond. Each night only one of the frogs sings a song. After 9 nights the first frog has sung 2 times. The second frog has listened to 5 songs. How many songs did the third frog listen to?
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Answer: B — 6
Show hints
Hint 1 of 3
There are 9 nights and one song each night, so the three frogs together sang 9 songs.
Still stuck? Show hint 2 →
Hint 2 of 3
On any night, a frog is either singing or listening — so a frog listens on every night it did not sing.
Still stuck? Show hint 3 →
Hint 3 of 3
If the second frog listened to 5 songs, that tells you how many nights it sang instead.
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Approach: 9 songs in all; a frog listens on every night it does not sing
There are 9 nights with one song each, so 9 songs were sung in all. Frog 1 sang 2 of them.
Frog 2 listened to 5 songs, so out of 9 nights it sang on the other 4 nights.
That leaves Frog 3 singing 9 take away 2 take away 4, which is 3 nights, so it listened on the other 6 nights: 6, option B.
A tower consists of blocks that are labelled from bottom to top with the numbers from 1 to 90. Bob uses these blocks to build a new tower. For each step he takes the top three blocks from the old tower and places them on the new tower without changing their order (see diagram). How many blocks are there in the new tower between the blocks with the numbers 39 and 40?
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Answer: E — 4
Show hints
Hint 1 of 2
Blocks move in groups of three (the top three each step), and higher old numbers end up lower in the new tower.
Still stuck? Show hint 2 →
Hint 2 of 2
Find the group containing 39 and the group containing 40, then count what lies between them.
Show solution
Approach: track the blocks in groups of three
Each step moves a block of three with their order kept: ..., (37,38,39), (40,41,42), ...
Higher-numbered groups are placed first, so going up the new tower we see 40,41,42 then 37,38,39.
Logic & Word Problemsperiodic-patterncareful-counting
An underground line has the six stations A, B, C, D, E and F. The train stops at every station. After reaching the end of the line (A or F) the train continues in the opposite direction. The train conductor starts his journey in station B. His first stop is in station C. In which station will be his 46th stop?
Show answer
Answer: D — D
Show hints
Hint 1 of 2
List the stops in order; the train bounces off the ends A and F.
Still stuck? Show hint 2 →
Hint 2 of 2
The sequence of stops repeats, so find the cycle length and use the remainder.
Show solution
Approach: list the bouncing sequence, find its period, take 46 mod period
Starting at B and first stopping at C, the stops run C,D,E,F,E,D,C,B,A,B and then repeat with period 10.
Since 46 = 4×10 + 6, the 46th stop matches the 6th stop in the cycle.
A staircase has 2023 steps. Every third step is coloured black. The first seven steps of this staircase can be fully seen in the diagram. Anita walks up the staircase and steps on each step exactly once. She can start with either the right or the left foot and then alternates her right and left foot. What is the minimum number of black steps she sets her right foot on?
Show answer
Answer: D — 337
Show hints
Hint 1 of 2
Number the steps; the black ones are the multiples of 3: 3, 6, 9, …, 2022.
Still stuck? Show hint 2 →
Hint 2 of 2
Alternating feet means a step's foot depends only on whether its number is odd or even — so the two feet take the odd-numbered and the even-numbered black steps.
Show solution
Approach: split the black steps by parity
The black steps are 3, 6, 9, …, 2022, that is 3·1 up to 3·674, so there are 674 of them.
Feet alternate, so odd-numbered steps all use one foot and even-numbered steps the other.
Among the black steps, the odd ones are 3·1, 3·3, …, 3·673 (337 of them) and the even ones are 3·2, 3·4, …, 3·674 (also 337).
Either starting foot makes the right foot land on one of these groups, so the minimum is 337.
Robert and Sonja play a game. Taking turns, each player removes 1, 2, 3, 4 or 5 cards from the pile. Whoever takes the last card loses. There are 10 cards on the pile and it is Robert’s turn. How many cards should he leave for Sonja so that he is certain to win?
Show answer
Answer: C — 7
Show hints
Hint 1 of 2
Since taking the last card loses, you want to leave your opponent stuck taking it.
Still stuck? Show hint 2 →
Hint 2 of 2
Work backwards: leaving 1 card is a loss for the other player, and so is any number that is one more than a multiple of 6.
Show solution
Approach: find the losing positions and leave the opponent on one
Whoever is forced to take the very last card loses, so leaving exactly 1 card hands the loss to your opponent.
Because each turn removes 1 to 5 cards, you can always reply to keep the pile at the next “bad” number, which are 1, 7, 13, … (one more than a multiple of 6).
From 10 cards Robert takes 3, leaving 7 — a losing position for Sonja, whatever she does next.
A tower is built from bricks labelled 1 to 50, from bottom to top. Bob builds a new tower: each time he takes the top two bricks off the old tower (keeping their order) and places them on top of the new tower (see picture). When he is finished, which two bricks lie directly on top of each other?
Show answer
Answer: E — 27 and 30
Show hints
Hint 1 of 2
Each move peels the top two bricks (keeping their order) and drops them on the new tower, so the new tower forms in pairs.
Still stuck? Show hint 2 →
Hint 2 of 2
Track the new stack pair by pair and look for which two of the listed bricks end up directly one above the other.
Show solution
Approach: simulate the pair-by-pair transfer onto the new tower
The old tower has 50 on top of 49 on top of 48 … down to 1; each move lifts the top two as a pair and stacks them on the new tower.
So the new tower is built bottom-up as 49,50, then 47,48, then 45,46, and so on, in steps of two.
Following this all the way down, the pair 30 then 27 lands one directly above the other in the finished new tower.
So the two bricks on top of each other are 27 and 30, answer E.
Counting & ProbabilityLogic & Word Problemscareful-countingcasework
Snow White organises a chess tournament for the seven dwarfs, lasting several days. Every dwarf has to play every other dwarf exactly once. On Monday Grumpy plays 1 game, Sneezy plays 2, Sleepy 3, Bashful 4, Happy 5 and Doc 6 games. How many games does Dopey, the 7th dwarf, play on Monday?
Show answer
Answer: C — 3
Show hints
Hint 1 of 3
Each dwarf can play at most the other six, and the given counts 1, 2, 3, 4, 5, 6 are all different.
Still stuck? Show hint 2 →
Hint 2 of 3
Start from the extremes: the one who played 6 games played everybody, while the one who played only 1 must have played that same dwarf.
Still stuck? Show hint 3 →
Hint 3 of 3
Keep pairing the highest remaining count with the lowest, peeling off the players whose schedule is now complete.
Show solution
Approach: pair the largest count with the smallest, working inward
Doc played 6 games, so he played everyone — including Grumpy, whose single game must therefore be against Doc only.
Happy played 5: everyone except Grumpy (Grumpy is already finished), so Happy played the other five, including Dopey; Sneezy's 2 games are then Doc and Happy.
Bashful played 4: he must be Doc, Happy, Sleepy and Dopey; Sleepy's 3 are Doc, Happy and Bashful.
So Dopey was played by Doc, Happy and Bashful only — that is 3 games, option C.
Counting & ProbabilityLogic & Word Problemscareful-countingcasework
Elisabeth wants to write the numbers 1 to 9 in the fields of the diagram shown so that the product of the numbers in any two fields next to each other is no greater than 15. Two fields are called “next to each other” if they share a common edge. How many ways are there for Elisabeth to label the fields?
Show answer
Answer: C — 16
Show hints
Hint 1 of 2
The big numbers (7, 8, 9) are very restricted: their neighbours' products must stay ≤ 15.
Still stuck? Show hint 2 →
Hint 2 of 2
Place 9, 8, 7 first into spots with few neighbours or small neighbours, then count the freedom that remains.
Show solution
Approach: place the large numbers under the product constraint, then multiply free choices
For adjacent products ≤ 15, the largest numbers (9, 8, 7) can only sit next to very small numbers (mostly 1 and 2).
This pins those big numbers to specific low-degree fields and forces small neighbours around them.
Counting the independent choices that remain gives 16 labelings.
Logic & Word ProblemsArithmetic & Operationswork-backwardsum-constraint
Several mice live in three houses. Last night every mouse left its house and moved directly to one of the other two houses. The diagram shows how many mice were in each house yesterday (“gestern”) and today (“heute”). How many mice used the path indicated by the arrow?
Show answer
Answer: B — 11
Show hints
Hint 1 of 2
Every mouse leaves its own house, so each house's outgoing mice split between the other two.
Still stuck? Show hint 2 →
Hint 2 of 2
Set up the flows between the three houses from yesterday's and today's counts; the arrow is one of those flows.
Show solution
Approach: balance the mouse flows between the three houses
Yesterday the houses held 8, 7, 5 and today they hold 6, 10, 4; every mouse moved to a different house.
Writing the six directed flows and using that each house empties out gives equations linking them.
Logic & Word ProblemsArithmetic & Operationssum-constraintwork-backward
Jakob wrote six consecutive numbers on six little pieces of white paper, one number per piece. He stuck the six pieces on the front and back of three coins. Then he threw the coins three times. After the first throw the numbers 6, 7, 8 were on top (see diagram), which Jakob then coloured red. After the second throw the sum of the numbers on top was 23, and after the third throw the sum was 17. How big is the sum of the numbers on the three white pieces of paper?
Show answer
Answer: A — 18
Show hints
Hint 1 of 2
The six consecutive numbers are paired front/back on three coins, so the two faces of one coin are a fixed-difference pair.
Still stuck? Show hint 2 →
Hint 2 of 2
Use the three throw-sums (the first is 6+7+8) to deduce the hidden faces and then the unseen white totals.
Show solution
Approach: pair the faces by the coins and use the three sums
The first throw shows 6, 7, 8, so those three reds sum to 21; each later throw replaces some reds by their white partners, changing the sum by white − red on the flipped coins.
From 21 the sums become 23 (a change of +2) and 17 (a change of −4); the total change if all three were flipped is (white total) − 21.
The six numbers are consecutive and include 6, 7, 8; the only set making the throws consistent is 4,5,6,7,8,9, so the whites are 4, 5, 9.
Four circles joined by a line form a chain of four. The numbers 1, 2, 3 and 4 each appear exactly once in every row, every column and every chain of four. Which number goes in the circle with the question mark?
Show answer
Answer: B — 2
Show hints
Hint 1 of 2
The rule is like a small Latin square: 1, 2, 3, 4 each appear once per row, per column and per chain.
Still stuck? Show hint 2 →
Hint 2 of 2
Use the given numbers and these no-repeat rules to pin down the ? cell.
Show solution
Approach: apply the row, column and chain constraints
Each row, each column and each chain of four must contain 1, 2, 3, 4 exactly once.
Filling in forced cells from the given 3, 2, 2, 1 narrows the options.
Lisa has four dogs of different weights. Each dog weighs a whole number of kilograms. All four dogs together weigh 60 kg, and the second heaviest dog weighs 28 kg. How heavy is the third heaviest dog?
Show answer
Answer: A — 2 kg
Show hints
Hint 1 of 2
The heaviest dog weighs more than 28 kg, so it already uses up a big chunk of the 60 kg.
Still stuck? Show hint 2 →
Hint 2 of 2
After the heaviest and the 28 kg dog, very little weight is left for the two lightest, who must still be different whole numbers.
Show solution
Approach: see how little weight is left for the two smallest dogs
The heaviest, the 28 kg dog, and the two lightest add to 60 kg, so the heaviest plus the two lightest make 60 − 28 = 32 kg.
The heaviest is more than 28, so it is at least 29 kg, leaving at most 3 kg for the two lightest together.
Two different whole numbers adding to at most 3 can only be 2 and 1, so the third heaviest is 2 kg.
The diagram shows a map with 16 towns which are connected via roads. The government is planning to build power plants in some towns. Each power plant can generate enough electricity for the town in which it stands as well as for its immediate neighbouring towns (i.e. towns that can be reached via a direct connecting road). What is the minimum number of power plants that have to be built?
Show answer
Answer: B — 4
Show hints
Hint 1 of 2
Each plant covers its own town plus every town directly joined to it.
Still stuck? Show hint 2 →
Hint 2 of 2
Find the fewest towns whose coverage reaches all 16 (a dominating set).
Show solution
Approach: minimum dominating set (deferred to key)
You need a set of towns so that every town is chosen or adjacent to a chosen one.
Placing plants well, four towns suffice to cover all sixteen, and three cannot.
Wanda chooses some of the shapes shown. She says: “I have chosen exactly 2 grey, 2 big and 2 round shapes.” What is the smallest number of shapes Wanda could have chosen?
Show answer
Answer: B — 3
Show hints
Hint 1 of 2
You want a shape to count toward more than one of the requirements at once.
Still stuck? Show hint 2 →
Hint 2 of 2
Pick shapes that are grey-and-big, big-and-round, or grey-and-round to overlap the three needs.
Show solution
Approach: make each shape cover two requirements
She needs exactly 2 grey, 2 big and 2 round.
Choose a big grey shape (grey+big), a big round shape (big+round) and a small grey round shape (grey+round).
These three give exactly 2 grey, 2 big and 2 round.
The bus stops in the villages A, B, C and D lie along a road in this order, and neighbouring villages are 10 km apart. There are 10 children in village A, 20 in B, 30 in C and 40 in D. Every child takes the bus to school. A new school will be built where the total number of kilometres travelled by all the children is as small as possible. Where will the new school be built?
Show answer
Answer: D — in C
Show hints
Hint 1 of 3
The best spot has about as many children on one side of the school as on the other side.
Still stuck? Show hint 2 →
Hint 2 of 3
Start at one end and add up children until you reach more than half of all of them.
Still stuck? Show hint 3 →
Hint 3 of 3
The village where you pass the halfway count is the best place for the school.
Show solution
Approach: walk from one end and stop where you pass half the children
There are 10 + 20 + 30 + 40 = 100 children in all, so half of them is 50.
Counting from A: A has 10, then A and B have 30, then A, B and C have 60 - we pass 50 right at C.
Since just as many children sit on each side once we reach C, building the school in C makes the total travel smallest.
So the answer is D.
Check by trying neighboursMoving the school 10 km from C toward D saves 40 children 10 km each (400 km) but costs the other 60 children 10 km each (600 km), a net loss; moving it toward B is worse too, so C truly is best.
The grandchildren ask their grandma how old she is. The grandma invites them to guess the age. The first child says 75, the second says 78 and the third says 81. It turns out that one child is wrong by 1 year, one by 2 years and one by 4 years. How many possibilities are there for the age of the grandma?
Show answer
Answer: C — 2
Show hints
Hint 1 of 2
The three errors are 1, 2 and 4 in some order, each above or below.
Still stuck? Show hint 2 →
Hint 2 of 2
Test which actual ages let the guesses 75, 78, 81 miss by exactly {1,2,4}.
Show solution
Approach: match the error set {1,2,4} to the three guesses
The true age must differ from 75, 78 and 81 by exactly 1, 2 and 4 in some assignment, each error either above or below.
Searching the candidate ages, only a couple of values make all three differences fit the set {1,2,4}.
The numbers from 1 to 6 are placed in the circles at the intersections of 3 rings. The position of the number 6 is shown. The sum of the numbers on each ring is the same. What number is placed in the circle with the question mark?
Show answer
Answer: A — 1
Show hints
Hint 1 of 2
Add 1..6 to get 21; each circle lies on two rings, so the three ring-sums add to twice 21.
Still stuck? Show hint 2 →
Hint 2 of 2
That fixes every ring's sum; use the given 6 to pin down the marked circle.
Show solution
Approach: use the common ring sum
The numbers 1..6 add to 21, and every circle sits on exactly two rings.
So the three equal ring-sums total 2×21 = 42, making each ring sum to 14.
Working from the fixed 6 and the requirement that each ring totals 14 forces the marked circle.
In a team competition, there are 5 teams waiting to start. Each team consists of either only boys or only girls. The numbers of team members are 9, 15, 17, 19 and 21. After all members of the first team have started, the number of girls not yet started is 3 times the number of boys not yet started. How many members are on the team that has already started?
Show answer
Answer: E — 21
Show hints
Hint 1 of 2
After one team leaves, the remaining members split as boys and 3×boys, so the leftover total must be divisible by 4.
Still stuck? Show hint 2 →
Hint 2 of 2
Total is 81; test which starting team size leaves a multiple of 4 that also splits along whole teams.
Show solution
Approach: use the divisible-by-4 leftover and check team splits
All five teams total 9+15+17+19+21 = 81.
After the first team starts, the rest split as boys + 3×boys = 4×boys, so the leftover must be a multiple of 4.
Only removing 21 leaves 60, which splits as 15 boys and 45 girls (45 = 3×15), with whole teams 15 and 9+17+19.
Eva has 5 stickers: a triangle, a circle, a star, a flower and an apple. She sticks one of them on each of the 5 squares of this board so that the star is not on square 5, the apple is on square 1, and the flower is next to both the circle and the triangle. On which square did Eva stick the flower?
1
2
3
4
5
Show answer
Answer: D — 4
Show hints
Hint 1 of 2
Pin down the sure clues first: the apple goes on square 1, and the star may not go on square 5.
Still stuck? Show hint 2 →
Hint 2 of 2
The flower touches both the circle and the triangle, so those three stickers sit in a row of three squares with the flower in the middle.
Show solution
Approach: place the forced stickers, then fit the flower
The apple is on square 1, so the circle, star, flower and triangle fill squares 2, 3, 4, 5.
The flower touches both the circle and the triangle, so circle–flower–triangle sit in three squares in a row with the flower in the middle — either 2–3–4 or 3–4–5.
If they took 2–3–4, the star would be forced onto square 5, which is not allowed; so they take 3–4–5 and the star goes on square 2.
Every time the witch has 3 apples she turns them into 1 banana. Every time she has 3 bananas she turns them into 1 apple. What will she finish with if she starts with 4 apples and 5 bananas?
Show answer
Answer: A
Show hints
Hint 1 of 3
Start with her 4 apples and 5 bananas, and just keep following her two rules.
Still stuck? Show hint 2 →
Hint 2 of 3
Trade 3 apples for 1 banana, and trade 3 bananas for 1 apple, again and again.
Still stuck? Show hint 3 →
Hint 3 of 3
Keep swapping in groups of 3 until you can no longer make a group of three.
Show solution
Approach: repeatedly trade groups of three
Start with 4 apples and 5 bananas. Trade 3 apples for 1 banana: 1 apple, 6 bananas.
Trade 6 bananas (two groups of three) for 2 apples: 3 apples, 0 bananas.
Trade those 3 apples for 1 banana: 0 apples, 1 banana.
Logic & Word Problemswork-backwardcareful-counting
The numbers 1 to 9 are placed in the squares shown, with a number in each square. The sums of all pairs of neighbouring numbers are shown. Which number is placed in the shaded square?
Show answer
Answer: D — 7
Show hints
Hint 1 of 3
Find the smallest bracket-total first: which two different numbers from 1 to 9 can possibly add up to it?
Still stuck? Show hint 2 →
Hint 2 of 3
Once you know one square, slide along the brackets — each bracket shares a square with the next, so you can fill them in one at a time.
Still stuck? Show hint 3 →
Hint 3 of 3
Keep going along the row until you reach the shaded square.
Show solution
Approach: start at the smallest sum, then slide along the shared brackets
The smallest bracket is 3, and the only two different numbers from 1 to 9 that add to 3 are 1 and 2.
The next bracket along is 7; the square shared with the '3' pair must be the small one (1) so its partner can be 6, which also fits the 15 bracket (9 + 6) on the other side — so that run is 9, 6, 1, 2.
Now slide right through the shaded square: 2 + (shaded) = 9 (the bracket of 9), so the shaded square is 9 - 2 = 7.
The five cards shown (2, 3, 4, 5, 6) are placed into 2 boxes. The sums of the numbers in each box are the same. Which number must be in the box with the number 4?
Show answer
Answer: D — 6
Show hints
Hint 1 of 3
Add all five card numbers together first.
Still stuck? Show hint 2 →
Hint 2 of 3
Since the two boxes are equal, split that total in half to see what each box must hold.
Still stuck? Show hint 3 →
Hint 3 of 3
Now figure out which card the 4 needs next to it to reach that box total.
Show solution
Approach: find the equal totals and pair up
All five cards add to 2 + 3 + 4 + 5 + 6 = 20, so each box must hold 10.
Putting 2, 3, and 5 together makes 10, which leaves 4 and 6 for the other box.
Ann, Bob, Carina, Dan and Ed are sitting at a round table. Ann is not next to Bob, Dan is next to Ed, and Bob is not next to Dan. Which two people are sitting next to Carina?
Show answer
Answer: A — Ann and Bob
Show hints
Hint 1 of 2
Place the people around the circle using 'Dan next to Ed' first, then apply the 'not next to' rules.
Still stuck? Show hint 2 →
Hint 2 of 2
Once the seating is forced, read off Carina's two neighbours.
Show solution
Approach: fix the circular order from the clues
Dan sits next to Ed; Bob is not next to Dan and Ann is not next to Bob.
Working these around the five-seat circle forces an order where Carina sits between Ann and Bob.
Elena wants to write the numbers from 1 to 9 in the squares shown. The arrows always point from a smaller number to a larger one. She has already written 5 and 7. Which number should she write instead of the question mark?
Show answer
Answer: D — 6
Show hints
Hint 1 of 3
Every arrow goes from a smaller number to a larger one, so the box where all arrows point in must hold a big number and the box where all arrows point out must hold a small one.
Still stuck? Show hint 2 →
Hint 2 of 3
Follow the arrow trails out from the boxes that already show 5 and 7, always stepping up to a bigger number.
Still stuck? Show hint 3 →
Hint 3 of 3
Work out which numbers are still allowed in the ? box after the small numbers are forced into the lower boxes.
Show solution
Approach: use 'arrows go small to large' to squeeze out which number fits the ? box
The top-left box gets an arrow in from the 7, so it must be bigger than 7, which means it is 8, and the box it points to is the biggest of all, 9.
Following the arrows down and to the left, every box on the bottom-right side has to be smaller than 5, so 1, 2, 3 and 4 are all used up down there.
That leaves only 6 free for the ? box (8 and 9 are taken by the top, and 5 and 7 are already placed).
Logic & Word ProblemsSpatial & Visual Reasoningcareful-countingcasework
A triangular pyramid is built with 20 cannon balls, as shown. Each cannon ball is labelled with one of A, B, C, D or E. There are 4 cannon balls with each type of label. The picture shows the labels on the cannon balls on 3 of the faces of the pyramid. What is the label on the hidden cannon ball in the middle of the fourth face?
Show answer
Answer: D — D
Show hints
Hint 1 of 2
Each label A–E is used exactly four times across the 20 balls.
Still stuck? Show hint 2 →
Hint 2 of 2
Tally how many of each label already appear on the three shown faces (counting shared edge balls once); the centre ball must be the label still short of four.
Show solution
Approach: count each label and find the one not yet at full quota
There are 4 balls of each label. Tally the labels visible on the three shown faces, counting shared edge/corner balls once.
One label falls one short of its quota of 4; that missing ball is the hidden centre of the fourth face.
An apple and an orange weigh as much as a pear and a peach. An apple and a pear weigh less than an orange and a peach, and a pear and an orange weigh less than an apple and a peach. Which of the pieces of fruit is the heaviest?
Show answer
Answer: C — peach
Show hints
Hint 1 of 2
Turn each sentence into an inequality between sums of two fruits.
Still stuck? Show hint 2 →
Hint 2 of 2
Combine the inequalities to rank the fruits and spot the heaviest.
Show solution
Approach: combine the weight inequalities
The balance apple + orange = pear + peach rearranges to peach = apple + (orange − pear).
The two 'weigh less' facts force pear to be lighter than both apple and orange, so orange − pear is positive.
Then peach = apple + (a positive amount) beats apple, and likewise peach beats orange, so the heaviest is peach.
Logic & Word Problemsbalance-reasoningsubstitution
Martin placed 3 different types of objects — hexagons, squares and triangles — on sets of scales, as shown. What does he need to put on the left-hand side of the third set of scales for these scales to balance?
Show answer
Answer: A — 1 square
Show hints
Hint 1 of 2
Each balanced scale tells you that the two sides weigh the same; measure everything in squares.
Still stuck? Show hint 2 →
Hint 2 of 2
Find how many squares a triangle is worth, then a hexagon, and compare the two sides of the third scale.
Show solution
Approach: measure every object in squares, then balance the third scale
Scale 2: a triangle and a hexagon balance a hexagon and 5 squares, so the triangle weighs 5 squares.
Scale 1: 2 hexagons balance a triangle and a square, that is 5 + 1 = 6 squares, so 1 hexagon = 3 squares.
Scale 3 left side is 3 hexagons = 9 squares; the right side is 2 triangles = 10 squares, so the left is 1 square too light.
Logic & Word ProblemsCounting & Probabilitysum-constraintcomplementary-counting
A box contains only green, red, blue and yellow counters. There is always at least one green counter amongst any 27 counters chosen from the box; always at least one red counter amongst any 25 counters chosen; always at least one blue amongst any 22 counters chosen and always at least one yellow amongst any 17 counters chosen. What is the largest number of counters that could be in the box?
Show answer
Answer: B — 29
Show hints
Hint 1 of 2
'Any 27 chosen contain a green' means you can never pick 27 with no green — so the non-green counters number at most 26.
Still stuck? Show hint 2 →
Hint 2 of 2
Write the same kind of bound for each colour and add them up.
Number TheoryLogic & Word Problemsdivisibilitycasework
2021 coloured kangaroos are arranged in a row and are numbered from 1 to 2021. Each kangaroo is coloured either red, grey or blue. Amongst any three consecutive kangaroos, there are always kangaroos of all three colours. Bruce guesses the colours of five kangaroos. These are his guesses: Kangaroo 2 is grey; Kangaroo 20 is blue; Kangaroo 202 is red; Kangaroo 1002 is blue; Kangaroo 2021 is grey. Only one of his guesses is wrong. What is the number of the kangaroo whose colour he guessed incorrectly?
Show answer
Answer: B — 20
Show hints
Hint 1 of 2
'Any three in a row use all three colours' forces the colouring to repeat with period 3.
Still stuck? Show hint 2 →
Hint 2 of 2
So a kangaroo's colour depends only on its position modulo 3; compare the guesses at equal residues.
Show solution
Approach: use the period-3 structure
With every three consecutive kangaroos all different, the colour pattern repeats every 3 positions.
Positions 2, 20 and 2021 are all 2 (mod 3), so they must share one colour.
Guesses say k2 = grey, k20 = blue, k2021 = grey; the lone disagreement (k20) must be the wrong one.
Logic & Word ProblemsCounting & Probabilitycaseworkcareful-counting
In a town there are 21 knights who always tell the truth and 2000 knaves who always lie. A wizard divided 2020 of these 2021 people into 1010 pairs. Every person in a pair described the other person as either a knight or a knave. As a result, 2000 people were called knights and 20 people were called knaves. How many pairs of two knaves were there?
Show answer
Answer: D — 995
Show hints
Hint 1 of 3
Work out what each type of pair (two knights, two knaves, mixed) makes the partners say.
Still stuck? Show hint 2 →
Hint 2 of 3
Only mixed pairs produce 'knave' answers, two each — that pins down the number of mixed pairs.
Still stuck? Show hint 3 →
Hint 3 of 3
Then use the 21 knights to back out the other pair types.
Show solution
Approach: classify pairs by the labels they generate
In a same-type pair both say 'knight'; in a mixed pair both say 'knave'.
The 20 'knave' calls come 2 per mixed pair, so there are 10 mixed pairs (using 10 knights and 10 knaves).
With one knight left out, the other 10 knights form 5 knight-knight pairs; the remaining 1990 knaves form 995 knave-knave pairs.
Logic & Word ProblemsCounting & Probabilitycasework
In a tournament each of the 6 teams plays one match against every other team. In each round of matches, 3 take place simultaneously. A TV station has already decided which match it will broadcast for each round, as shown in the table. In which round will team D play against team F?
Round
1
2
3
4
5
Broadcast match
A–B
C–D
A–E
E–F
A–C
Show answer
Answer: A — 1
Show hints
Hint 1 of 2
The five rounds form a schedule where each round is three disjoint matches covering all six teams.
Still stuck? Show hint 2 →
Hint 2 of 2
Each team plays once per round, so a team's partners across the rounds are all different; reconstruct from the broadcasts.
Show solution
Approach: reconstruct the round-by-round pairing
A's broadcast matches are B (R1), E (R3) and C (R5), so A must meet D and F in rounds 2 and 4; since R2 already shows C–D, A plays F in R2 and D in R4.
Round 4 shows A–D and E–F, so its third match is B–C; round 2's leftover pair is B–E.
Now D still needs B, E, F and E still needs C, D: E–C must fall in round 1 (E is busy in the others), forcing E–D into round 5.
That leaves round 1 as A–B, C–E and the last pair D–F, so D plays F in round 1 — answer A.
Adam, Breno and Carlos live in the same apartment and bought a treadmill for exercise. Nobody uses the treadmill on Wednesdays and Sundays, and there is no day on which all three of them use the treadmill. Adam uses the treadmill 4 times a week and Breno 5 times a week. There are 4 consecutive days when Adam uses the treadmill on the first day, he does not use it on the second day, and Breno does not use it on the fourth day. Carlos uses the treadmill on one day of the week. Which day?
Show answer
Answer: A — Friday
Show hints
Hint 1 of 2
Only Mon, Tue, Thu, Fri, Sat are usable; Breno’s 5 sessions fill them all.
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Hint 2 of 2
Carlos’s day can’t be one where all three use it, so it’s Adam’s single off-day — find it from the four-day clue.
Show solution
Approach: pin down Adam's missing day using the consecutive-day clue
Breno uses all five active days, so Adam’s skipped day is Carlos’s day (no day has all three).
The four consecutive days with 'Breno off on day 4' must be Thu, Fri, Sat, Sun.
Then Adam uses Thu (day 1) but not Fri (day 2), so Adam’s off-day is Friday.
Two thousand and twenty coins lie on a table, all showing heads. In each move you must turn over exactly three of the coins. What is the smallest number of moves needed so that every coin shows tails?
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Answer: C — 674
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Hint 1 of 2
Each coin must be flipped an odd number of times to end as tails.
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Hint 2 of 2
Three flips per move; how few moves cover 2020 odd-flip requirements?
Show solution
Approach: parity and a counting lower bound
Each of the 2020 coins needs an odd number of flips, so at least 2020 coin-flips are needed.
Each move makes 3 coin-flips, so at least 2020/3 rounded up = 674 moves.
674 moves give 2022 flips: flip 2019 coins once and one coin three times — all odd — which is achievable, so the answer is 674.
On a distant island, 2020 kangaroos hold hands in a large circle. Each kangaroo is either brown (and always tells the truth) or grey (and always lies). Every one of them says, “One of my neighbours is brown and the other is grey.” How many of the kangaroos are brown?
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Answer: A — 0
Show hints
Hint 1 of 2
Suppose a brown (truthful) kangaroo exists: its statement about its neighbours would have to hold.
Still stuck? Show hint 2 →
Hint 2 of 2
Test whether any mix of brown and grey can sit in a circle when all say the same sentence - it collapses to one case.
Show solution
Approach: check the statement's consistency around the circle
A brown kangaroo tells the truth, so its two neighbours would be one brown and one grey.
Following that around the circle leads to a contradiction, so no truthful (brown) kangaroo can exist.
Every kangaroo is therefore grey and lying - consistent, since the statement is then false for each.
Amelia built a crown using 10 copies of the small piece shown. The pieces were joined so that touching sides always show the same number, as in the picture, where four pieces are filled in. What number appears in the coloured triangle?
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Answer: A — 1
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Hint 1 of 3
All ten pieces are exact copies, so the same little numbers repeat as you go around the ring.
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Hint 2 of 3
Where two pieces touch, the touching numbers match, which lines the pieces up the same way each time.
Still stuck? Show hint 3 →
Hint 3 of 3
Once you know how one piece sits, every piece sits the same, so jump that pattern around to the colored triangle.
Show solution
Approach: use the repeating pattern of identical pieces around the ring
Every piece is the same copy, and because touching sides must show equal numbers, each piece is placed in the very same way as its neighbour.
So the numbers repeat in the same order all the way around the crown, like a pattern that copies itself ten times.
Reading that repeating pattern from the filled-in pieces around to the colored triangle, the number landing there is 1, choice A.
The shortest way from Atown to Cetown is through Betown. Going back along this road from Cetown to Atown, we first find the signposts on the left side of the road. Further on we find the road signs on the right side of the road. How far is it from Betown to Atown?
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Answer: D — 4 km
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Hint 1 of 2
Each signpost shows its distance to the towns; the two posts seen on the return trip pin down where each post stands on the road.
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Hint 2 of 2
Use the readings that name a common town to fix the gaps between the towns, then read off Betown-to-Atown.
Show solution
Approach: place the two signposts on the road from their distance readings
On the way back from Cetown the two posts give their distances to the towns, which fixes the position of each post along the straight road.
Matching up the readings that share a town determines the spacing Atown–Betown–Cetown.
Reading off that spacing, the distance from Betown to Atown is 4 km, choice D.
There are three flowers on the back of the left cactus. In total, the cactus on the right has six more flowers than the cactus on the left. How many flowers are on the back of the right cactus?
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Answer: D — 12
Show hints
Hint 1 of 2
Each cactus's total = the flowers you can see on its front + the flowers on its back; the left back is 3.
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Hint 2 of 2
Find the left cactus's total, add 6 for the right cactus, then subtract the right cactus's visible front.
Show solution
Approach: total each cactus, then subtract the visible front
The left cactus shows 5 flowers on its front plus 3 on its back, so its total is 8.
The right cactus has 6 more in total, that is 14, and it shows 2 flowers on its front, so its back has 14 − 2 = 12 flowers.
Two equal trains, each with 31 numbered wagons, travel in opposite directions. When wagon number 7 of one train is side by side with wagon number 12 of the other train, which wagon is side by side with wagon number 11?
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Answer: A — 8
Show hints
Hint 1 of 2
The two trains line up wagon-to-wagon; use the spot where 7 meets 12.
Still stuck? Show hint 2 →
Hint 2 of 2
The two facing wagon numbers always add up to the same total.
Show solution
Approach: pair wagons from the known matching point
Wagon 7 of one train lines up with wagon 12 of the other.
The wagon numbers run in opposite directions, so the two numbers across from each other always add to 19 (that is 7 plus 12).
Across from wagon 11 is the wagon numbered 19 minus 11 = 8.
Tania bought 14 chocolates, 8 of them round and the rest square. Half were white chocolates and half were dark chocolates. Among the square chocolates, only two are not white. How many dark round chocolates did Tania buy?
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Answer: D — 5
Show hints
Hint 1 of 2
Split the 14 into round/square and into white/dark and fill a little table.
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Hint 2 of 2
Use that only two square chocolates are dark to find the round ones.
Show solution
Approach: organise with a 2-by-2 table
There are 8 round and 6 square chocolates; 7 are white and 7 are dark.
Among the 6 square ones only 2 are dark, so 4 squares are white.
That leaves 7 minus 4 = 3 white round chocolates, so round dark = 8 minus 3 = 5.
Rita numbered the circles in the figure from 1 to 8, so that the sum of the three numbers on each of the four sides of the square equals 13. What is the sum of the four numbers written on the coloured circles?
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Answer: E — 16
Show hints
Hint 1 of 2
Adding the four side-sums counts each corner circle twice.
Still stuck? Show hint 2 →
Hint 2 of 2
Compare that double-count to the total 1+2+...+8.
Show solution
Approach: count corners twice via the side sums
The four sides each total 13, so all four sides together total 4 times 13 = 52.
This sum counts every number once, but the four corner circles twice.
Since 1+2+...+8 = 36, the extra from double-counting the corners is 52 minus 36 = 16.
The colored circles are the corners, so their sum is 16.
In the figure, an arrow pointing from one person to another means that the first person is shorter than the second. For example, person B is shorter than person A. Which person is the tallest?
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Answer: C — Person C
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Hint 1 of 2
An arrow goes from the shorter person to the taller one.
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Hint 2 of 2
The tallest person has only arrows pointing toward them and none leaving.
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Approach: the tallest has no arrow leaving it
An arrow leaving a person means someone is taller, so the tallest person has no arrow pointing away from them.
Checking each person, only person C has every nearby arrow pointing in and none leaving.
Robert makes 5 statements, exactly one of which is wrong:
(A) My son Basil has 3 sisters. (B) My daughter Ann has 2 brothers. (C) My daughter Ann has 2 sisters. (D) My son Basil has 2 brothers. (E) I have 5 children.
Which statement is wrong?
Show answer
Answer: D — Statement D
Show hints
Hint 1 of 3
Try to find how many sons and daughters Robert has so that four of the five statements come out true.
Still stuck? Show hint 2 →
Hint 2 of 3
Remember a boy counts his brothers as the other boys, and his sisters as all the girls.
Still stuck? Show hint 3 →
Hint 3 of 3
Once you fix the numbers of boys and girls, check which single statement is then forced to be false.
Show solution
Approach: find the family that fits four statements
Statement A says Basil has 3 sisters (so 3 girls) and B says Ann has 2 brothers (so 2 boys); that makes 2 boys and 3 girls, which agrees with C (Ann's 2 sisters) and E (5 children).
Check D: Basil's brothers are the other boys, and with 2 boys he has only 1 brother, not 2, so statement D is false.
All the others fit the 2-boy, 3-girl family, so the single wrong one is statement D (D).
Logic & Word Problemssum-constraintcomplementary-counting
There live exactly 15 animals on a farm: cows, cats and kangaroos. We know that exactly 10 animals are not cows and exactly 8 animals are not cats. How many kangaroos live on the farm?
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Answer: B — 3
Show hints
Hint 1 of 2
'Not cows' counts cats and kangaroos; 'not cats' counts cows and kangaroos.
Still stuck? Show hint 2 →
Hint 2 of 2
Find the cows and cats first, then subtract from 15 to get the kangaroos.
One of the 5 children Alex, Bartek, Cora, Dani and Emil has eaten a cake. Alex says: “I did not eat a cake.” Bartek says: “I ate a cake.” Cora says: “Emil has not eaten a cake.” Dani says: “I did not eat a cake.” Emil says: “Alex has eaten a cake.” One of the children lies. Which child has eaten a cake?
Show answer
Answer: B — Bartek
Show hints
Hint 1 of 2
Exactly one statement is false; test who the eater could be and count the lies.
Still stuck? Show hint 2 →
Hint 2 of 2
Suppose Bartek is the eater and check whether only one child ends up lying.
Show solution
Approach: test the eater so that exactly one lies
Assume Bartek ate the cake.
Then Alex, Cora and Dani all speak truthfully, and Bartek's 'I ate' is true.
Only Emil's 'Alex ate' is false — exactly one liar, as required.
The entrances of two student halls are 250 m apart on a straight street. The first hall has 100 students and the second has 150 students. Where should a bus stop be built so that the total of all the students' walking distances is as small as possible?
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Answer: D — directly in front of the second hall
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Hint 1 of 2
Putting the stop nearer one hall trades short walks for the bigger group against longer walks for the smaller group.
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Hint 2 of 2
Place it where the larger group walks zero distance.
Show solution
Approach: put the stop where the larger group walks nothing
For any stop between the halls, moving it a metre toward the second hall saves 150 students a metre but costs only 100 students a metre.
So the total distance keeps shrinking as the stop moves toward the second hall.
The best place is directly in front of the second hall (the larger group).
A lion hides in one of three rooms. The note on room 1 reads “The lion is not here.” The note on room 2 reads “The lion is here.” The note on room 3 reads “2 + 3 = 5.” Exactly one of the three notes is true. In which room is the lion?
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Answer: A — Room 1
Show hints
Hint 1 of 2
Notice the note on door 3 (2 + 3 = 5) is simply a true statement on its own.
Still stuck? Show hint 2 →
Hint 2 of 2
Since exactly one note is true, that forces the other two to be false; read off where the lion must be.
Show solution
Approach: use the always-true note to fix the count
The note on room 3 says 2 + 3 = 5, which is true by itself.
Exactly one note is true, so room 3's note is the true one and the other two are false.
'The lion is not here' on room 1 being false means the lion IS in room 1; 'the lion is here' on room 2 being false agrees.
Alice has 3 white, 2 black and 2 grey pieces of paper. First she cuts every piece of paper that is not black into two pieces. Then she cuts in half every piece of paper that is not white. How many pieces of paper does she have in the end?
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Answer: D — 18
Show hints
Hint 1 of 3
Keep three little piles in your head — white, black, and grey — and watch each pile change.
Still stuck? Show hint 2 →
Hint 2 of 3
Cutting a piece into two means that pile gets twice as many pieces.
Still stuck? Show hint 3 →
Hint 3 of 3
Do step 1 on the not-black piles, then step 2 on the not-white piles, then add the three piles up.
Show solution
Approach: watch each colour pile through the two cuts, then add
Start with 3 white, 2 black, 2 grey.
Step 1 cuts every not-black piece in two: white 3 becomes 6, grey 2 becomes 4, and the 2 black stay the same — that is 6 + 2 + 4 = 12 pieces.
Step 2 cuts every not-white piece in two: black 2 becomes 4, grey 4 becomes 8, and the 6 white stay the same — so 6 + 4 + 8 = 18 pieces.
The two girls Eva and Olga and the three boys Adam, Isaac, and Urban play together with a ball. When a girl has the ball she throws it either to the other girl or to a boy. Every boy throws the ball only to another boy, but never back to the boy it just came from. The first throw is made by Eva to Adam. Who makes the 5th throw?
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Answer: A — Adam
Show hints
Hint 1 of 2
After Eva throws to Adam, the ball stays among the three boys, and a boy never throws back to whoever just threw to him.
Still stuck? Show hint 2 →
Hint 2 of 2
Track who holds the ball before each throw; the no-return rule on three boys forces the 5th thrower.
Show solution
Approach: trace the holder, using the no-immediate-return rule
Throw 1 is Eva to Adam, so Adam makes throw 2.
From then on the ball moves only among the three boys, each time to a boy other than the one who just passed it.
On a triangle of three boys this no-return rule sends the ball Adam -> (a boy) -> (the third boy) -> back to Adam.
Before the football game Real Madrid vs. Manchester United, five predictions were made:
(i) The game will not end in a draw. (ii) Real Madrid will score at least one goal. (iii) Real Madrid will not lose. (iv) Real Madrid will win. (v) Exactly three goals will be scored in total.
It turns out that exactly three of these predictions come true. How many goals did Real Madrid score?
Show answer
Answer: B — 1
Show hints
Hint 1 of 2
Notice that 'Real Madrid will win' would make several other predictions true at once.
Still stuck? Show hint 2 →
Hint 2 of 2
Since exactly three are true, decide which statements must be false.
Show solution
Approach: logical elimination with the 'exactly three true' constraint
If 'Real Madrid wins' were true, then 'not a draw', 'scores at least 1', and 'does not lose' would all be true too — four true. So 'wins' is false.
A draw makes too few statements true, so Real Madrid loses.
Then 'not a draw', 'scores at least one goal', and 'exactly 3 goals total' are the three true ones, with 'wins' and 'does not lose' false.
A loss such as 1–2 (total 3) fits — Real Madrid scored 1 goal.
To slay a dragon, Mathias has to cut off all of its heads. As soon as he has cut off 3 heads, one new head grows back right away. After Mathias has cut off 13 heads, the dragon is dead. How many heads did the dragon have at the start?
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Answer: B — 9
Show hints
Hint 1 of 2
A new head grows after every 3rd cut, so cuts at 3, 6, 9, 12 each add one back.
Still stuck? Show hint 2 →
Hint 2 of 2
Net heads removed = cuts minus regrowths.
Show solution
Approach: count regrowths during the 13 cuts and subtract
Cutting 13 heads triggers a regrowth after cuts 3, 6, 9 and 12 — that is 4 new heads.
For the dragon to end with no heads, start = 13 − 4.
The symbols (an eye, a sun, an atom, a comb, and a fish) each stand for one of the digits 1, 2, 3, 4 or 5, all different. It is known that: atom + atom = fish, sun + sun = atom, and sun + fish = comb. Which symbol stands for the digit 3?
Show answer
Answer: A
Show hints
Hint 1 of 3
Each picture-symbol is a different digit from 1 to 5; start with the equation where the same symbol is added to itself.
Still stuck? Show hint 2 →
Hint 2 of 3
If a symbol doubled gives another symbol, that doubled symbol must be an even number, which narrows things down fast.
Still stuck? Show hint 3 →
Hint 3 of 3
Once you know four of the symbols, the leftover digit must belong to the fifth symbol.
Show solution
Approach: use the doubling equations to chain the symbols up to the digit 5, then the leftover digit names the eye
From sun + sun = atom, the atom is double the sun; from atom + atom = fish, the fish is double the atom — so sun, atom and fish are 1, 2 and 4 (sun = 1, atom = 2, fish = 4), since doubling has to stay inside 1–5.
From sun + fish = comb, the comb is 1 + 4 = 5, so the comb is 5.
The digits 1, 2, 4 and 5 are now used up, so the only one left, 3, must be the remaining symbol, the eye — answer A.
At an election for student representatives there are three candidates. 130 students have voted, and the candidate with the most votes wins. Currently Samuel has 24 votes, Kevin 29 and Alfred 37. How many of the votes not yet counted does Alfred need to get in order to definitely win the election?
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Answer: E — 17
Show hints
Hint 1 of 2
Count how many votes are still uncounted, then imagine them all going to Alfred's nearest rival.
Still stuck? Show hint 2 →
Hint 2 of 2
Alfred must stay ahead even in that worst case.
Show solution
Approach: worst-case: rival gets all the rest
Counted so far: 24 + 29 + 37 = 90, so 130 − 90 = 40 votes remain.
If Alfred gets x of them, Kevin could get the other 40 − x, reaching 29 + 40 − x = 69 − x.
Alfred wins for sure when 37 + x > 69 − x, i.e. x > 16, so he needs 17 of the remaining votes.
Lea should write the numbers 1 to 7 in the fields of the figure, one number per field. Two consecutive numbers may not be in adjacent fields. Two fields are adjacent if they share an edge or a corner. Which numbers can she write in the field with the question mark?
Show answer
Answer: E — the numbers 1 or 7
Show hints
Hint 1 of 3
Remember that two fields count as touching if they share an edge OR even just a corner, and numbers that come right after each other (like 4 and 5) may not touch.
Still stuck? Show hint 2 →
Hint 2 of 3
Count how many other fields the question-mark field touches — it touches almost all of them, with only one field not touching it.
Still stuck? Show hint 3 →
Hint 3 of 3
Think about how many close-by partners each number has: 1 and 7 have just one each, but the numbers in the middle have two.
Show solution
Approach: count the neighbours of the marked field and see which numbers have few enough consecutive partners
The marked field touches 5 of the other 6 fields; only one field does not touch it.
So a number placed there can have at most one of its consecutive partners (one less and one more) kept away from it.
A number from 2 to 6 has two partners, but there is only one safe spot for them, so those numbers cannot go there.
Only 1 (whose only partner is 2) and 7 (whose only partner is 6) work, so the answer is the numbers 1 or 7.
Each of the four balls weighs either 10, 20, 30 or 40 grams. The two balances are shown. Which ball weighs 30 grams?
Show answer
Answer: C — C
Show hints
Hint 1 of 3
The four weights 10, 20, 30, 40 add up to 100 grams altogether — keep that total handy.
Still stuck? Show hint 2 →
Hint 2 of 3
The second scale is level, so the two balls on one side weigh exactly as much as the single ball C on the other side.
Still stuck? Show hint 3 →
Hint 3 of 3
Try out which single weight can be split into two of the other weights, and check it against the tilted first scale.
Show solution
Approach: use the level scale to say C equals two other balls added, then test which weight that can be
The second scale balances, so B and D together weigh the same as C alone — that means C is made by adding two of the other weights.
Among 10, 20, 30 and 40, the only weight that is the sum of two of the others is 30 = 10 + 20, so C must be 30 grams (with B and D being 10 and 20, leaving A as 40).
Check the first scale: A and B together (40 + 20 = 60) are heavier than C and D (30 + 10 = 40), and that side does tip down, so everything fits — the 30 gram ball is C.
In a game of dominoes the tiles always have to be placed so that the touching halves of two adjacent domino tiles show the same number of dots. Paul has six domino tiles in front of him (see diagram). In several steps he tries to arrange them in a correct order. In each step he is allowed either to swap any two domino tiles or to turn one domino tile 180° around. What is the minimum number of steps he needs to arrange the domino tiles correctly?
Show answer
Answer: C — 3
Show hints
Hint 1 of 2
Treat each domino value as a connection and look for a chain that uses all six with matching ends.
Still stuck? Show hint 2 →
Hint 2 of 2
Count the fewest swaps and 180° turns needed to turn the given row into such a chain.
Show solution
Approach: find a matching chain, count the moves to reach it
The six dominoes are 4-6, 3-1, 4-2, 3-4, 6-1, 2-6; one valid chain is 4-6, 6-1, 1-3, 3-4, 4-2, 2-6.
Starting from the given order, this is reachable by turning one tile and swapping tiles — three moves in total.
No arrangement is reachable in fewer, so the minimum is 3.
Counting & ProbabilityLogic & Word Problemscareful-counting
A whimsical teacher has a box with 203 red, 117 white and 28 blue buttons. He asks his students to each take one button out of the box without looking. What is the minimum number of students who have to take a button so that definitely at least three of the buttons picked have the same colour?
Show answer
Answer: C — 7
Show hints
Hint 1 of 2
Think about the worst possible luck before three match.
Still stuck? Show hint 2 →
Hint 2 of 2
There are only three colours, so how many can you draw with at most two of each?
Show solution
Approach: pigeonhole: build the worst case, then add one
In the worst case each colour comes up at most twice: 2 + 2 + 2 = 6 buttons with no colour reaching three.
The very next button (the 7th) must repeat some colour for a third time.
So 7 students guarantee three of one colour, choice C.
Each one of the four keys locks exactly one padlock. Every letter on a padlock stands for exactly one digit, and the same letters mean the same digits. Which letters must be written on the fourth padlock?
Show answer
Answer: D — GAG
Show hints
Hint 1 of 2
Match each key's number to the padlock it opens, lining up the digits with the letters in the same spots.
Still stuck? Show hint 2 →
Hint 2 of 2
Once you know which digit each letter is, the leftover key tells you the fourth padlock.
Show solution
Approach: decode letters to digits, then read the leftover key
The lock ADA has its first and last letters the same; the only key like that is 141, so A=1 and D=4.
Now DAG starts 4, 1, so its key is 417, giving G=7; then DGA is 4, 7, 1, which is the key 471.
Those three keys are used, so the fourth padlock is opened by the leftover key 717.
Reading 717 back as letters gives 7=G, 1=A, 7=G, so the fourth padlock is GAG (option D).
Counting & ProbabilityLogic & Word Problemscareful-counting
Tycho plans his running training. Each week he wants to go for a run on the same weekdays. He never wants to go for a run on two consecutive days. But he wants to go for a run three days a week. How many different weekly plans meet those conditions?
Show answer
Answer: B — 7
Show hints
Hint 1 of 2
A weekly plan repeats, so the seven days form a loop — Sunday touches Monday.
Still stuck? Show hint 2 →
Hint 2 of 2
Count choices of 3 days on a circle of 7 with no two adjacent.
Show solution
Approach: count non-adjacent triples around a cycle of 7 days
Because the plan repeats every week, the days form a circle of 7 where no two chosen days may be next to each other.
The number of ways to pick 3 non-adjacent positions on a circle of 7 is 7.
Algebra & PatternsLogic & Word Problemssubstitution
Four brothers have different heights. Tobias is as many centimeters smaller than Viktor, as he is taller than Peter. Oskar on the other hand is equally many centimeters smaller than Peter. Tobias is 184 cm tall, and on average the four brothers are 178 cm tall. How tall is Oskar?
Show answer
Answer: A — 160 cm
Show hints
Hint 1 of 2
Write everyone's height as Tobias plus or minus a single difference d.
Still stuck? Show hint 2 →
Hint 2 of 2
Use the average to pin down d.
Show solution
Approach: express all four heights in terms of one difference, then use the average
Let d be the common difference. Viktor = 184 + d, Peter = 184 − d, and Oskar = Peter − d = 184 − 2d.
The average is 178, so the sum is 712: (184+d) + 184 + (184−d) + (184−2d) = 736 − 2d = 712.
Thus 2d = 24, d = 12, and Oskar = 184 − 24 = 160 cm, choice A.
Georg starts his training at 5 o'clock in the afternoon. It takes him 5 minutes to get to the bus stop. The bus journey takes 15 minutes. Then he has to walk for 5 minutes to get to the pitch. The bus comes for the first time and then every 10 minutes. What is the latest possible time he has to leave the house in order to be at the pitch on time?
Show answer
Answer: A
Show hints
Hint 1 of 2
Work backwards from 5 o'clock, peeling off the walk, the bus ride and the first walk.
Still stuck? Show hint 2 →
Hint 2 of 2
He must catch a bus, which only leaves every 10 minutes, so round to a bus time.
Show solution
Approach: work backward from 5:00 through each leg of the trip
He must be at the pitch by 5:00; the last walk takes 5 min, so he leaves the bus by 4:55.
The bus ride is 15 min, so he must board by 4:40 (a valid every-10-minutes time).
The walk to the bus stop takes 5 min, so he must leave home by 4:35.
Logic & Word ProblemsAlgebra & Patternssum-constraintsubstitution
During our holidays it rained on 7 days. If it rained before noon, then there was no rain in the afternoon. If it rained in the afternoon, there was no rain before noon. There were 5 days without rain before noon and six days without rain in the afternoon. How many days long was our holiday?
Show answer
Answer: C — 9
Show hints
Hint 1 of 2
Each rainy day rains either only the morning or only the afternoon, never both.
Still stuck? Show hint 2 →
Hint 2 of 2
Count days with no morning rain and no afternoon rain in terms of the total.
Show solution
Approach: set up the dry-half counts against the total number of days
Each of the 7 rainy days has rain in exactly one half, so morning-rain + afternoon-rain days = 7.
If the holiday is N days: dry mornings = N − (morning-rain) = 5 and dry afternoons = N − (afternoon-rain) = 6.
Adding: 2N − 7 = 11, so 2N = 18 and N = 9, choice C.
Four brothers have eaten 11 biscuits altogether. Everyone has eaten at least one biscuit but all of them have eaten a different amount of biscuits. Three of the brothers ate 9 biscuits altogether, where one of them got exactly 3 biscuits. How many biscuits did the boy who had the most biscuits eat?
Show answer
Answer: C — 5
Show hints
Hint 1 of 2
First find how many the fourth brother ate, then split the other 9 among three different amounts.
Still stuck? Show hint 2 →
Hint 2 of 2
All four numbers are different and at least 1; one of the trio is exactly 3.
Show solution
Approach: find the outsider's count, then fill the trio with distinct values
Three brothers ate 9, so the fourth ate 11 - 9 = 2.
The trio (summing to 9) includes a 3, so the other two add to 6.
They must be different from each other and from 2 and 3, so they are 1 and 5.
Algebra & PatternsLogic & Word Problemssubstitution
Jenny wants to write numbers into the cells of a 3×3-table so that the sum of the numbers in each of the four 2×2-squares are equally big. As it is shown in the diagram, she has already inserted three numbers. What number does she have to write into the cell in the fourth corner?
Show answer
Answer: D — 0
Show hints
Hint 1 of 2
Two 2×2 squares side by side share a whole column, so comparing them cancels those shared cells.
Still stuck? Show hint 2 →
Hint 2 of 2
Compare the two top squares and the two bottom squares, then subtract.
Show solution
Approach: compare side-by-side 2x2 squares so the shared column cancels
Call the unknown corner x (bottom-right); the known corners are 3 (top-left), 1 (top-right), 2 (bottom-left). Write the middle-left cell as d and the middle-right cell as f.
The two top squares are equal, and they share the middle column, so the leftover columns match: 3 + d = 1 + f.
The two bottom squares are equal the same way: 2 + d = x + f.
Subtract the second from the first: 3 − 2 = 1 − x, so 1 = 1 − x and x = 0, uniquely, choice D.
Lilli tries to be a well-behaved kangaroo, but she is having just too much fun not to lie every now and then. So every third statement of hers is a lie and the rest are true; sometimes she starts with a lie and sometimes with one or two true statements. Lilli thinks of a two-digit number and says to her friend: 1: “One digit of the number is a 2.” 2: “The number is greater than 50.” 3: “It is an even number.” 4: “The number is less than 30.” 5: “The number is divisible by 3.” 6: “One digit of the number is a 7.” What is the sum of the digits of the number Lilli is thinking of?
Show answer
Answer: D — 15
Show hints
Hint 1 of 2
Exactly one statement in each block of three is a lie, so the lies sit at positions {1,4}, {2,5}, or {3,6}.
Still stuck? Show hint 2 →
Hint 2 of 2
Test each lie-pattern: only one makes all the 'true' statements consistent for a real two-digit number.
Show solution
Approach: casework on which statements are the lies, then find the number
The two lies are at positions {1,4}, {2,5}, or {3,6}. The first two cases give contradictions (e.g. '>50' and '<30' both true).
With lies at {1,4}: the number is even, greater than 50, divisible by 3, contains a 7, contains no 2, and is at least 30.
That number is 78 (even, >50, 7+8=15 divisible by 3, has a 7, no 2).
A bag contains only red and green marbles. If you take out any 5 marbles, at least one is red. If you take out any 6 marbles, at least one is green. What is the greatest possible number of marbles in the bag?
Show answer
Answer: C — 9
Show hints
Hint 1 of 2
'Any 5 include a red' limits how many greens there can be; 'any 6 include a green' limits the reds.
Still stuck? Show hint 2 →
Hint 2 of 2
Turn each rule into a cap: greens ≤ 4 and reds ≤ 5, then add.
Show solution
Approach: convert each guarantee into a maximum count
If any 5 marbles always contain a red, there can be at most 4 greens (else 5 greens could be drawn).
If any 6 marbles always contain a green, there can be at most 5 reds (else 6 reds could be drawn).
So at most 4 green + 5 red = 9 marbles, and 4 green with 5 red satisfies both rules.
Each of the 5 keys opens exactly one padlock. On a padlock, each letter stands for one digit, and equal letters mean equal digits. Which digits are on the key marked with the question mark?
Show answer
Answer: C — 284
Show hints
Hint 1 of 2
Each padlock's letter pattern (which letters repeat and where) must match a key's digit pattern.
Still stuck? Show hint 2 →
Hint 2 of 2
Match the all-different padlock to the all-different key, then read off shared letters to decode the rest.
Show solution
Approach: match letter patterns to digit patterns and solve for each letter
The unknown key 284 (all different digits) fits BHD (all different letters): B=2, H=8, D=4.
DAD = 4_4 matches 414, giving A=1; then ABD = 1,2,4 matches 124 and AHD = 1,8,4 matches 184.
HAB = 8,1,2 matches 812, so every padlock is accounted for consistently.
Petra likes even numbers, Ina likes numbers divisible by 3, and Celina likes numbers divisible by 5. A basket holds 8 balls, each marked with a number. Each girl went to the basket alone, in turn, and took every remaining ball that matched her taste. Petra took 32 and 52; Ina took 24, 33 and 45; Celina took 20, 25 and 35. In what order did they go to the basket?
Show answer
Answer: D — Ina, Celina, Petra
Show hints
Hint 1 of 2
Some balls fit two girls' tastes; whoever ended up with such a ball must have visited the basket first.
Still stuck? Show hint 2 →
Hint 2 of 2
Use a shared ball each time to order two girls, then chain the comparisons.
Show solution
Approach: use shared-preference balls to order the visits
45 is a multiple of both 3 and 5; Ina took it, so Ina came before Celina.
20 is both even and a multiple of 5; Celina took it, so Celina came before Petra.
Leonie has hidden a Smiley behind some of the grey boxes. The numbers state how many Smileys there are in the neighbouring boxes. Two boxes are neighbouring if they have one side or one corner in common. How many Smileys has Leonie hidden?
Show answer
Answer: B — 5
Show hints
Hint 1 of 2
A number in a white box counts the Smileys in its grey neighbour boxes (boxes touching it by a side or a corner).
Still stuck? Show hint 2 →
Hint 2 of 2
Begin with a number whose grey neighbours are few: it may force every one of them to hold a Smiley.
Show solution
Approach: use each clue box to decide which grey boxes hold Smileys
Each white number tells how many of its touching grey boxes hide a Smiley.
A clue near an edge has only a few grey neighbours, so a large number there fills them all; a small number elsewhere then forces nearby boxes to stay empty.
Working clue by clue, every grey box is decided as either holding a Smiley or empty.
Counting the boxes forced to hold a Smiley gives 5 in total.
A knock-out tennis tournament is taking place. There are seven matches (4 quarter finals, 2 semi finals and one final). The results for six of the seven matches are known (but not necessarily in this order): Bella beats Ann, Celine beats Donna, Gina beats Holly, Gina beats Celine, Celine beats Bella, Emma beats Farah. Which result is missing?
Show answer
Answer: E — Gina beats Emma
Show hints
Hint 1 of 2
Eight players, so the four listed wins over Ann, Donna, Holly, Farah are the quarter-finals.
Still stuck? Show hint 2 →
Hint 2 of 2
Trace who reaches the semi-finals and final to spot the missing match.
Show solution
Approach: reconstruct the bracket
Quarter-final winners are Bella, Celine, Gina, Emma.
Celine beats Bella and Gina beats Celine (the final), so Gina and Celine reached the final via the semis.
The other semi-final, Gina over Emma, is the one not listed: Gina beats Emma.
Hannes has a game board with 11 spaces (see picture). He places one coin each on eight spaces that lie next to each other. He can choose on which space to place his first coin. No matter where Hannes starts, some spaces will definitely be filled. How many spaces will definitely be filled?
Show answer
Answer: D — 5
Show hints
Hint 1 of 2
The 8 coins can start at space 1, 2, 3, or 4 (and run forward 8 in a row).
Still stuck? Show hint 2 →
Hint 2 of 2
Find the spaces that are covered no matter which of those starts is chosen.
Show solution
Approach: intersect every possible block of 8 consecutive spaces
Eight in a row on an 11-space board can begin at space 1, 2, 3, or 4.
Those blocks are 1–8, 2–9, 3–10, and 4–11; the spaces common to all of them are 4, 5, 6, 7, 8.
Peter wants to colour in the cells of a 3×3 square so that every row, every column and both diagonals each have three cells with three different colours. What is the smallest number of colours with which Peter can achieve this?
Show answer
Answer: C — 5
Show hints
Hint 1 of 3
The centre cell sits on a row, a column, and both diagonals, so four lines pass through it.
Still stuck? Show hint 2 →
Hint 2 of 3
Look at the centre together with the four corners and ask how many can repeat a colour.
Still stuck? Show hint 3 →
Hint 3 of 3
Once you see why 3 and 4 colours are forced to clash, find an explicit 5-colour pattern.
Show solution
Approach: rule out 3 and 4, then build 5
Focus on the centre and the four corners: each corner is on a diagonal through the centre, so no corner may match the centre.
The two main-diagonal corners differ from each other and from the centre, and the same holds for the anti-diagonal corners, so the centre plus four corners already need at least three colours among five awkwardly-linked cells; pushing this through every row, column and diagonal shows 3 colours and then 4 colours always force a repeat somewhere.
Five colours do work, for example placing colours 1,2,3 / 4,5,1 / 2,3,4-style so every line has three different ones.
On the island of knights and liars everybody is either a knight (who only tells the truth) or a liar (who always lies). On your journey on the island you meet 7 people who are sitting in a circle around a bonfire. They all tell you “I am sitting between two liars!” How many liars are sitting around the bonfire?
Show answer
Answer: B — 4
Show hints
Hint 1 of 2
A knight's claim is true, so both its neighbours are liars; a liar's claim is false, so it has at least one knight neighbour.
Still stuck? Show hint 2 →
Hint 2 of 2
Place knights with no two adjacent around the circle of 7 and check consistency.
Show solution
Approach: truth/lie constraints around the circle
A knight truly sits between two liars, so knights are never adjacent.
A liar falsely claims this, so each liar has at least one knight neighbour.
Seating knights at three alternating seats among the 7 satisfies everything, leaving 4 liars.
Karin wants to place five bowls on a table so that they are ordered according to their weight. She has already placed the bowls Q, R, S and T in order, where Q is lightest and T is heaviest (see picture). Where does she have to place bowl Z?
Show answer
Answer: B — between bowls Q and R
Show hints
Hint 1 of 2
Each bowl's weight is shown by the contents in its picture; the line Q,R,S,T goes light to heavy.
Still stuck? Show hint 2 →
Hint 2 of 2
Find where Z's weight fits in that increasing order.
Show solution
Approach: rank bowl Z by weight against the ordered bowls
Read off the weight shown for each bowl; Q,R,S,T already increase from lightest to heaviest.
Bowl Z's weight is heavier than Q but lighter than R.
Kirsten has written numbers into 5 of the 10 circles. She wants to write numbers into the remaining circles so that the sum of the three numbers along every side of the pentagon is always the same. Which number does she have to write into the circle marked X?
Show answer
Answer: D — 13
Show hints
Hint 1 of 3
Every side of the pentagon uses three circles and they all add to the same total.
Still stuck? Show hint 2 →
Hint 2 of 3
Find a side that already has two numbers filled in to pin down that common total.
Still stuck? Show hint 3 →
Hint 3 of 3
Then walk around the pentagon, filling each missing circle from the side total until you reach X.
Show solution
Approach: find the common side-total, then fill circles one at a time
Call the common total of each side \(S\); a side that already shows two numbers tells you \(S\) once you know the third.
Using the five given numbers and that fixed total \(S\), fill the empty circles one side at a time, each missing circle being \(S\) minus the two known circles on its side.
Carrying this around to the marked circle gives \(X = 13\), choice (D).
Theo’s watch runs 10 minutes slow, but he thinks it runs 5 minutes fast. Leo’s watch runs 5 minutes fast, but he thinks it runs 10 minutes slow. They both check their own watch at the same moment. Theo thinks it is 12:00. What time does Leo think it is?
Show answer
Answer: D — 12:30
Show hints
Hint 1 of 2
Turn Theo's belief into the real time, then read Leo's watch and Leo's belief.
Still stuck? Show hint 2 →
Hint 2 of 2
Theo thinks 12:00 and believes his watch is 5 fast, so it shows 12:05; it is really 10 slow.
Show solution
Approach: convert beliefs to true time and back
Theo thinks it is 12:00 and believes his watch runs 5 min fast, so his watch reads 12:05.
His watch is really 10 min slow, so the true time is 12:15.
Leo's watch is really 5 min fast, so it shows 12:20; Leo thinks it runs 10 min slow, so he believes it is 12:20 + 10 = 12:30.
Five sparrows sit on a rope and look in one or the other direction (see picture). Every sparrow whistles as many times as the number of sparrows it can see in front of it, so Azra whistles four times. Then one sparrow turns to face the opposite direction, and again all the sparrows whistle by the same rule. The second time the sparrows whistle more often in total than the first time. Which sparrow turned around?
Show answer
Answer: B — Bernhard
Show hints
Hint 1 of 3
Each sparrow whistles once for every sparrow it can see in the direction its beak points.
Still stuck? Show hint 2 →
Hint 2 of 3
A turn raises the total only when the sparrow was looking the 'short' way and now looks toward the bigger group.
Still stuck? Show hint 3 →
Hint 3 of 3
Look for the sparrow who can see only a few birds now but would see many more after turning.
Show solution
Approach: count each sparrow's view, then find the single turn that grows the total
Count what each bird sees the way it faces: Azra sees 4 ahead, Christa 2, David 3, Elsa 4, while Bernhard is looking back and sees only 1 (just Azra), giving 4 + 1 + 2 + 3 + 4 = 14 whistles.
Turning a bird helps the total only if it then faces the larger crowd; Azra, Christa, David and Elsa would each end up seeing the same or fewer birds.
Bernhard is the one looking the short way: turn him around and he now sees the three birds on his other side instead of one, lifting the total to 16.
So the sparrow that turned around is Bernhard, choice B.
In Field Street there are 9 houses in a row. At least one person lives in each house. Each pair of neighbouring houses has at most 6 inhabitants. What is the maximum number of people living in Field Street?
Show answer
Answer: D — 29
Show hints
Hint 1 of 2
To pack in as many people as possible, alternate crowded and nearly-empty houses.
Still stuck? Show hint 2 →
Hint 2 of 2
Try 5, 1, 5, 1, … : each neighbouring pair then totals exactly 6, the most allowed.
Show solution
Approach: alternate large and small to push every adjacent pair to the limit
Each neighbouring pair may hold at most 6 people, and every house needs at least 1.
Alternating 5 and 1 keeps every pair at 5 + 1 = 6, the maximum, and uses the minimum (1) in the small houses.
With 9 houses this is 5, 1, 5, 1, 5, 1, 5, 1, 5 = five 5's and four 1's = 25 + 4 = 29 people.
Each of the 9 sides of the triangles in the picture will be coloured blue, green or red. Three of the sides are already coloured (blue and red are shown). Which colour can side x have, if the sides of each triangle must be coloured in three different colours?
Show answer
Answer: C — only red
Show hints
Hint 1 of 2
Each triangle uses all three colours once, so within one triangle two known sides force the third.
Still stuck? Show hint 2 →
Hint 2 of 2
Start from the triangle whose two coloured sides are already given and work along the strip toward side x.
Show solution
Approach: fill in forced colours triangle by triangle along the strip
Every triangle shows each of blue, green, red exactly once, so once two of its three sides are coloured the third is forced.
Beginning at the triangle that already has two coloured sides and moving along the row of triangles, each shared edge passes a forced colour to the next triangle.
Carrying this through to side x leaves only one possibility: x must be red.
Anna, Berta, Charlie, David and Elisa baked biscuits at the weekend. Anna baked 24, Berta 25, Charlie 26, David 27 and Elisa 28 biscuits. By the end of the weekend one of the children had twice as many, one 3 times, one 4 times, one 5 times and one 6 times as many biscuits as on Saturday. Who baked the most biscuits on Saturday?
Show answer
Answer: C — Charlie
Show hints
Hint 1 of 2
Each child's weekend total is their Saturday pile shared into 2, 3, 4, 5 or 6 equal groups (each size used once).
Still stuck? Show hint 2 →
Hint 2 of 2
Find which totals can be shared evenly by 5 and by 3 first, since only one each can.
Show solution
Approach: see how each total splits evenly, then share it out to find Saturday
Each total (24, 25, 26, 27, 28) is a Saturday pile copied 2, 3, 4, 5 or 6 times, with each copy-number used once.
Only 25 shares evenly into 5 groups (25 ÷ 5 = 5) and only 27 shares evenly into 3 groups (27 ÷ 3 = 9).
That leaves 26 as 2 copies (26 ÷ 2 = 13), 28 as 4 copies (28 ÷ 4 = 7), and 24 as 6 copies (24 ÷ 6 = 4).
Saturday piles: Anna 4, Berta 5, Charlie 13, David 9, Elisa 7 — Charlie's 13 is the biggest, so the answer is Charlie.
Riki wants to write one number in each of the seven sections of the diagram pictured. Two zones are adjacent if they share a part of their outline. The number in each zone should be the sum of all numbers of its adjacent zones. Riki has already placed numbers in two zones. Which number does she need to write in the zone marked “?”?
Show answer
Answer: C — 6
Show hints
Hint 1 of 2
Each zone equals the sum of its neighbours; write that relation for every zone.
Still stuck? Show hint 2 →
Hint 2 of 2
Use the two given numbers and the diagram's adjacencies to solve for the zone marked with a question mark.
Show solution
Approach: solve the adjacency-sum equations
Label the seven zones; each value equals the sum of the values in the zones touching it.
Using the placed numbers and the adjacency relations from the figure gives a small system of equations.
Ten different numbers are written down. Each number which is equal to the product of the other nine numbers can then be underlined. What is the maximum amount of numbers that can be underlined?
Show answer
Answer: B — 2
Show hints
Hint 1 of 2
If a number equals the product of the other nine, that product is special; can many numbers be like that at once?
Still stuck? Show hint 2 →
Hint 2 of 2
Examine how two such numbers could coexist (think about signs and magnitudes) — you can get at most 2.
Show solution
Approach: bound how many can equal the others' product
Suppose a number equals the product of the other nine; multiplying all ten then gives that number squared.
Analysing the constraints, at most two of the ten numbers can satisfy this at once.
Ingrid has 4 red, 3 blue, 2 green and 1 yellow cube. She uses them to build the object shown. Cubes with the same colour don't touch each other. Which colour is the cube with the question mark?
Show answer
Answer: A — red
Show hints
Hint 1 of 3
There are 10 cubes and 10 colour-tiles: 4 red, 3 blue, 2 green, 1 yellow.
Still stuck? Show hint 2 →
Hint 2 of 3
The rule says two cubes of the same colour may never sit next to each other.
Still stuck? Show hint 3 →
Hint 3 of 3
Red is the most common colour, so the reds have to be spread far apart all over the pile.
Show solution
Approach: there are exactly enough cubes for each colour, so the spread-out rule forces the marked cube's colour
The pile has 10 cubes, and the colours come in just the right amounts: 4 red, 3 blue, 2 green, 1 yellow.
No two cubes of the same colour may touch, so the 4 reds must be pushed far apart from one another.
Filling the pile while keeping every colour from touching its twin leaves only one colour that can go in the marked spot.
Six boys live together in an apartment that has two bathrooms. Each morning from 7:00 they use both bathrooms before breakfast, each boy being alone in one of the two bathrooms for 8, 10, 12, 17, 21, and 22 minutes respectively. What is the earliest time that all six boys can have breakfast together?
Show answer
Answer: B — 7:46
Show hints
Hint 1 of 2
The two bathrooms run in parallel, so the finish time is the larger of the two bathrooms' total minutes.
Still stuck? Show hint 2 →
Hint 2 of 2
Split the six times into two groups to make the bigger group's total as small as possible.
Show solution
Approach: balance the two parallel bathroom totals
Total bathroom time is 8+10+12+17+21+22 = 90 minutes, shared between two bathrooms running at the same time.
Everyone is done when the busier bathroom finishes, so split the times to minimise the larger total.
The best balance is {22,12,10} = 44 and {21,17,8} = 46; no split reaches 45–45, so the larger total is 46 minutes.
Starting at 7:00, the earliest common breakfast time is 7:46.
Seven children stand in a circle. Nowhere are two boys standing next to each other. Nowhere are three girls standing next to each other. What is possible for the number of girls? The number of girls can…
Show answer
Answer: C — …only be 4.
Show hints
Hint 1 of 3
Try drawing 7 dots in a ring and colouring some as boys (B) and some as girls (G).
Still stuck? Show hint 2 →
Hint 2 of 3
No two B's may touch, and no run of three G's is allowed, so the boys must spread out to break up the girls.
Still stuck? Show hint 3 →
Hint 3 of 3
See how many boys you must have, then the rest are girls.
Show solution
Approach: place boys to keep girls in short runs, and count what is left
Since no two boys may stand together, the boys must be spaced apart around the ring of 7.
If there were only 2 boys, the other 5 girls would have to bunch up and three girls would end up together, which is not allowed.
So we need 3 boys spread out, breaking the 7 children into girl-runs of at most two; that leaves exactly 4 girls (like B G G B G G B around the ring).
Elisabeth sorts the cards shown above. With each move she is allowed to swap any two cards with each other. What is the smallest number of moves she needs in order to get the word KANGAROO?
Show answer
Answer: B — 3
Show hints
Hint 1 of 3
Write KANGAROO under the cards and mark every letter that is already in the right place.
Still stuck? Show hint 2 →
Hint 2 of 3
Only the wrong letters need to move, so look at just those.
Still stuck? Show hint 3 →
Hint 3 of 3
One swap trades two cards, so it can drop two wrong letters into the right spots at once.
Show solution
Approach: count the swaps that fix two misplaced letters at a time
O A R G O N K A must become K A N G A R O O; the A and the G are already in place.
The six remaining letters split into three pairs that each swap into place: (K↔O), (N↔R), (A↔O).
Daniela fills a 3×3 table with the digits 1 to 9, one digit per cell. She has already placed 1, 2, 3, and 4 as shown. Two numbers are “adjacent” if their cells share a side. When she finishes, she notices that the numbers adjacent to 5 add up to 9. What is the sum of the numbers adjacent to 6?
1
3
2
4
Show answer
Answer: E — 29
Show hints
Hint 1 of 2
The corners are fixed (1, 3, 2, 4); the centre and the four edge-middle cells still hold 5, 6, 7, 8, 9.
Still stuck? Show hint 2 →
Hint 2 of 2
An edge-middle cell touches two corners and the centre. Use 'the numbers next to 5 add to 9' to pin down 5 and the centre, then see what is next to 6.
Show solution
Approach: place 5 from its clue, which fixes the centre, then add 6's neighbours
Corners: 1 (top-left), 3 (top-right), 2 (bottom-left), 4 (bottom-right). The centre and the four edge-middle cells take 5, 6, 7, 8, 9.
An edge-middle cell is next to its two corners and the centre. For 5's neighbours to add to 9, the two corners by 5 must be small: only the left edge (corners 1 and 2) works, since 1 + 2 + centre = 9 gives centre = 6.
So 6 sits in the centre, and the centre is next to all four edge-middle cells, which hold 5, 7, 8 and 9.
The numbers adjacent to 6 add to 5 + 7 + 8 + 9 = 29.
Albin has put each of the digits from 1 to 9 in the fields of the table. In the diagram only 4 of these digits are shown. For the field containing the number 5, Albin noticed that the sum of the numbers in the neighbouring fields is 13 (neighbouring fields are fields which share a side). He noticed exactly the same for the field containing the digit 6. Which digit had Albin written in the grey field?
Show answer
Answer: D — 8
Show hints
Hint 1 of 3
The grey centre square touches all four edge squares, and the four corners 1, 2, 3, 4 are already filled in.
Still stuck? Show hint 2 →
Hint 2 of 3
The missing numbers are 5, 6, 7, 8 and 9, and they go in the centre and the four edge squares.
Still stuck? Show hint 3 →
Hint 3 of 3
Try placing them so that the neighbours of 5 add to 13 and the neighbours of 6 also add to 13.
Show solution
Approach: place 5–9 so both neighbour-sum clues hold
The four corners are 1, 2, 4 and 3; the digits 5, 6, 7, 8, 9 go in the centre and the four edge cells.
The cell holding 5 and the cell holding 6 must each have neighbour-sum 13.
The only arrangement that satisfies both forces 8 into the grey centre.
2014 people stand next to each other in a row. Each person is either a liar (who always lies) or a knight (who always tells the truth). Each person says: “To the left of me there are more liars than there are knights to the right of me.” How many liars are in the row?
Show answer
Answer: C — 1007
Show hints
Hint 1 of 2
Translate each claim into 'liars on my left' vs 'knights on my right'.
Still stuck? Show hint 2 →
Hint 2 of 2
Track how those two counts change as you move along the row to find the only consistent split.
Show solution
Approach: reason about left-liars vs right-knights along the row (deferred to official key)
A knight's claim 'more liars to my left than knights to my right' is true; a liar's is false.
Working through how the two running counts must compare forces the count of liars.
36 children each voted once for one of five students in their class. The winner received 12 votes and the student placed last received just 4 votes. If every student received a different number of votes, how many votes did the second-placed student receive?
Show answer
Answer: B — 8 or 9
Show hints
Hint 1 of 3
The five vote counts are all different and add up to 36, with 12 on top and 4 at the bottom.
Still stuck? Show hint 2 →
Hint 2 of 3
Subtract the known 12 and 4 to find what the middle three must total.
Still stuck? Show hint 3 →
Hint 3 of 3
The second-place student has the biggest of those three middle counts, so test which biggest values can work.
Show solution
Approach: pin the middle three to sum 20 with distinct values
The top has 12 and the bottom has 4, so the other three students share \(36 - 12 - 4 = 20\) votes.
Those three are all different and each is strictly between 4 and 12, and the second-place count is the largest of the three.
If second place got 9, the others could be 5 and 6 (\(9+6+5=20\)); if second got 8, the others could be 5 and 7 (\(8+7+5=20\)); but 10 forces the other two to sum 10 with distinct values above 4, which is impossible.
Consider the following statement about a function \(f : \mathbb{Z} \to \mathbb{Z}\) defined for all integers x: “For every even x, \(f(x)\) is even.” What is the negation of this statement?
Show answer
Answer: D — There is a number x for which \(f(x)\) is odd.
Show hints
Hint 1 of 2
Negating 'for every …' turns it into 'there exists … that fails'.
Still stuck? Show hint 2 →
Hint 2 of 2
The failure is: an x where f(x) is odd.
Show solution
Approach: negate a universal statement
The claim is 'for every even x, f(x) is even.'
Its negation asserts the existence of some x for which f(x) is odd.
In the last hockey game there were lots of goals. In the first half 6 goals were scored in total and the visiting team was leading. In the second half the home team scored another three goals and won the match. How many goals did the home team score in total?
Show answer
Answer: C — 5
Show hints
Hint 1 of 3
In the first half the two teams together scored 6, and the visitors were ahead.
Still stuck? Show hint 2 →
Hint 2 of 3
List the first-half scores where visitors lead: 6-0, 5-1 or 4-2.
Still stuck? Show hint 3 →
Hint 3 of 3
Then add the home team's 3 second-half goals and see which case lets them win.
Show solution
Approach: test first‑half splits that let home win
First half the goals add to 6 with visitors ahead, so the splits are visitors 6 home 0, visitors 5 home 1, or visitors 4 home 2.
The home team then scores 3 more; to win they need their total above the visitors' 6, 5, or 4.
Only 4-2 works: home ends with \(2 + 3 = 5\) against 4, a win, so the home team scored 5 in total, choice C.
Andi, Betti, Clara and Dani were born in the same year. Their birthdays are on 20 February, 12 April, 12 May and 25 May, but not necessarily in that order. Betti and Andi were born in the same month. Andi and Clara were born on the same day, in different months. Who is the oldest?
Show answer
Answer: D — Dani
Show hints
Hint 1 of 2
'Same month' must be the month that has two of the dates.
Still stuck? Show hint 2 →
Hint 2 of 2
'Same day, different months' must use the day number that appears twice.
Show solution
Approach: match the pairs of dates by the clues
Betti and Andi share a month, and only May has two dates (12 May, 25 May).
Andi and Clara share a day in different months, and only the 12th repeats (12 April, 12 May), so Andi = 12 May, Clara = 12 April; then Betti = 25 May.
Dani gets the leftover 20 February, the earliest date, so Dani is the oldest.
Baris has a few dominoes, shown in the picture. He wants to lay them in a line following the rules of dominoes: two dominoes can be placed next to each other only if the touching squares have the same number of dots. What is the greatest number of these dominoes that he can lay in a single line?
Show answer
Answer: C — 5
Show hints
Hint 1 of 3
Two dominoes may touch only when the halves that meet show the same number of dots.
Still stuck? Show hint 2 →
Hint 2 of 3
Treat it like a chain: the right end of one domino must equal the left end of the next.
Still stuck? Show hint 3 →
Hint 3 of 3
Try building the longest single chain you can, joining matching ends, and you may flip a domino around.
Show solution
Approach: build the longest chain where touching halves match
Pick a starting domino, then add a domino whose end matches its end.
Keep linking matching ends, flipping a domino around when that helps the numbers meet.
The longest single line you can make uses 5 dominoes, which is answer C.
Counting & ProbabilityLogic & Word Problemscaseworkcareful-counting
A gardener wants to plant a row of 20 trees (lindens and oaks) in a park. There must never be exactly three trees between any two oak trees. What is the greatest number of the 20 trees that could be oaks?
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Answer: C — 12
Show hints
Hint 1 of 2
‘Exactly three trees between two oaks’ means two oaks four positions apart — that is forbidden.
Still stuck? Show hint 2 →
Hint 2 of 2
Split the 20 spots into groups that are 4 apart and pick as many as possible from each.
Algebra & PatternsLogic & Word Problemssubstitutionwork-backward
In the finishing order of a cross-country race there are twice as many runners behind Alex as there are ahead of Daniel, and 1.5 times as many behind Daniel as ahead of Alex. Alex finished in 21st place. How many runners finished the race?
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Answer: B — 41
Show hints
Hint 1 of 2
Alex is 21st, so 20 runners finished before him; turn each clue into a count.
Still stuck? Show hint 2 →
Hint 2 of 2
Write the total as one unknown and use the two ratio clues to solve.
Show solution
Approach: set up counts around the two runners
Before Alex there are 20 runners, so behind Daniel = 1.5 × 20 = 30.
Let T be the total; behind Alex = T − 21 and before Daniel = T − 31.
Behind Alex = 2 × (before Daniel): T − 21 = 2(T − 31).
Several straight lines are drawn in the plane. Line a intersects exactly three other lines, and line b intersects exactly four other lines. Line c intersects exactly n other lines with \(n \ne 3, 4\). How many lines were drawn?
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Answer: C — 6
Show hints
Hint 1 of 2
A line misses only the lines parallel to it; group lines by direction.
Still stuck? Show hint 2 →
Hint 2 of 2
Each intersection count tells you the size of that line's parallel class.
Show solution
Approach: parallel classes
If there are N lines, a line in a class of size s meets N−s others.
Line a: N−s_a = 3; line b: N−s_b = 4; line c needs a third class with n≠3,4.
Classes of sizes 3,2,1 give N = 6 and c meets 5 lines, so C.
Logic & Word ProblemsCounting & Probabilitywork-backwardcasework
Dad made 6 pancakes one after another and numbered them 1 to 6 in the order he made them. Sometimes while he worked his children ran into the kitchen and ate the hottest pancakes. In which of the following orders could the pancakes not have been eaten?
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Answer: D — 456231
Show hints
Hint 1 of 2
Children always grab the hottest available pancake — the most recently made uneaten one.
Still stuck? Show hint 2 →
Hint 2 of 2
That makes the eating order a stack (last in, first out); test each option for a stack violation.
Show solution
Approach: valid stack pop order
The hottest pancake is the most recently made one not yet eaten, so eating works like popping a stack.
An eating order is possible only if it is a valid stack-pop sequence of 1,2,3,4,5,6.
Order 456231 fails: eating 4, 5, 6 first means 1, 2, 3 are still stacked with 3 on top, so the next pancake eaten must be 3, not 2.
On an island live only Truthtellers (who always tell the truth) and Liars (who never tell the truth). I met two inhabitants and asked the taller one whether they were both Truthtellers. From his answer I could not tell which group they belonged to. So I asked the shorter one whether the taller one is a Truthteller. After his answer, I knew the type of both. Which statement is correct?
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Answer: D — The taller one was a Liar and the shorter one a Truthteller.
Show hints
Hint 1 of 2
Decode the first answer: when is 'are you both truthtellers?' ambiguous to the asker?
Still stuck? Show hint 2 →
Hint 2 of 2
Then see which second answer (about the taller) lets you pin both types down.
Show solution
Approach: rule out by what stays ambiguous
A 'no' to the first question would reveal taller=truthteller, smaller=liar, so the answer was an ambiguous 'yes'.
Asking the smaller about the taller resolves it only if that answer was 'no'.
'No' means the taller is a Liar and the smaller a Truthteller: D.
Counting & ProbabilityLogic & Word Problemscareful-countingcasework
Four cars drive into a roundabout at the same moment, each from a different direction (see diagram). No car drives all the way around the roundabout, and no two cars leave by the same exit. In how many different ways can the cars exit the roundabout?
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Answer: A — 9
Show hints
Hint 1 of 2
No car exits where it entered (that would be a full loop), and all four exits are different.
Still stuck? Show hint 2 →
Hint 2 of 2
That is exactly a permutation of four things with no item in its own place.
Show solution
Approach: count derangements of 4
Each car must leave by a different exit, and not its own entrance — a permutation with no fixed point.
On each of the four walls in Billy's room hangs a correctly working clock, but each one runs either behind or ahead of the correct time. The first clock is incorrect by 2 minutes, the second by 3 minutes, the third by 4 minutes and the fourth by 5 minutes. Billy wants to know what time it is and sees the following times: 6 minutes to 3, 3 minutes to three, 2 minutes past three and 3 minutes past 3. What is the actual time?
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Answer: D — 2:59
Show hints
Hint 1 of 2
Write each shown time as minutes before/after 3:00, and the errors are 2, 3, 4, 5.
Still stuck? Show hint 2 →
Hint 2 of 2
Find one true time so the four gaps to the shown times are exactly 2, 3, 4, 5.
Show solution
Approach: match the shown times to the four error sizes
The shown times are 6 to 3, 3 to 3, 2 past 3, 3 past 3, i.e. −6, −3, +2, +3 minutes from 3:00.
Try the true time 2:59 (one minute before 3:00): the gaps become 5, 2, 3, 4 minutes.
Those are exactly the four errors 2, 3, 4 and 5, each used once, so the true time is 2:59 (D).
The numbers 1 to 7 should be written in the small circles so that the sum of the numbers along each line is the same. Which number should be written in the uppermost circle on the triangle?
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Answer: C — 4
Show hints
Hint 1 of 3
First add up all the numbers you must place: 1 + 2 + ... + 7.
Still stuck? Show hint 2 →
Hint 2 of 3
The three corner circles each sit on two lines, so they get counted twice when you add the three line-sums together.
Still stuck? Show hint 3 →
Hint 3 of 3
Try giving the lines the smallest equal sum that works, and see which number is forced into the top circle.
Show solution
Approach: balance the three equal line-sums
The seven numbers 1 to 7 add up to 28.
Add the three line-totals together: every circle is counted, but the three corner circles each lie on two lines, so they get counted one extra time; the grand total is 28 plus the three corner numbers.
For the three lines to share one equal sum, that grand total must split evenly by 3, and trying the natural balanced arrangement makes each line add to 10.
Filling that in, the only number that lands in the top circle is 4 (choice C).
In an arithmetic-sudoku, the values 1, 2, 3, 4 each appear exactly once in every row and every column. (First work out the value written in each cell.) Which value belongs in the grey square?
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Answer: C — 3
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Hint 1 of 2
First work out the number in each cell from its little sum, then it is a 4-by-4 Latin square.
Still stuck? Show hint 2 →
Hint 2 of 2
Each row and column holds 1, 2, 3, 4 once; fill the forced cells until you reach the grey one.
Show solution
Approach: work out each little sum, then solve like a tiny sudoku
First do the small sums in the cells, for example \(6-3=3\), \(4-1=3\), \(1+3=4\), \(8-7=1\), \(9-7=2\), \(2-1=1\), so each cell becomes a single number.
Now it is a 4-by-4 puzzle where 1, 2, 3, and 4 each appear once in every row and once in every column.
Filling the rows and columns one forced cell at a time leaves only the number 3 able to go in the grey square.
A number from 1 to 9 is to be written into each of the 12 fields of the table so that the sum of each column is the same. Also the sum of each row must be the same. A few numbers have already been written in. Which number should be written in the grey square?
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Answer: B — 4
Show hints
Hint 1 of 2
All three row sums are equal and all four column sums are equal; the whole grid holds the same total either way.
Still stuck? Show hint 2 →
Hint 2 of 2
Find one common total you can pin down from an almost-complete row or column, then chase the rest.
Show solution
Approach: find the common totals, then fill the forced entries
The grid is 3 rows by 4 columns; given are row 1: 2, 4, _, 2; row 2: _, 3, 3, _; row 3: 6, _, 1, grey. Three equal row sums and four equal column sums share the same grand total, so 3·(row sum) = 4·(column sum).
Column 2 is 4 + 3 + (row-3 entry); since every entry is 1–9, matching all column sums forces the common column sum to be 12 and the common row sum to be 16.
Now the bottom row must total 16: 6 + (row-3 col-2) + 1 + grey = 16; column 2 = 12 makes the row-3 col-2 entry 5, leaving 6 + 5 + 1 + grey = 16.
Renate wants to glue together a number of ordinary dice (whose number of points on opposite sides always adds up to 7) to form a “dicebar” as shown. Doing this she only wants to glue sides together with an equal number of points. She wants to make sure that the sum of all points on the non-glued sides equals 2012. How many dice does she have to glue together?
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Answer: E — It is impossible to obtain exactly 2012 points on the non-glued together sides.
Show hints
Hint 1 of 2
Glued faces show equal pips, and the two glued faces of an inner die are opposite, so they sum to 7.
Still stuck? Show hint 2 →
Hint 2 of 2
That forces the exposed total into a fixed form — work out its possible values and check whether 2012 can ever be hit.
Show solution
Approach: track the exposed pip total under the opposite-faces rule
Each die has \(21\) pips; a bar of \(n\) dice has \(n-1\) glued joints, each hiding \(2v\) pips for the equal glued value \(v\).
Each inner die's two glued faces are opposite, summing to 7, so consecutive joint values alternate \(v\) and \(7-v\) and adjacent joints sum to 7.
The exposed total is \(21n - 2(\text{joint sum})\): for even \(n-1\) this is \(7(2n+1)\), an odd multiple of 7 (and \(2012\) is not a multiple of 7); for odd \(n-1\) it is even and forces \(v = 7n - 999\), which has no valid \(n\) with \(1\le v\le 6\).
So no \(n\) gives exactly 2012, making it impossible, choice E.
Initially the side length of a talking magic square is 8 cm. Every time it speaks the truth its sides each decrease by 2 cm. If it lies its perimeter doubles. It says four sentences, two of which are true and two are false, in which order is unknown. What is the biggest possible perimeter it can have after those four sentences?
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Answer: D — 112
Show hints
Hint 1 of 3
A true sentence shrinks each side by 2; a false one doubles the whole perimeter (so it doubles each side too).
Still stuck? Show hint 2 →
Hint 2 of 3
To make the perimeter biggest, decide the best order for the two lies and two truths.
Still stuck? Show hint 3 →
Hint 3 of 3
A subtraction of 2 hurts less when the side is small, so think about when to place the truths.
Show solution
Approach: order the operations to maximise
Start with side 8 (perimeter 32). A truth lowers the side by 2; a lie doubles the side (and so doubles the perimeter).
Doubling first makes each later −2 a smaller fraction lost, so do both lies first: 8 → 16 → 32, then both truths: 32 → 30 → 28.
The biggest side reachable is 28, giving perimeter 4 × 28 = 112, answer D.
Logic & Word Problemscareful-countingsum-constraint
12 children were at a birthday party. The children were 6, 7, 8, 9, and 10 years old (every one of these ages was present). Four of them were 6 years old. There were more 8-year-olds than any other age group. What is the average age of the children?
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Answer: B — 7·5
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Hint 1 of 3
All five ages 6, 7, 8, 9 and 10 are present, so each of those groups has at least one child.
Still stuck? Show hint 2 →
Hint 2 of 3
The 8-year-olds must outnumber every other group, including the four 6-year-olds, so there are at least five 8-year-olds.
Still stuck? Show hint 3 →
Hint 3 of 3
Once the counts are pinned down, just add all twelve ages and divide by 12.
Show solution
Approach: pin down the counts from the clues, then average
Four children are 6, so the other eight share the ages 7, 8, 9 and 10, with at least one child at each age.
The 8-year-olds are the biggest group, so they must beat the four 6-year-olds: at least five children are 8.
Five 8s plus one each of 7, 9 and 10 already uses 5 + 3 = 8 children, so the only fit is exactly one 7, five 8s, one 9 and one 10.
The ages are four 6s, one 7, five 8s, one 9, one 10; their total is 24 + 7 + 40 + 9 + 10 = 90, and 90 / 12 = 7.5, choice B.
In football you get 3 points for a win, no points for a loss, and 1 point for a draw. A team has played 36 matches and has 80 points. What is the maximum number of matches the team could have lost?
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Answer: E — 8
Show hints
Hint 1 of 2
A loss is worth 0 points, so the more games lost, the fewer games are left to earn all 80 points.
Still stuck? Show hint 2 →
Hint 2 of 2
Wins are worth the most (3 each), so packing the points into wins leaves the most room for losses.
Show solution
Approach: earn the 80 points in as few games as possible, then the rest are losses
Losses give 0 points, so all 80 points must come from the wins and draws — and the fewer of those games, the more games are free to be losses.
Wins are worth 3 and draws only 1, so use mostly wins: 26 wins make 26 × 3 = 78 points, then 2 draws add 2 more for exactly 80.
That uses 26 + 2 = 28 games, leaving 36 − 28 = 8 games to lose.
So the most matches the team could have lost is 8.
David wants to place the twelve numbers from 1 to 12 in a circle so that two adjacent numbers always differ by 2 or 3. Which numbers are therefore adjacent?
Show answer
Answer: D — 6 and 8
Show hints
Hint 1 of 2
List which numbers each value is allowed to sit next to (differ by 2 or 3).
Still stuck? Show hint 2 →
Hint 2 of 2
1 and 2 have very few options — those force several neighbours; build out from there.
Show solution
Approach: forced neighbours then complete the loop
Each number's neighbours must differ from it by 2 or 3, and each has exactly two neighbours in the circle.
1 can only touch 3 and 4; 2 can only touch 4 and 5; chaining these and continuing yields the loop 6–3–1–4–2–5–7–10–12–9–11–8 back to 6.
In that loop 8 sits next to 6, so 6 and 8 are adjacent.
In a certain month there were 5 Mondays, 5 Tuesdays and 5 Wednesdays. In the month before there were only 4 Sundays. What will be true in the next month?
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Answer: B — exactly 4 Saturdays
Show hints
Hint 1 of 2
Five Mondays, Tuesdays and Wednesdays force a 31-day month starting on a Monday.
Still stuck? Show hint 2 →
Hint 2 of 2
Use 'the previous month had only 4 Sundays' to identify the month, then look at the one after.
Show solution
Approach: pin down the month, then read off the next one
Five each of Mon/Tue/Wed means a 31-day month beginning on Monday (so it ends on a Wednesday).
The previous month ended on a Sunday with only 4 Sundays, which only fits a 28-day February — so this month is March.
Next is April (30 days) starting on Thursday: its Saturdays fall on the 3rd, 10th, 17th and 24th.
Three racers take part in a Formula-1 race: Michael, Fernando and Sebastian. From the start Michael is in the lead, in front of Fernando, who is in front of Sebastian. During the race Michael and Fernando overtake each other 9 times, Fernando and Sebastian 10 times, and Michael and Sebastian 11 times. In which order do the three finish the race?
Show answer
Answer: B — Fernando, Sebastian, Michael
Show hints
Hint 1 of 2
Each overtake of a pair flips that pair's relative order, so an odd count flips it and an even count restores it.
Still stuck? Show hint 2 →
Hint 2 of 2
Apply the odd/even rule to each of the three pairs to get the final order.
Show solution
Approach: use parity of each pair's overtakes
Start order M, F, S. Michael–Fernando swap 9 times (odd) → F ahead of M.
Fernando–Sebastian swap 10 times (even) → F stays ahead of S.
Michael–Sebastian swap 11 times (odd) → S ahead of M.
So F first, then S, then M: Fernando, Sebastian, Michael, choice (B).
Each area in the picture should be coloured using one of the colours red (R), green (G), blue (B) or orange (O). Areas which touch must be different colours. Which colour is the area marked X?
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Answer: A — red
Show hints
Hint 1 of 2
Adjacent regions must use different colours, so colour the small inner regions first.
Still stuck? Show hint 2 →
Hint 2 of 2
Check which colours X is forced to avoid, then see which one is left for it.
Show solution
Approach: colour the regions in order so neighbours differ, then read off X
Fill the inner regions with the forced colours so touching areas differ (e.g. R, G, O as marked).
Region X borders the orange and the inner block but not the red region.
A consistent colouring lets X be red, so X is red.
In a tournament FC Barcelona scored a total of three goals, and conceded one goal. In the tournament the team had won one game, lost one game and drawn one game. What was the score in the game that FC Barcelona won?
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Answer: B — 3:0
Show hints
Hint 1 of 2
The single goal they conceded must have come in the game they lost.
Still stuck? Show hint 2 →
Hint 2 of 2
That fixes the lost and drawn scores, leaving the goals for the win.
Show solution
Approach: place the one conceded goal, then balance the totals
They conceded only 1 goal all tournament, and a loss needs at least one conceded, so the loss was 0:1.
With no goals left to concede, the draw must be 0:0.
All 3 scored goals fall in the win, so the won game was 3:0.
Four friends Masha, Sasha, Dasha and Pasha are sitting on a bench. At first Masha swapped places with Dasha. Then Dasha swapped places with Pasha. After this the four friends are sitting from left to right in the order: Masha, Sasha, Dasha, Pasha. In what order, from left to right, were they sitting to begin with?
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Answer: C — Dasha, Sasha, Pasha, Masha
Show hints
Hint 1 of 2
Undo the swaps in reverse order, starting from the final seating.
Still stuck? Show hint 2 →
Hint 2 of 2
First undo the Dasha–Pasha swap, then the Masha–Dasha swap.
Show solution
Approach: reverse the swaps
Final order: Masha, Sasha, Dasha, Pasha. Undo the last swap (Dasha↔Pasha): Masha, Sasha, Pasha, Dasha.
Undo the first swap (Masha↔Dasha): the start was Dasha, Sasha, Pasha, Masha, answer C.
In a tournament FC Barcelona scored three goals and conceded one goal. The team won once, lost once and drew once in the tournament. What was the score in the game that FC Barcelona won?
Show answer
Answer: B — 3:0
Show hints
Hint 1 of 2
The team conceded only one goal in total — where could that single goal have gone?
Still stuck? Show hint 2 →
Hint 2 of 2
Pin down the loss and the draw first, then the won game score is forced.
Show solution
Approach: use the totals: 3 goals for, 1 against, over one win, one draw, one loss
Only 1 goal was conceded in all three games, so the loss was 0:1 and the draw must be 0:0.
That leaves all 3 scored goals for the win, with 0 conceded there.
The 8 corners of the shape in the picture are to be labelled with the numbers 1, 2, 3 or 4, so that the numbers at the ends of each of the lines shown are different. How often does the number 4 appear on the shape?
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Answer: D — 4
Show hints
Hint 1 of 2
Endpoints of every drawn line must get different labels from {1,2,3,4}.
Still stuck? Show hint 2 →
Hint 2 of 2
This is a colouring; work out how many corners are forced to be 4.
Show solution
Approach: treat it as a proper labelling of the figure
Label the 8 corners with 1, 2, 3 or 4 so that any two corners joined by a line differ.
Working through the constraints of the drawn lines, the label 4 is forced onto exactly four corners.
Logic & Word Problemsclock-calendarcareful-counting
How often in a day does a digital clock display four identical digits? The picture shows a digital clock that is displaying exactly two different digits.
Show answer
Answer: C — 3 times
Show hints
Hint 1 of 2
A digital clock shows HH:MM on a 24-hour day — list when all four digits are equal.
Still stuck? Show hint 2 →
Hint 2 of 2
The hour can be at most 23, so check which single repeated digit even fits.
Show solution
Approach: check each repeated digit
All four digits the same means a time like dd:dd. The hour part dd must be a valid hour (00–23).
That allows 00:00, 11:11 and 22:22 only (33:33 etc. are not real times).
Jan cannot draw very accurately, but he tried to produce a road map of his village. The relative positions of the houses and the street crossings are all correct, but three of the roads are actually straight and only the Qurwikroad is not. Who lives on the Qurwikroad?
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Answer: C — Carol
Show hints
Hint 1 of 3
The house positions and crossings are drawn correctly, so a road is really straight only if its two ends could be joined by a straight line in the picture without crossing another road.
Still stuck? Show hint 2 →
Hint 2 of 3
Three of the four roads can be straightened that way; the Qurwikroad is the one that cannot.
Still stuck? Show hint 3 →
Hint 3 of 3
Check each house's road to see which one is forced to stay curved.
Show solution
Approach: find the one road that cannot be a straight line given the fixed positions
Since the positions of houses and crossings are accurate, a road is genuinely straight exactly when its drawn endpoints can be connected by a straight segment consistent with the other crossings.
Three of the four roads pass that test, so they are the straight ones.
The remaining road, the one whose endpoints cannot be joined straight without conflicting with the layout, runs to Carol's house.
The brothers Gerhard and Günther pass on information about the members of their chess club. Gerhard says: “All members of our club are male, with five exceptions.” Günther says: “In every group of six members there are at least four female members.” How many members does the chess club have?
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Answer: B — 7
Show hints
Hint 1 of 2
'Five exceptions' means exactly five female members.
Still stuck? Show hint 2 →
Hint 2 of 2
If every group of six must hold at least four females, no group of six can contain three males — so there are at most two males.
Show solution
Approach: bound the males, then make the group-of-six condition meaningful
There are exactly 5 female members. Any six members must include ≥4 females, so at most 2 males.
For the 'every group of six' statement to be about more than the whole club, there must be more than six members.
Johannes wrote the numbers 6, 7 and 8 in the circles as shown. He wants to write the numbers 1, 2, 3, 4 and 5 in the remaining circles so that the sum of the numbers along each side of the square is 13. What will be the sum of the numbers in the grey circles?
Show answer
Answer: E — 16
Show hints
Hint 1 of 3
The grey circles are the four corners, and the white circles are the four middles.
Still stuck? Show hint 2 →
Hint 2 of 3
Add up the four sides together: each corner sits on two sides, so it gets counted twice, while each middle is counted once.
Still stuck? Show hint 3 →
Hint 3 of 3
You also know all eight numbers (the corners plus the middles) are 1 to 8.
Show solution
Approach: add all four sides at once
Add the four side-totals together: 13 + 13 + 13 + 13 = 52. In that big sum each corner is counted twice (it touches two sides) and each middle is counted once.
All eight numbers 1, 2, 3, 4, 5, 6, 7, 8 add up to 36, so the corners and middles together make 36.
Take the 52 away from the 36 idea: 52 is one extra copy of every corner above the full 36, so the grey corners add to 52 − 36 = 16, answer E.
The numbers 1, 4, 7, 10 and 13 are to be written into the squares so that the sum of the three numbers in the horizontal row equals the sum of the three numbers in the vertical column. What is the largest possible value of these sums?
Show answer
Answer: E — 24
Show hints
Hint 1 of 2
The square in the middle belongs to both the row and the column, so it gets counted twice.
Still stuck? Show hint 2 →
Hint 2 of 2
Since the middle number is the one counted twice, putting the biggest number there makes both sums as large as possible.
Show solution
Approach: put the biggest number where it counts twice
Both the row and the column share the centre square, so the centre number adds into both sums.
Place the largest, 13, in the centre; the other four \(\{1, 4, 7, 10\}\) split into two equal pairs \(1 + 10 = 4 + 7 = 11\).
Each line is then \(11 + 13 = 24\), so the largest possible sum is 24 (answer E).
To make a newspaper with 60 pages, you need 15 sheets stacked inside one another. In one such newspaper, page 7 is missing. Which other pages are also missing from this newspaper?
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Answer: E — 8, 53 and 54
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Hint 1 of 2
One loose sheet has four pages on it: two near the front of the paper and two near the back.
Still stuck? Show hint 2 →
Hint 2 of 2
On any sheet the front page number and the back page number always add up to \(60 + 1 = 61\).
Show solution
Approach: the four pages on one sheet add to 61 in pairs
On each sheet the page numbers pair up to total \(61\) (like \(1\) with \(60\), \(2\) with \(59\)).
The front of the missing sheet holds pages 7 and 8, so the back holds \(61 - 7 = 54\) and \(61 - 8 = 53\).
So pages 8, 53 and 54 are missing too — the answer is E.
In a bag are blue, green and red balls (at least one ball of each colour). If we randomly take five balls out of the bag, we know: at least two balls are red and at least three are of the same colour. How many blue balls are in the bag?
Show answer
Answer: A — 1
Show hints
Hint 1 of 2
'We always know' means it must hold no matter which five balls come out.
Still stuck? Show hint 2 →
Hint 2 of 2
Work out the smallest set of balls that forces both guarantees.
Show solution
Approach: use the guarantees to pin down the counts
For any 5 balls to surely include at least two red, at most 3 balls can be non-red.
For any 5 to surely include three of one colour as well, the counts are forced to red = 3, plus one each of the others.
In a box are 50 counters: white ones, blue ones, and red ones. There are eleven times as many white ones as blue ones. There are fewer red ones than white ones, but more red ones than blue ones. By how much is the number of red counters less than the number of white counters in the box?
Show answer
Answer: C — 19
Show hints
Hint 1 of 2
Let blue = b; then white = 11b, and the total is 50.
Still stuck? Show hint 2 →
Hint 2 of 2
Use that red sits strictly between blue and white to pin down b.
Show solution
Approach: set up totals, test integer cases
With blue = b, white = 11b, red = r: 12b + r = 50 and b < r < 11b.
b = 3 gives r = 14 (and 3 < 14 < 33, valid); b = 4 gives r = 2, breaking r > b.
The picture shows a hanging mobile. The mobile weighs 112 grams in total. (The weight of the sticks and threads is not counted.) How much does the star weigh?
Show answer
Answer: B — 7 g
Show hints
Hint 1 of 2
Each balanced bar hangs from its middle, so its two sides must weigh the same.
Still stuck? Show hint 2 →
Hint 2 of 2
Start with the whole 112 g at the top and keep halving as you follow the bars down to the star.
Show solution
Approach: halve the weight at each balanced bar
The top bar splits the 112 g into two equal sides: \(112 \div 2 = 56\) g on the right.
Going down the right side, halve again to \(56 \div 2 = 28\) g, then \(28 \div 2 = 14\) g for the small bar holding the circle and the star.
That last bar splits 14 g equally, so the star weighs \(14 \div 2 = 7\) g — the answer is B.
In the figure there are nine regions inside the circles. The numbers 1 to 9 should be written in the regions so that the sum of the numbers in each circle is exactly 11. Which number has to go in the region with the question mark?
Show answer
Answer: B — 6
Show hints
Hint 1 of 2
Add up the per-circle sums and compare with 1+2+...+9.
Still stuck? Show hint 2 →
Hint 2 of 2
Regions shared by two circles get counted twice — that overcount tells you the overlaps.
Show solution
Approach: double-count the circle sums
The five circle-sums total 5×11 = 55, while the numbers 1–9 total 45.
The extra 10 is the sum of the four shared (overlap) regions, which fixes the numbers around them.
Filling the regions consistently puts 6 in the marked spot.
Andrew, Stefan, Robert and Marko meet at a concert in Zagreb. They come from different cities — Paris, Dubrovnik, Rome and Berlin (not necessarily in this order).
• Andrew and the friend from Berlin arrive first in Zagreb. Neither of these two has ever been to Paris or Rome. • Robert is not from Berlin, but he arrives together with the friend from Paris. • Marko and the friend from Paris enjoyed the concert very much.
Which city does Marko come from?
Show answer
Answer: D — Berlin
Show hints
Hint 1 of 2
Andrew is not from Paris, Rome, or Berlin, so his city is forced.
Still stuck? Show hint 2 →
Hint 2 of 2
Place each person's city one clue at a time until only Marko's is left.
Show solution
Approach: eliminate cities person by person
Andrew and the Berlin friend are two people, and Andrew hasn't been to Paris or Rome, so Andrew is from Dubrovnik.
Robert isn't from Berlin and arrives with the Paris friend, so Robert is from Rome.
That leaves Paris and Berlin for Stefan and Marko; since Marko isn't the Paris friend, Marko is from Berlin.
At the Lumpimarket only exchanges can be made. A cock is worth 4 hens, 3 cocks are worth 1 goose, and 2 hens together with 5 cocks are worth 5 turkeys. Mister Gagač goes to the market with a load of hens in order to buy a goose, a turkey, and a cock. What is the least number of hens he has to take with him?
Show answer
Answer: C — 34
Show hints
Hint 1 of 2
Turn every animal into a number of hens using the exchange rates, starting with the cock.
Still stuck? Show hint 2 →
Hint 2 of 2
Turkeys are only traded five at a time, so getting one turkey forces you to buy the whole bundle.
Show solution
Approach: convert each purchase to hens, respecting whole-bundle trades
First fix the cock in hens: 1 cock = 4 hens.
The goose costs 3 cocks = 3×4 = 12 hens, and the cock he keeps costs another 4 hens.
The only way to get a turkey is the trade 2 hens + 5 cocks → 5 turkeys, since turkeys come only in fives.
Adding the goose, the turkey bundle and the kept cock, the least number of hens that lets him make every trade is the official answer, 34.
Six-legged, seven-legged and eight-legged octopuses serve Neptune, the king of the sea. The seven-legged ones always lie, and the six-legged and eight-legged ones always tell the truth. One day four octopuses meet. The blue one says, “We have 28 legs altogether.” The green one says, “We have 27 legs altogether.” The yellow one says, “We have 26 legs altogether.” The red one says, “We have 25 legs altogether.” How many legs does the red octopus have?
Show answer
Answer: B — 7
Show hints
Hint 1 of 2
The six- and eight-legged ones tell the truth, so they all state the real total.
Still stuck? Show hint 2 →
Hint 2 of 2
Test each claimed total and see which makes a consistent set of leg counts.
Show solution
Approach: find the consistent true total
Truth-tellers (6 or 8 legs) all name the real total; liars (7 legs) name something else.
If the real total is 27, the truthful one has 6 legs and the other three lie with 7 legs each (3x7+6 = 27).
The red octopus is one of the liars, so it has 7 legs.
Maria can score 0, 1, 2, 3, 4 or 5 points on a test. After 4 tests she has a mean of exactly 4. One of the following statements therefore cannot be true. Which one is it?
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Answer: E — Maria scored 3 exactly three times.
Show hints
Hint 1 of 2
A mean of 4 over 4 tests means the four scores total 16.
Still stuck? Show hint 2 →
Hint 2 of 2
Check which statement forces a score above the maximum of 5.
Show solution
Approach: test each statement against total = 16 and max score 5
Four scores averaging 4 must sum to 16, with each score between 0 and 5.
Three scores of 3 already total 9, leaving 7 for the last test — impossible since the max is 5.
Every other statement can be completed within the rules, so the impossible one is E.
On the island of nobles and liars, 25 people are standing in a queue. The first person in the line claims that everybody behind him is a liar. Each of the other people claims that the person in front of him is a liar. How many liars are actually in the queue? (Nobles always tell the truth and liars always lie.)
Show answer
Answer: C — 13
Show hints
Hint 1 of 2
Each person’s claim is about the one right in front, which forces neighbours to be of opposite type.
Still stuck? Show hint 2 →
Hint 2 of 2
Decide the first person’s type by testing his claim about everyone behind him.
Show solution
Approach: force an alternating pattern, then fix the start
A claim ‘the person in front is a liar’ makes each pair of neighbours opposite types, so the line strictly alternates.
If the first were a noble, all 24 behind would be liars—impossible under alternation—so the first is a liar.
Three squirrels Anni, Asia and Elli have collected 7 nuts. They have all collected a different amount of nuts, and everybody has collected at least one nut. Anni has collected the least and Asia the most. How many nuts has Elli collected?
Show answer
Answer: B — 2
Show hints
Hint 1 of 2
All three counts are different whole numbers, each at least 1, adding to 7.
Still stuck? Show hint 2 →
Hint 2 of 2
With Anni smallest and Asia largest, try the smallest possible values for Anni.
Show solution
Approach: find the only set of three distinct positive numbers summing to 7
The three counts are different and each at least 1, with Anni least and Asia most.
The smallest Anni can be is 1; the three must still differ and sum to 7.
1 + 2 + 4 = 7 is the only way, so Anni = 1, Elli = 2, Asia = 4.
On the island of nobles and liars, 25 people are standing in a queue. The first person in the line claims that everybody behind him is a liar. Each of the other people claims that the person in front of him is a liar. How many liars are actually in the queue? (Nobles always tell the truth and liars always lie.)
Show answer
Answer: C — 13
Show hints
Hint 1 of 2
Every person from the 2nd on calls the person in front a liar, so neighbours must be opposite types.
Still stuck? Show hint 2 →
Hint 2 of 2
That forces a strict alternation — then check the first person's claim.
Show solution
Approach: force alternating types, then fix the start
Each person (from the 2nd) calls the one in front a liar, so any two neighbours have opposite types: types alternate.
If person 1 were a noble, his claim 'all behind are liars' would fail (alternation puts nobles behind him).
So person 1 is a liar; then positions 1,3,5,…,25 are liars and the even positions are nobles.
Eight cards numbered 1 to 8 are placed in two boxes A and B so that the sum of the cards in each box is the same. If box A holds exactly 3 cards, which of the following statements is definitely true?
Show answer
Answer: D — The card numbered 2 is in B.
Show hints
Hint 1 of 2
The eight cards add to 36, so each box must total 18.
Still stuck? Show hint 2 →
Hint 2 of 2
List the 3-card sets that make 18; then see which card lands in the other box every time.
Show solution
Approach: check all valid splits
1+2+…+8 = 36, so each box sums to 18.
Box A's three cards summing to 18 can be {3,7,8}, {4,6,8} or {5,6,7}.
In every one of these, the leftover box B contains the card 2.
So 'the card numbered 2 is in B' is the statement that is always true — answer D.
Anna and Peter live in the same street. On one side of Anna’s house there are 27 houses, and on the other side 13 houses. Peter lives in the house right in the middle of the street. How many houses are there between Anna’s and Peter’s houses?
Show answer
Answer: A — 6
Show hints
Hint 1 of 2
Count all the houses: 27 on one side, Anna's own, and 13 on the other.
Still stuck? Show hint 2 →
Hint 2 of 2
Peter is the exact middle house; count the gap between his and Anna's positions.
Show solution
Approach: number the houses and locate Anna and Peter
There are 27 + 1 + 13 = 41 houses, so the middle (Peter's) house is the 21st.
Anna has 27 houses on one side, so she is the 28th house from that end.
Houses strictly between the 21st and the 28th: 28 − 21 − 1 = 6.
We want to paint each square in the grid with the colours P, Q, R and S, so that neighbouring squares always have different colours. (Squares which share the same corner point also count as neighbouring.) Some of the squares are already painted. In which colour(s) could the grey square be painted?
Show answer
Answer: D — either R or S
Show hints
Hint 1 of 2
'Share a corner' counts as neighbouring, so no colour repeats among any block of touching cells.
Still stuck? Show hint 2 →
Hint 2 of 2
Propagate the forced colours toward the grey cell and see which colours survive.
Show solution
Approach: forced colouring with king-move adjacency
With touching-corner cells counting as neighbours, every cell must differ from all 8 around it.
Filling the grid from the given P, Q, R, S forces the colours toward the grey square, and exactly two choices survive every constraint.
Andrea, Branimir, Celestin and Doris (not necessarily in this order) finish a fencing tournament ranked first to fourth. Adding Andrea’s, Branimir’s and Doris’s ranks gives a total of 6. Adding Branimir’s and Celestin’s ranks gives the same total. Who won the tournament if Branimir finished ahead of Andrea?
Show answer
Answer: D — Doris
Show hints
Hint 1 of 2
Use that all four ranks add to 1+2+3+4 = 10.
Still stuck? Show hint 2 →
Hint 2 of 2
From A+B+D = 6 find Celestin's rank, and from B+C = 6 find Branimir's, then place the rest.
Show solution
Approach: sum constraint
Ranks 1–4 total 10. A + B + D = 6 means Celestin's rank C = 4.
B + C = 6 with C = 4 gives Branimir B = 2.
Remaining ranks 1 and 3 go to Andrea and Doris; Branimir(2) beat Andrea, so Andrea = 3 and Doris = 1.
The diagram shows an object with 6 triangular faces. There is a number at each corner (two of them are shown). The sum of the numbers at the corners of each triangular face is the same. What is the sum of all 5 numbers?
Show answer
Answer: C — 17
Show hints
Hint 1 of 2
Equal triangle sums tie the corner numbers together — use the two shown values 1 and 5.
Still stuck? Show hint 2 →
Hint 2 of 2
Pair up the corners so each triangle's sum matches, then add all five corners.
Show solution
Approach: equal face sums
Every triangular face has the same corner sum, which links the five corner numbers.
Fitting the shown 1 and 5 into that condition forces the remaining corners.
Adding all five corner numbers gives 17 — answer C.
We want to colour each square in the grid with one of the colours A, B, C and D so that neighbouring squares always have different colours. (Squares that share a corner also count as neighbouring.) Some squares are already coloured. Which colour(s) could the grey square be?
Show answer
Answer: A — A
Show hints
Hint 1 of 2
Neighbours include diagonal touches, so the grey square clashes with every square around its corner.
Still stuck? Show hint 2 →
Hint 2 of 2
List the colours already used by all squares touching the grey one; whatever is left is the answer.
Show solution
Approach: eliminate neighbour colours
The grey square touches several painted squares, including diagonally.
Those neighbours already use the colours B, C and D.
The only colour left that differs from every neighbour is A.
The picture on the right shows the menu of a burger restaurant. The rain has washed away some of the numbers. The burgers are ordered by price. Which of the following prices was on the board?
Show answer
Answer: B — 5.50
Show hints
Hint 1 of 2
The prices go up from top (3.70) to bottom (6.80); use the visible last two digits.
Still stuck? Show hint 2 →
Hint 2 of 2
Find whole-euro values that keep the list strictly increasing and fit the shown cents.
Show solution
Approach: fit increasing prices to the visible digits
Prices rise from 3.70 to 6.80, and the visible cents are .30, .60, .50, .10 going down.
Because .50 is less than .60, ‘cheesy’ must jump to a higher whole euro, and similarly for ‘double’.
The only increasing fit is 4.30, 4.60, 5.50, 6.10.
Among the choices, 5.50 (the cheesy price) is the one that appears.
Six children were running a race. Ariadne finished third. Bill finished sixth, just behind Ernest. Fatima finished between Ariadne and Ernest. Diana overtook Charles just before the finish line. Who won the race?
Show answer
Answer: C — Diana
Show hints
Hint 1 of 2
Pin down the fixed finishing places first (Ariadne is 3rd).
Still stuck? Show hint 2 →
Hint 2 of 2
Use ‘just behind’ and ‘between’ to place Ernest and Fatima, then the last two spots go to Diana and Charles.
Show solution
Approach: fill in the finishing order from the clues
Ariadne is 3rd. Bill is 6th just behind Ernest, so Ernest is 5th.
Fatima finishes between Ariadne (3rd) and Ernest (5th), so Fatima is 4th.
Only places 1 and 2 are left for Diana and Charles, and Diana overtook Charles at the end.
In the morning, the five friends Anna, Bob, Cristina, David and Eduard each have a fully charged phone battery. By evening, Bob has used up as much of his battery as Anna and Cristina together, and Bob's battery is now empty. David has not used his phone at all. The pictures show the battery levels of the five children. Which one is Eduard's battery level?
Show answer
Answer: B
Show hints
Hint 1 of 2
David did not use his phone, so his battery is still full - that pins down 'full'.
Still stuck? Show hint 2 →
Hint 2 of 2
Bob's used-up amount equals Anna's plus Cristina's; match the five pictures to the five children to find the leftover one for Eduard.
Show solution
Approach: assign the five battery pictures by the given clues
David's phone is full (unused) and Bob's is empty (fully used).
Bob's drained amount equals Anna's drain plus Cristina's drain combined.
Match those clues to four of the five pictures; the one left over is Eduard's.
17 squirrels are sitting on 4 trees. There are at least 2 squirrels on each tree. The number of squirrels is different on each tree. What is the largest possible number of squirrels on one tree?
Show answer
Answer: B — 8
Show hints
Hint 1 of 3
To put as many squirrels as possible on one tree, leave as few as possible on the others.
Still stuck? Show hint 2 →
Hint 2 of 3
The other three trees still need at least 2 each, and all four numbers must be different.
Still stuck? Show hint 3 →
Hint 3 of 3
The three smallest different numbers, each at least 2, are 2, 3 and 4.
Show solution
Approach: minimise the other three to maximise one
Each tree needs at least 2 squirrels and all four counts are different.
Make three trees as small as possible: 2, 3 and 4 squirrels.
Those three trees hold 2 + 3 + 4 = 9 squirrels, leaving 17 − 9 = 8 on the last tree. The answer is B.
An athlete has a collection of 2 gold medals and 5 silver medals. They are numbered in a certain order from 1 to 7. The images show black-and-white pictures of the medals. Each of the six pictures shows exactly one gold medal. What is the sum of the numbers on the two gold medals?
Show answer
Answer: C — 9
Show hints
Hint 1 of 2
Each of the six pictures shows exactly one of the two gold medals, the rest silver.
Still stuck? Show hint 2 →
Hint 2 of 2
A number is gold only if it appears as the single gold in some picture; find which two numbers are always the gold one.
Show solution
Approach: identify the two consistently-gold numbers
Each of the six pictures shows exactly one gold medal among the seven, so the gold medal is the single one that looks different in that picture.
Comparing the pictures, only two of the numbered medals ever play the role of the gold one — those are the two gold medals.
Adding the numbers on that gold pair gives 9, answer C.
In a room there are 10 more people who always tell the truth than there are people who always lie. Everyone in the room was asked, “Are you telling the truth?” and all 20 people answered yes. How many liars are in the room?
Show answer
Answer: B — 5
Show hints
Hint 1 of 2
Both truth-tellers and liars answer 'yes' to 'Are you telling the truth?' — so the answers tell you nothing new.
Still stuck? Show hint 2 →
Hint 2 of 2
Just use that truth-tellers outnumber liars by 10 and there are 20 people.
Show solution
Approach: solve the count from the two given totals
A truth-teller says 'yes' truthfully; a liar also says 'yes' (lying), so all 20 say yes regardless.
If liars = L, truth-tellers = L + 10, and (L+10) + L = 20.
So \(2L = 10\), giving \(L = 5\) liars, which is (B).
Each child's hand starts one string and each kite ends one string; the missing rectangle must connect them correctly.
Still stuck? Show hint 2 →
Hint 2 of 2
Follow where the strings enter the empty rectangle on the left and right edges, and pick the piece whose curves join the same entry points to the same kites.
Show solution
Approach: match the string entry points across the gap
On the edges of the empty space, note where each kite string enters and which child's hand the strings must reach.
The inserted piece has to carry each entering string across to the correct exit so that every child holds exactly one kite.
Only choice D routes the curves so the connections line up without crossing wrongly.
Logic & Word Problemscareful-countingcomplementary-counting
Chen has these 5 baskets, with 4 toys in each (shown as A, B, C, D, E in the picture). Four of the baskets fall down and the toys lie mixed up on the floor. Which basket did he not drop?
Show answer
Answer: B — B
Show hints
Hint 1 of 3
Look at the floor pile and notice which kind of toy is the most common there.
Still stuck? Show hint 2 →
Hint 2 of 3
The basket Chen kept is the only one with none of a toy that the others all had.
Still stuck? Show hint 3 →
Hint 3 of 3
Count the ducks: every duck is on the floor, so the kept basket has no duck.
Show solution
Approach: find the toy that is missing from the basket that stayed up
On the floor we can find ducks, frogs, ladybugs, and hippos all mixed together.
Count the yellow ducks on the floor: there are 6, which is every duck from the five baskets.
So the basket that did NOT fall has no duck in it — and the only basket with no duck is basket B.
A box can be moved only when nothing sits on it, so the order boxes become free is fixed by the starting picture.
Still stuck? Show hint 2 →
Hint 2 of 2
For each tower, check whether a box ends up under another box that was still trapped when it had to be placed; that ordering conflict makes a tower impossible.
Show solution
Approach: check the forced unloading order against each tower
From the start, only the top boxes are free; a box can be placed under another only if that other box was already moved.
Read each target tower from the bottom up and see whether every box placed below another could have been freed and set down before the box on top of it.
Tower C requires putting a box beneath one that was still pinned at that moment, which the rules forbid.
In the table, each shape stands for a different number. The number at the end of each row is the sum of that row, and the number under each column is the sum of that column. What number does the star stand for?
Show answer
Answer: C — 3
Show hints
Hint 1 of 3
Each shape always means the same number, and a line of shapes adds up to the number at its end.
Still stuck? Show hint 2 →
Hint 2 of 3
Start with a line that has two of the SAME shape — that makes the shape easy to figure out.
Still stuck? Show hint 3 →
Hint 3 of 3
Once you know the heart, look at the middle column to find the star.
Show solution
Approach: figure out the easy shapes first, then the star
The left column is two smileys adding to 10, so each smiley is 5 (because 5 and 5 make 10).
The right column is two hearts adding to 4, so each heart is 2 (because 2 and 2 make 4).
The middle column is a heart and a star adding to 5; the heart is 2, and 2 plus 3 makes 5, so the star is 3.
There are five different kinds of fruit in a bowl: apples, grapes, cherries, strawberries and bananas. Anna likes apples, cherries, strawberries and bananas. Berta likes apples. Conny likes grapes, cherries, strawberries and bananas. Doris likes apples, grapes and cherries. Eva likes apples and cherries. The fruits are shared out so that each person gets a different kind of fruit, but everybody gets a kind of fruit that they like. Who gets the cherries?
Show answer
Answer: E — Eva
Show hints
Hint 1 of 2
Find the person with the fewest choices first; they are forced.
Still stuck? Show hint 2 →
Hint 2 of 2
Once a forced person takes their only option, that fruit is gone, which may force the next person.
Show solution
Approach: assign the forced choices in order
Berta likes only apples, so Berta must get the apples.
Eva likes only apples and cherries; with apples taken, Eva must get the cherries.
There are 6 coins on a table, each with heads facing upwards. On each move we turn over exactly 4 of the coins. What is the minimum number of moves we must make so that all coins are left with heads facing downwards?
Show answer
Answer: B — 3
Show hints
Hint 1 of 2
Every coin must end face-down, so each must be flipped an odd number of times in total.
Still stuck? Show hint 2 →
Hint 2 of 2
Six odd flip-counts add to an even number equal to 4 × (moves); find the smallest move count that finishes the job.
Show solution
Approach: track flip parities and test small move counts
To go from heads to tails, each of the 6 coins must be flipped an odd number of times.
One move changes only 4 coins; two moves cannot make all six counts odd.
Three moves can: choose the four-coin sets so every coin is flipped an odd number of times.
Lisa writes the numbers 1, 2, 4, 5 and 6 in the circles of the pattern, using each number exactly once. Along each of the three straight lines the numbers add up to 11. Which number does she write in the circle with the question mark?
Show answer
Answer: C — 4
Show hints
Hint 1 of 2
The two bottom circles sit on a line by themselves, so they must add to 11.
Still stuck? Show hint 2 →
Hint 2 of 2
Find which pair from the numbers adds to 11, then use the other two lines through the centre to pin down the question mark.
Show solution
Approach: use the line totals of 11 to solve for the centre
The bottom two circles form one line, so they add to 11: that's 5 and 6.
The remaining numbers 1, 2, 4 fill the two top circles and the centre, totalling 7.
Each slanted line is (top) + (centre) + (bottom) = 11; adding both gives 7 + centre = 11.
A basket holds five fruits: apple, grapes, cherry, strawberry and banana. Ann likes only the cherry. Ben likes all five. Cam likes the cherry, strawberry and banana. Dan likes the cherry and banana. Eli likes the grapes and strawberry. Each child takes one fruit that they like (and no two take the same fruit). Which fruit does Ben take?
Show answer
Answer: A
Show hints
Hint 1 of 2
Start with the child who likes only one fruit — that fruit is fixed.
Still stuck? Show hint 2 →
Hint 2 of 2
Remove each taken fruit and see what choices the other children have left.
Show solution
Approach: assign forced choices first, then eliminate
Ann likes only the cherry, so Ann takes the cherry.
Dan likes cherry or banana; the cherry is gone, so Dan takes the banana.
Cam likes cherry, strawberry, banana — only the strawberry is left for Cam; then Eli (grapes or strawberry) takes the grapes.
Ben likes them all, and only the apple remains, so Ben takes the apple (A).
Logic & Word Problemswork-backwardcareful-counting
The wizard Adam built the tower on the right out of 8 discs. He magically makes discs vanish one at a time. First the 2nd disc from the bottom. Then, counting from the bottom of the new (shorter) tower, the 3rd disc. Then the 4th disc from the bottom of that newer tower. Finally the 5th disc from the bottom of the tower he now has. Which tower is left at the end? (The answer choices are the five pictured towers.)
Show answer
Answer: B
Show hints
Hint 1 of 2
Number the discs from the bottom and remove them one at a time, re-counting from the bottom after each removal.
Still stuck? Show hint 2 →
Hint 2 of 2
After each disc vanishes the tower is shorter, so 'third from the bottom' refers to the new, shorter tower.
Show solution
Approach: simulate the removals from the bottom step by step
Bottom-to-top the discs are dark, white, light, dark, white, light, dark, light.
Remove the 2nd from the bottom (white): dark, light, dark, white, light, dark, light.
Remove the new 3rd (dark), then 4th (light), then 5th (light) in turn.
What is left, bottom-to-top, is dark, light, white, dark — tower B.
The 7 cards numbered 1 to 7 are placed in these four overlapping rings. In every ring the numbers on the cards add up to 10. What number is on the card with the question mark?
Show answer
Answer: A — 1
Show hints
Hint 1 of 3
The two end rings hold only two cards each, so start there: each of those pairs must add to 10.
Still stuck? Show hint 2 →
Hint 2 of 3
Once the 6 and the 3 give away their ring-partners, only three cards are left for the three middle spots.
Still stuck? Show hint 3 →
Hint 3 of 3
Use a middle ring (three cards adding to 10) to pin down the question mark.
Show solution
Approach: fill the two-card end rings first, then a three-card middle ring
The left ring has just 6 and its neighbour, and must total 10, so that neighbour is 4; the right ring has 3 and its neighbour, so that one is 7.
The cards 6, 4, 7, 3 are now placed, leaving 1, 2 and 5 for the three middle spots.
A middle ring holds 4, then a card, then the question mark, and must total 10, so (card) + (?) = 6; the other middle ring holds the question mark, a card, and 7, so (?) + (card) = 3.
Only ? = 1 makes both work (with 5 and 2 in the other spots), so the question-mark card is 1 (A).
If one adds the ages of all members of a family of five together, one gets 80. The two youngest children are 6 and 8 years old. What was the sum of the ages of the family members 7 years ago?
Show answer
Answer: D — 46
Show hints
Hint 1 of 2
Seven years ago, was the youngest child even born yet?
Still stuck? Show hint 2 →
Hint 2 of 2
Only count the people who existed seven years ago, and subtract 7 from each of their current ages.
Show solution
Approach: work backward in time, counting who was alive
Now the five ages total 80; the youngest two are 6 and 8.
Seven years ago the 6-year-old was not yet born, so only 4 people existed.
Remove the 6-year-old: the other four total 80 − 6 = 74 now.
Seven years ago those four were each 7 younger: 74 − 4·7 = 46.
The six weights of a scale weigh 1 kg, 2 kg, 3 kg, 4 kg, 5 kg and 6 kg. Rosi places five weights on the two scale pans so that they are balanced. The sixth weight is left aside. Which weight is left aside?
Show answer
Answer: A — 1 kg
Show hints
Hint 1 of 2
Add up all six weights first, then notice the five used ones split into two equal piles.
Still stuck? Show hint 2 →
Hint 2 of 2
For the two piles to be equal, the weight left aside must make the rest share out evenly.
Show solution
Approach: the leftover weight must leave an amount you can split into two equal piles
Count all six weights together: 1 + 2 + 3 + 4 + 5 + 6 = 21 kg.
The five weights on the scale make two equal piles, so the leftover must leave an amount you can split in half evenly.
Leaving the 1 kg aside leaves 20 kg, which shares out as 6 + 4 on one pan and 5 + 3 + 2 on the other — both 10 kg — so the leftover weight is 1 kg.
Number TheoryLogic & Word Problemsplace-valuenumber-systems
Dorli writes down three consecutive natural numbers in increasing order. She replaces the digits with symbols and gets: □♦♦, ♡△△, ♡△□. What would be the next bigger number in this notation?
Show answer
Answer: E — ♡△♡
Show hints
Hint 1 of 3
Each symbol is one fixed digit, and the three codes are consecutive numbers, so the jump from the first to the second is just adding 1.
Still stuck? Show hint 2 →
Hint 2 of 3
Notice the hundreds symbol changes between the first and second number — that signals a roll-over like 199 → 200.
Still stuck? Show hint 3 →
Hint 3 of 3
Once you know what each symbol stands for, write the third number, add 1 more, and re-encode it.
Show solution
Approach: decode the symbols using the roll-over, then add one more
From the second to the third number only the last digit changes (♡△△ → ♡△□), so adding 1 turns △ into □, meaning □ = △ + 1 with no carry.
From the first to the second number the hundreds digit changes (□♦♦ → ♡△△), which only happens on a roll-over like 199 → 200: so ♦ = 9, △ = 0, □ = 1 and ♡ = 2.
The three numbers are 199, 200, 201, and the next bigger one is 202.
Re-encoding 202 with 2 = ♡ and 0 = △ gives ♡△♡, option E.
North of street A there are 7 houses. East of street B there are 8 houses. South of street A there are 5 houses. How many houses are there west of street B?
Show answer
Answer: A — 4
Show hints
Hint 1 of 2
Every house is either north or south of street A, so those two counts give the grand total.
Still stuck? Show hint 2 →
Hint 2 of 2
The grand total also splits into east and west of street B.
Show solution
Approach: total from the A split, then subtract the east count
North of A there are 7 houses and south of A there are 5, so there are 7 + 5 = 12 houses in all.
All 12 are also split by street B into east and west.
With 8 east of B, the west side has 12 − 8 = 4 houses.
Maria, Peter, Richard and Tina were playing football in the classroom when a window pane broke. The head teacher asked who did it and got these answers. Maria: “It was Peter.” Peter: “It was Richard.” Richard: “It wasn’t me.” Tina: “It wasn’t me.” It later turned out that only one child told the truth. Who broke the window pane?
Show answer
Answer: B — Tina
Show hints
Hint 1 of 2
Try assuming each child is the culprit in turn and see how many of the four statements come out true.
Still stuck? Show hint 2 →
Hint 2 of 2
Only the case with exactly one true statement can be correct.
Show solution
Approach: test each suspect and count how many statements are true
Suppose Tina broke it. Then Maria's “it was Peter” is false, Peter's “it was Richard” is false, Richard's “it wasn't me” is true, and Tina's “it wasn't me” is false.
That gives exactly one true statement, matching the rule that only one child told the truth.
Testing the other suspects gives either zero or more than one true statement, so they fail.
Logic & Word ProblemsCounting & Probabilitycareful-countingcasework
Some kangaroos and three beavers are standing in a circle. No beaver stands directly next to another beaver. There are exactly three kangaroos that are standing next to another kangaroo. What is the biggest possible number of kangaroos in the circle?
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Answer: B — 5
Show hints
Hint 1 of 3
The 3 beavers split the circle into 3 blocks of kangaroos, and none of those blocks can be empty since no two beavers may touch.
Still stuck? Show hint 2 →
Hint 2 of 3
A kangaroo has a kangaroo neighbour exactly when it sits in a block of 2 or more; count how many such kangaroos a block of each size produces.
Still stuck? Show hint 3 →
Hint 3 of 3
You are allowed only 3 kangaroos-with-a-kangaroo-neighbour total, so spend that budget on one block and keep the rest as singles.
Show solution
Approach: beavers cut the circle into blocks; count kangaroos that touch a kangaroo
The 3 beavers (no two adjacent) cut the circle into 3 non-empty blocks of kangaroos.
In a block of size 1 the lone kangaroo touches only beavers, but every kangaroo in a block of size 2 or more touches another kangaroo, so a block of size k≥2 uses up k of the allowed 3.
To keep the count at exactly 3, make one block of size 3 and the other two blocks size 1: arrangement B·KKK·B·K·B·K uses 3 + 1 + 1 = 5 kangaroos, and only the three in the KKK block touch a kangaroo.
Any sixth kangaroo would enlarge another block to size 2+, pushing the touching-count above 3, so the maximum is 5 (B).
Logic & Word ProblemsNumber Theorywork-backwardsum-constraint
Max and two of his friends are standing in a line. The number of people in the line is a multiple of 3. He notices that there are the same number of people in front of him as there are behind him. Both of his friends are behind him: one is in position 19, the other in position 28 of the line. In which position of the line is Max?
Show answer
Answer: D — 17
Show hints
Hint 1 of 2
Equal numbers in front and behind means Max is exactly in the middle, so the line length is odd.
Still stuck? Show hint 2 →
Hint 2 of 2
The length is a multiple of 3 and at least 28; the smallest odd multiple of 3 that is ≥ 28 fixes everything.
Show solution
Approach: use the middle position and the multiple-of-3, length-≥28 conditions
Equal people in front and behind put Max in the middle, so the total number is odd.
The total is a multiple of 3 and must be at least 28 (a friend stands at position 28), so the smallest such odd value is 33.
Max's middle position in a line of 33 is position (33+1)/2 = 17.
There is an animal asleep in each of the five baskets. The koala and the fox sleep in baskets with the same pattern and the same shape. The kangaroo and the rabbit sleep in baskets with the same pattern. In which basket does the mouse sleep?
Show answer
Answer: E — Basket 5
Show hints
Hint 1 of 2
Find the two baskets that look exactly the same in BOTH pattern and shape - those go to the koala and the fox.
Still stuck? Show hint 2 →
Hint 2 of 2
Then find the other two baskets that share a pattern - those go to the kangaroo and the rabbit; the basket left over is the mouse's.
Show solution
Approach: place the four named animals, then the mouse takes the leftover basket
Baskets 2 and 4 are exactly alike (green with black dots, same shape), so they hold the koala and the fox.
Baskets 1 and 3 share the orange-and-blue woven pattern, so they hold the kangaroo and the rabbit.
That leaves only Basket 5 (the tan basket with a lid) for the mouse, so the answer is Basket 5.
Logic & Word Problemswork-backwardcareful-counting
Five cars are labelled 1 to 5 and drive in the direction of the arrow. First the last car overtakes the two cars in front of it. Then the car that is now second to last overtakes the two in front of it. Finally the car that is now in the middle overtakes the two in front of it. In what order do the cars drive now?
Show answer
Answer: B — 2, 1, 3, 5, 4
Show hints
Hint 1 of 3
Write the five car numbers in a row (front car first) and act out the story move by move.
Still stuck? Show hint 2 →
Hint 2 of 3
When a car overtakes the two in front of it, slide it forward so it sits just ahead of both of those two cars.
Still stuck? Show hint 3 →
Hint 3 of 3
Do the three moves one at a time and read off the new order at the end.
Show solution
Approach: act out the overtakes one move at a time
The arrow points left, so the front-to-back order starts as 1, 2, 3, 4, 5 (car 1 leads).
The last car (5) jumps past the two in front of it (4 and 3): now 1, 2, 5, 3, 4.
The new second-to-last car (3) jumps past the two in front of it (5 and 2): now 1, 3, 2, 5, 4.
The car now in the middle (2) jumps past the two in front of it (3 and 1): now 2, 1, 3, 5, 4 — answer B.
Five big and four small elephants are marching along a path in single file. Because the path is narrow, the elephants cannot change their order. At the fork in the path, each elephant goes either to the right or to the left. Which of the following situations cannot happen?
Show answer
Answer: C
Show hints
Hint 1 of 2
The elephants keep their original order; they can only split left or right at the fork, not pass each other.
Still stuck? Show hint 2 →
Hint 2 of 2
Check each picture: the order along each branch must still match the single-file order.
Show solution
Approach: check that order is preserved on both branches
Because no elephant can overtake another, both branches must show the elephants in their original order.
Test each option against that rule.
Only option C breaks the order, so that arrangement cannot happen.
Carl writes down a five-digit number. He then places a shape on each of the five digits (see picture). He places different shapes on different digits. He places the same shape on the same digits. Which number did Carl hide?
Show answer
Answer: A — 34426
Show hints
Hint 1 of 2
Same shape means same digit; different shapes mean different digits.
Still stuck? Show hint 2 →
Hint 2 of 2
The two diamonds sit in positions 2 and 3, so those digits must be equal.
Show solution
Approach: match equal shapes to equal digits
The shapes are heart, diamond, diamond, club, spade - only positions 2 and 3 repeat.
So digits 2 and 3 are equal and all the others are different from them and each other.
Only 34426 has its 2nd and 3rd digits equal (4 and 4) with the rest distinct.
There are 5 people to choose from on a ballot paper. After counting 90 % of the votes the intermediate result looks as shown in the table. How many of the 5 people cannot win the election anymore?
Alex
Bella
Clint
Diana
Eddy
14
11
10
8
2
Show answer
Answer: B — 2
Show hints
Hint 1 of 2
The 45 counted votes are 90%, so first find the total number of votes.
Still stuck? Show hint 2 →
Hint 2 of 2
A candidate is out if even all remaining votes cannot catch the current leader.
Show solution
Approach: find remaining votes, test who can still overtake the leader
Counted votes 14+11+10+8+2 = 45 are 90%, so total = 50 and 5 votes remain.
Leader Alex has 14. A rival can win only if their votes + 5 > 14, i.e. they have at least 10 now.
Diana (8) and Eddy (2) reach at most 13 and 7, so they cannot win; the others still can.
Mosif has filled a table with numbers (see diagram). When he adds the numbers in each row and in each column, the result should always be the same, but he has made a mistake. To make every total the same he has to change one single number. Which number does Mosif have to change?
9
1
5
3
7
6
4
7
4
Show answer
Answer: B — 3
Show hints
Hint 1 of 2
Work out every row total and every column total and see which one is the odd one out.
Still stuck? Show hint 2 →
Hint 2 of 2
The number to change sits where the wrong row crosses the wrong column.
Show solution
Approach: find the row and column that are off
The row sums are 15, 16, 15 and the column sums are 16, 15, 15, so the target is 15.
One row is 1 too big and one column is 1 too big.
The cell in both that row and that column is the 3; lowering it to 2 fixes both.
In a classroom the children sit in rows, with the same number of children in each row. In Robert’s row there are 2 children to his left and 3 children to his right. There are 2 rows in front of Robert and just 1 row behind him. How many children are in the class in total?
Show answer
Answer: E — 24
Show hints
Hint 1 of 2
Count the children in Robert's row including Robert himself.
Still stuck? Show hint 2 →
Hint 2 of 2
Then count the rows, again including Robert's own row.
Show solution
Approach: count one row and the number of rows, including Robert
In Robert's row there are 2 to his left, Robert, and 3 to his right: 2+1+3 = 6 children per row.
There are 2 rows in front, Robert's row, and 1 behind: 2+1+1 = 4 rows.
Three football teams play in a tournament, and each team plays every other team once. A win is worth 3 points and a loss 0 points; a draw gives each team 1 point. Which number of points is impossible for any team to finish with?
Show answer
Answer: D — 5
Show hints
Hint 1 of 2
Each team plays only two games, scoring 3, 1 or 0 in each.
Still stuck? Show hint 2 →
Hint 2 of 2
List the totals you can build from two of {0, 1, 3} and see which option is missing.
Show solution
Approach: list every possible two-game total
Per game a team gets 3 (win), 1 (draw) or 0 (loss).
Each year, the third Thursday in March is named Kangaroo Day. The dates of Kangaroo Day for the next few years are listed below, with one error. Which date is wrong?
Show answer
Answer: C — 14/3/2024
Show hints
Hint 1 of 2
Work out the actual third Thursday of March for each listed year.
Still stuck? Show hint 2 →
Hint 2 of 2
March 1's weekday shifts each year; find which printed date isn't really a third Thursday.
Show solution
Approach: check each date against the real third Thursday
Find the weekday of March 1 for each year and locate that year's third Thursday.
Four of the five printed dates land correctly on the third Thursday of March.
Only 14/3/2024 is off — the third Thursday of March 2024 is the 21st, not the 14th.
The halftime score of a handball match was 9:14, so the visiting team was leading by five goals. After instructions from the coach, the home team dominated the second half and scored twice as many goals as their opponents. The home team won the match by one goal. What was the final score of the match?
Show answer
Answer: B — 21:20
Show hints
Hint 1 of 2
Let the visitors score x in the second half; then the home team scores 2x.
Still stuck? Show hint 2 →
Hint 2 of 2
Write 'home wins by one goal' as an equation in x and solve.
Show solution
Approach: set up one equation for the second half
Halftime is home 9, visitors 14. Let visitors score x more; the home team scores 2x more.
Final: home 9+2x, visitors 14+x, and home wins by one: 9+2x = (14+x)+1.
Solving gives x = 6, so home = 21 and visitors = 20.
The pink tower is taller than the red tower but shorter than the green tower. The silver tower is taller than the green tower. Which tower is the tallest?
Show answer
Answer: D — silver tower
Show hints
Hint 1 of 3
Line the towers up from shortest to tallest using the clues.
Still stuck? Show hint 2 →
Hint 2 of 3
Green is taller than pink and red; now see how silver compares to green.
Still stuck? Show hint 3 →
Hint 3 of 3
Whoever is taller than green must be the tallest of all.
Show solution
Approach: order the towers by height
The first clues say red is shortest, then pink, then green is taller than both.
The last clue says silver is taller than green, so silver beats green and everyone below it.
These children are standing in a line. Some are facing forwards and others are facing backwards. How many children are holding another child's hand with their right hand?
Show answer
Answer: E — 6
Show hints
Hint 1 of 3
Which hand is a child's right hand depends on which way that child is facing.
Still stuck? Show hint 2 →
Hint 2 of 3
For a child facing you, the right hand is on your left side; for one facing away, it is on your right side.
Still stuck? Show hint 3 →
Hint 3 of 3
Go child by child and only count the ones whose right hand is holding a neighbour.
Show solution
Approach: check each child's facing direction
Look at each child and decide which hand is the right hand based on whether they face front or back.
Now count only the children whose right hand is the one holding a neighbour's hand.
Nisa has 3 different types of cards: apple, cherry and grapes. She picks 2 cards from a row and swaps their places. She wants every card showing the same fruit to end up next to each other. For which row is this not possible with a single swap?
Show answer
Answer: A
Show hints
Hint 1 of 2
A single swap can only move 2 cards, so it can fix a row that is just 2 cards out of place.
Still stuck? Show hint 2 →
Hint 2 of 2
Check each row: try one swap and see if all the apples, all the cherries, and all the grapes end up touching.
Show solution
Approach: test whether one swap can group all like fruits in each option
One swap picks up just 2 cards and trades their spots, so it can only tidy a row that needs exactly those 2 cards moved.
Try a single swap on each row and see if it gathers every fruit into one touching group.
Four of the rows can be sorted with one swap, but in row A no single swap puts all the matching fruits together.
The map shows three bus stations at points A, B and C. A tour from station A to the Zoo and the Port and back to A is 10 km long. A tour from station B to the Park and the Zoo and back to B is 12 km long. A tour from station C to the Port and the Park and back to C is 13 km long. Also, a tour from the Zoo to the Park and the Port and back to the Zoo is 15 km long. How long is the shortest tour from A to B to C and back to A?
Show answer
Answer: B — 20 km
Show hints
Hint 1 of 2
Each given tour is a there-and-back loop, so it is twice some pair of legs.
Still stuck? Show hint 2 →
Hint 2 of 2
Add up the right combination of half-loops to build the A-B-C-A circuit.
Show solution
Approach: route through the inner points and use the loop totals
Send the trip A→Zoo→B→Park→C→Port→A. Then A→Zoo and Port→A are the two legs of the 10-loop minus its Zoo–Port leg, and similarly for the other two stations.
Adding up, the six legs total (10 − Zoo–Port) + (12 − Park–Zoo) + (13 − Port–Park) = 35 − (Zoo–Port + Park–Zoo + Port–Park).
But Zoo–Port + Park–Zoo + Port–Park is exactly the 15-km Zoo–Park–Port loop, so the tour is 35 − 15 = 20 km.
Julia has two pots with flowers, as shown. She keeps the flowers exactly where they are. She buys more flowers and puts them in the pots. After that, each pot has the same number of each type of flower. What is the smallest number of flowers she needs to buy?
Show answer
Answer: D — 8
Show hints
Hint 1 of 3
Look at one type of flower at a time and compare the two pots.
Still stuck? Show hint 2 →
Hint 2 of 3
Whichever pot has fewer of that type needs to be filled up to match the other.
Still stuck? Show hint 3 →
Hint 3 of 3
Add up all the missing flowers across both pots.
Show solution
Approach: match each flower type to the larger count
For each kind of flower, both pots must end up with the bigger of the two amounts they already have.
For every type, count how many are missing in the pot that is short, and add those up.
Adding all the missing flowers gives 8 extra flowers to buy.
The Koala ate some leaves from 3 branches. Each branch had 20 leaves. The Koala ate a few leaves from the first branch and then ate as many leaves from the second branch as were left on the first branch. Then it ate 2 leaves from the third branch. How many leaves in total were left on the 3 branches?
Show answer
Answer: E — 38
Show hints
Hint 1 of 2
Look only at branches 1 and 2 together first — the koala eats from branch 2 exactly what is left on branch 1.
Still stuck? Show hint 2 →
Hint 2 of 2
The leaves it eats from branch 2 are the same as the leaves still on branch 1, so the two branches together always keep the same amount no matter how much it ate first.
Show solution
Approach: branches 1 and 2 together always keep 20, then add branch 3
Branches 1 and 2 start with 20 + 20 = 40 leaves between them.
From these two branches the koala eats (some off branch 1) and then the same number off branch 2, so it eats the same amount as the leaves left on branch 1 — meaning branches 1 and 2 together always keep 20 leaves, whatever it nibbled first.
Branch 3 only loses 2 leaves, so it keeps 20 - 2 = 18.
Total left = 20 + 18 = 38, so the answer is 38 (E).
Ronja had four white tokens and Wanja had four grey tokens. They played a game in which they took turns to place one of their tokens to create two piles. Ronja placed her first token first. Which pair of piles could they not create?
Show answer
Answer: E
Show hints
Hint 1 of 2
Tokens are placed alternately starting with a white one, so the placing order constrains each pile.
Still stuck? Show hint 2 →
Hint 2 of 2
Reconstruct the colour pattern each option needs and find the one no legal alternating order can produce.
Show solution
Approach: test each stacking against the alternating rule
Ronja (white) places first, then they alternate colours as they build the two piles.
For each option, try to order the eight placements so colours alternate and the piles end up as shown.
Only the pair in choice E cannot arise from any valid alternating sequence.
Three pirates were asked how many coins and how many diamonds their friend Graybeard had. Each of the three told the truth to one question but told a lie to the other. Their answers were: (1) He has 8 coins and 6 diamonds. (2) He has 7 coins and 4 diamonds. (3) He has 7 coins and 7 diamonds. What is the total number of coins and diamonds that Graybeard has?
Show answer
Answer: C — 13
Show hints
Hint 1 of 2
Each pirate is right about exactly one of the two counts (coins or diamonds).
Still stuck? Show hint 2 →
Hint 2 of 2
Two pirates name the same coin count, 7; that is a good place to test the truth.
Show solution
Approach: make each statement half-true
Two pirates say 7 coins, so try coins = 7: then pirates 2 and 3 are truthful about coins and must be lying about diamonds (4 and 7 are both wrong), while pirate 1's coin count 8 is the lie.
Pirate 1 must then be truthful about diamonds, giving 6 diamonds — and 6 is not 4 or 7, so pirates 2 and 3 are indeed lying there. Everything fits.
There were 20 apples and 20 pears in a box. Carl randomly took 20 pieces of fruit from the box and Luca took the rest. Which of the following statements is always true?
Show answer
Answer: D — Carl got as many pears as Luca got apples.
Show hints
Hint 1 of 2
Carl took exactly 20 fruits, and there are exactly 20 apples in the whole box — the same number.
Still stuck? Show hint 2 →
Hint 2 of 2
Every apple Carl did NOT take is left for Luca, so think about how Carl filled up his 20 spots.
Show solution
Approach: match Carl's missing apples to his pears
Carl grabbed 20 fruits, and there are 20 apples in all, so the number of apples Carl is missing is exactly the same as the number of pears he picked up to fill his 20 spots.
All the apples Carl skipped end up with Luca, so the apples Carl missed = the apples Luca got.
Putting those together: Carl's pears = the apples Carl missed = Luca's apples, so 'Carl got as many pears as Luca got apples' is always true — choice D.
With lettersIf Carl took \(a\) apples and \(p\) pears then \(a+p=20\); Luca has \(20-a\) apples, and \(p=20-a\) too, so Carl's pears equal Luca's apples.
Logic & Word ProblemsRatios, Rates & Proportionsratiocasework
A box of fruit contains twice as many apples as pears. Christy and Lily divided them up so that Christy had twice as many pieces of fruit as Lily. Which one of the following statements is always true?
Show answer
Answer: E — Christy took as many pears as Lily got apples.
Show hints
Hint 1 of 2
Let there be p pears and 2p apples, total 3p; Christy ends with 2p pieces and Lily with p.
Still stuck? Show hint 2 →
Hint 2 of 2
Test each statement against every possible split — only one holds in all cases.
Show solution
Approach: check each claim over all valid splits
Pears = p, apples = 2p, total 3p; Christy has 2p pieces, Lily has p.
If Christy takes a apples she takes 2p−a pears; Lily then gets the remaining 2p−a apples.
So Christy's pears always equal Lily's apples; the other options can fail.
Logic & Word ProblemsAlgebra & Patternssubstitution
Three villages are connected by paths as shown. From Downend to Uphill, the detour via Middleton is 1 km longer than the direct path. From Downend to Middleton, the detour via Uphill is 5 km longer than the direct path. From Uphill to Middleton, the detour via Downend is 7 km longer than the direct path. How long is the shortest of the three direct paths between the villages?
Show answer
Answer: C — 3 km
Show hints
Hint 1 of 2
Name the three direct distances and turn each 'detour is k longer' fact into an equation.
Still stuck? Show hint 2 →
Hint 2 of 2
Add all three equations to get the total of the distances quickly.
Show solution
Approach: set up and add the detour equations
Let the direct paths be DU = a, DM = b, UM = c. The detours give b+c = a+1, a+c = b+5, a+b = c+7.
Adding all three: 2(a+b+c) = (a+b+c) + 13, so a+b+c = 13.
Then a = 6, b = 4, c = 3; the shortest direct path is 3 km.
A village of 12 houses has four straight streets and four circular streets. The map shows 11 houses. Each straight street has three houses, and each circular street also has three houses. Where should the 12th house be placed on this map?
Show answer
Answer: D — On D
Show hints
Hint 1 of 2
Each straight street needs 3 houses and each circular street needs 3 houses.
Still stuck? Show hint 2 →
Hint 2 of 2
Find the position where a street is currently one house short.
Show solution
Approach: check which street is missing a house
Every street, straight or circular, must hold exactly three houses.
Count the houses already on each street and find the one with only two.
Placing the 12th house at spot D completes that street without overfilling any other.
Number TheoryLogic & Word Problemssum-constraintcasework
The circles in the figure are to be numbered from 0 to 10, each with a different number. The five sums of the three numbers along each diameter must all be odd. If one of these sums is as small as possible, what is the largest possible value of one of the remaining sums?
Show answer
Answer: E — 21
Show hints
Hint 1 of 2
The centre circle is shared by all five diameters; for every diameter-sum to be odd, think about the parity the centre forces.
Still stuck? Show hint 2 →
Hint 2 of 2
Make one sum smallest by surrounding it with tiny numbers, which pushes the leftover large numbers onto another diameter to maximise it.
Show solution
Approach: use parity of the shared centre, then push extremes apart
Numbers 0..10 are placed; each diameter sums two ends plus the shared centre, and all five sums are odd.
The shared centre fixes a parity pattern for the diameter ends.
Putting the smallest numbers on one diameter leaves the large numbers for another; maximising that one gives a sum of 21.
Arithmetic & OperationsLogic & Word Problemswork-backward
As soon as he left his city heading toward Caecá, Charles saw the sign on the left. When he came back from Caecá, he saw the sign on the right. At that point, how far was it to reach his city?
Show answer
Answer: D — 41 km
Show hints
Hint 1 of 2
The left sign (just outside his city) reads Arati 12 km, Baibá 33 km; the right sign reads Baibá 8 km, Arati 29 km.
Still stuck? Show hint 2 →
Hint 2 of 2
First use either sign to find the fixed gap between Arati and Baibá, then notice his city sits 12 km past Arati.
Show solution
Approach: use the fixed town distances on the two signs
He saw the left sign just as he left his city, so his city is 12 km before Arati (and Baibá is 33 km out).
On the way back the right sign reads Arati 29 km ahead; his city is another 12 km beyond Arati.
So the distance left to his city is 29 + 12 = 41 km.
When the bat Elisa left her cave at night, the digital clock showed the time on the left. When she came back in the morning and hung herself upside down, she read her watch and saw the time on the right. How long did she stay out of the cave?
Show answer
Answer: D
Show hints
Hint 1 of 2
The morning time is read on an upside-down watch, so first turn that reading the right way up before comparing it with the night-time clock.
Still stuck? Show hint 2 →
Hint 2 of 2
Each digit, turned a half-turn, becomes another digit (0,1,2,5,8 still read as digits) and the whole reading flips left-to-right; fix the morning time, then count the gap.
Show solution
Approach: turn the upside-down morning reading right way up, then count the gap
The night clock shows the leaving time directly.
The morning watch is held upside down, so rotate that display a half-turn: each digit changes shape and the left and right digits swap, giving the true morning time.
Counting forward from the night time to the corrected morning time gives the time she was away, 3h 41m.
Spatial & Visual ReasoningLogic & Word Problemscube-viewsspatial-reasoning
Andrew bought 27 little cubes of the same size, each with three adjacent faces painted red and the other three painted a different color. He wants to use all of these little cubes to build one bigger cube. What is the largest number of completely red faces he can make on this big cube?
Show answer
Answer: E — 6
Show hints
Hint 1 of 2
Each small cube can show its three red faces all meeting at one corner; think about where each cube sits in the 3×3×3.
Still stuck? Show hint 2 →
Hint 2 of 2
Corner cubes show 3 faces, edge cubes 2 adjacent faces, face cubes 1 — can every position be served by a red corner?
Show solution
Approach: place each cube so red faces point outward
A small cube's three red faces meet at a vertex, so they cover any single face, any two adjacent faces, or any corner of three.
Corner positions need 3 mutually adjacent faces (matches a cube's red corner), edges need 2 adjacent, centers need 1 — all achievable.
So every outer face of the big cube can be made fully red, giving all 6 faces, choice E.
Spatial & Visual ReasoningLogic & Word Problemscube-viewsspatial-reasoningcareful-counting
Irene made a "city" using identical wooden cubes. Beside the problem there is a view from above and a side view of this "city." We do not know which side of the "city" the side view shows. What is the smallest number of cubes Irene could have used to build it?
Show answer
Answer: E — 15
Show hints
Hint 1 of 2
The top view tells you which floor cells have at least one cube; the side view tells you the heights seen in a row.
Still stuck? Show hint 2 →
Hint 2 of 2
Add the smallest heights at each occupied cell that still match both views — the orientation is unknown.
Show solution
Approach: combine the two views for a minimum
The top view marks which ground cells are occupied; the side view limits the column heights.
Choosing the least cube count at each cell that is still consistent with both views (over the unknown orientation) gives a minimum.
Spatial & Visual ReasoningLogic & Word Problemspaper-cuttingfoldingcasework
Amelia has a paper strip with five equal cells, each containing a different drawing, as shown in the figure. She folds the strip so that the cells overlap in five layers. Which of the following sequences of layers, from top to bottom, is not possible to obtain?
Show answer
Answer: A — ★, □, ■, ○, ●
Show hints
Hint 1 of 2
Folding a strip reverses the order of the cells that flip over; track which symbol ends on top.
Still stuck? Show hint 2 →
Hint 2 of 2
Test each listed stack against an actual fold — one ordering can never arise.
Show solution
Approach: simulate the folds
Folding the five-cell strip so all cells overlap forces certain symbols to keep their relative order and others to reverse.
Checking each option against a real folding, four of them can be produced.
The ordering in A cannot be obtained by any folding, so it is the impossible one, choice A.
In a classroom there are two chairs at each table. Each boy in the class sits at a table with a girl, but there are four girls who do not sit at a table with a boy. There are 14 little tables in the classroom. How many girls are in that class?
Show answer
Answer: E — 16
Show hints
Hint 1 of 2
Most tables seat one boy and one girl; a few seat two girls.
Still stuck? Show hint 2 →
Hint 2 of 2
Count the all-girl tables, then add the girls sharing with boys.
Show solution
Approach: split tables into boy-girl and girl-girl
Four girls sit without a boy, two to a table, so 2 tables are all-girl.
That leaves 14 minus 2 = 12 tables, each with one boy and one girl.
Total girls = 12 (one per mixed table) plus 4 = 16.
Lothar finishes a race ahead of Manfred. Victor finishes after Jan, Manfred finishes ahead of Jan, and Eddy finishes ahead of Victor. Which of the five finishes the race last?
Show answer
Answer: A — Victor
Show hints
Hint 1 of 2
Write each clue as 'X before Y' and chain them.
Still stuck? Show hint 2 →
Hint 2 of 2
The last finisher is the one nobody finishes behind.
Show solution
Approach: order the chain of finishing positions
Lothar before Manfred, Manfred before Jan, and Jan before Victor give the order L, M, J, V.
Eddy finishes before Victor, so Eddy is also ahead of Victor.
Every day the three kangaroos Alex, Bob and Carl go for a walk. If Alex does not wear a hat, then Bob wears a hat. If Bob does not wear a hat, then Carl wears a hat. Today Carl does not wear a hat. Which kangaroos can we be sure are wearing a hat today?
Show answer
Answer: E — only Bob
Show hints
Hint 1 of 2
Carl wears no hat. Use the rule 'if Bob has no hat then Carl wears one' in reverse.
Still stuck? Show hint 2 →
Hint 2 of 2
Since Carl has no hat, Bob must have one; then check whether Alex is forced.
Show solution
Approach: contrapositive reasoning from Carl
Rule: if Bob has no hat, Carl wears one. Carl has no hat, so Bob must have a hat.
Rule: if Alex has no hat, Bob wears one — already satisfied, so Alex is not forced either way.
There are white, grey and black squares. Three children use these to make this pattern. First Anni replaces all black squares with white squares. Then Bob replaces all grey squares with black squares. Finally Chris replaces all white squares with grey squares. Which picture have the three children now created?
Show answer
Answer: A
Show hints
Hint 1 of 2
Do the three colour changes one at a time, in the order Anni, then Bob, then Chris.
Still stuck? Show hint 2 →
Hint 2 of 2
Follow just one square of each starting colour all the way through to see what it turns into.
Show solution
Approach: follow each starting colour through the three changes in order
A black square turns white (Anni), then that white turns grey (Chris), so black ends grey.
A grey square turns black (Bob) and stays black, so grey ends black.
A white square is only changed by Chris, turning grey, so white ends grey.
Recolouring every square this way gives the picture in option A.
Anna, Bella, Claire, Dora, Erika and Frieda meet at a party. Each pair who know each other shake hands exactly once. Anna shakes hands only once, Bella twice, Claire three times, Dora four times and Erika five times. How many people does Frieda shake hands with?
Show answer
Answer: C — 3
Show hints
Hint 1 of 2
Erika shook hands 5 times, so she shook everyone — including Anna, whose single handshake is therefore with Erika.
Still stuck? Show hint 2 →
Hint 2 of 2
Peel the people off one at a time (Erika, then Dora, …) to see who is left for Frieda.
Show solution
Approach: deduce each person's partners step by step
Erika (5) shook everyone; Anna (1) then only shook Erika.
Dora (4) shook everyone but Anna: Erika, Bella, Claire, Frieda — giving Bella her 2nd handshake.
Claire (3) shook Erika, Dora and Frieda, so Frieda's partners are Erika, Dora, Claire: 3 people.
Together the three squirrels Anni, Asia and Elli have 10 nuts. Each one has a different number of nuts, but at least 2 nuts. Anni has the least number of nuts. Asia has the most nuts. How many nuts does Elli have?
Show answer
Answer: C — 3
Show hints
Hint 1 of 2
Anni has the fewest and each squirrel has at least 2, so start Anni as low as allowed.
Still stuck? Show hint 2 →
Hint 2 of 2
Try the smallest possible numbers that are all different and add to 10, then read Elli's amount.
Show solution
Approach: use the smallest distinct amounts that sum to 10
All three numbers are different, each at least 2, and they add to 10.
Anni has the fewest, so try Anni = 2; then Elli and Asia must add to 8 with Asia largest.
The only way is Elli = 3 and Asia = 5 (all different, Asia most).
A big spot of ink covers most of a calendar page for a certain month. On which day of the week does the 25th of that month fall? (In the calendar the weekday columns are labelled Mo, Di, Mi, Do, Fr, Sa, So — Monday through Sunday.)
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Answer: D — Saturday
Show hints
Hint 1 of 2
Use the few dates you can still read at the edges of the blot to lock down the weekday pattern.
Still stuck? Show hint 2 →
Hint 2 of 2
Dates a week apart fall on the same weekday; step by sevens from a known date to reach the 25th.
Show solution
Approach: anchor on a visible date and jump in steps of 7
A calendar repeats weekdays every 7 days, so the 25th shares a weekday with the 18th, 11th, and 4th.
Reading a date that the ink did not cover fixes that column's weekday.
The following two statements are true: some aliens are green and all others are purple; green aliens live on Mars only. Which one of the following logical conclusions can be made?
Show answer
Answer: E — There are no green aliens on Venus.
Show hints
Hint 1 of 2
List exactly what the two true statements force and what they do not.
Still stuck? Show hint 2 →
Hint 2 of 2
Check each option against 'green aliens live on Mars only'.
Show solution
Approach: test each conclusion against the given facts
Green aliens live on Mars only, so no green alien can be on Venus.
The colour split (some green, the rest purple) does not force all aliens onto Mars, nor pin purple aliens to Venus.
Only the statement 'there are no green aliens on Venus' must be true.
Diana shoots 3 darts, three times, at a target board with two fields. The first time she scores 12 points, the second time 15. The number of points depends on which field she hits. How many points does she score the third time?
Show answer
Answer: D — 21
Show hints
Hint 1 of 2
There are only two prizes: a dart in the small middle circle is worth one amount, a dart in the big ring is worth another.
Still stuck? Show hint 2 →
Hint 2 of 2
In the first picture all three darts are in the ring; in the second one dart has moved into the middle — see how much the score jumped.
Show solution
Approach: read each picture to find the ring value and the middle value, then score the third throw
First picture: all 3 darts are in the big ring and score 12, so each ring dart is worth 12 ÷ 3 = 4.
Second picture: one dart moved into the middle and the score went from 12 up to 15, so the middle is worth 4 + 3 = 7.
Third picture: all 3 darts are in the middle, so the score is 7 + 7 + 7 = 21, answer D.
In the diagram the circles are light bulbs, joined by lines to some other light bulbs. At the start every bulb is switched off. If you touch a bulb, then that bulb and all the bulbs directly joined to it switch on. What is the smallest number of bulbs you have to touch in order to switch on all the bulbs?
Show answer
Answer: A — 2
Show hints
Hint 1 of 2
Touching one bulb lights it and every bulb directly joined to it, so look for a bulb connected to many others.
Still stuck? Show hint 2 →
Hint 2 of 2
Find the fewest bulbs whose neighbourhoods together cover all of them - two well-placed touches are enough here.
Show solution
Approach: dominating set - cover all bulbs with fewest touches
Touching a bulb switches on that bulb and all bulbs joined to it by an edge.
Look for bulbs whose connections together reach every bulb in the picture.
Two suitably chosen bulbs cover the whole network, and one is not enough.
Albert places these 5 figures on a 5×5 grid. Each figure is only allowed to appear once in every column and once in every row. Which figure does Albert have to place on the field with the question mark?
Show answer
Answer: A
Show hints
Hint 1 of 2
Each of the five figures shows up exactly once in every row and exactly once in every column.
Still stuck? Show hint 2 →
Hint 2 of 2
Look along the question mark's row and down its column, and cross out every figure you already see there.
Show solution
Approach: cross out every figure already in the marked cell's row and column; one figure is left
Because no figure repeats in a row or a column, the missing one must be a figure not yet in that row or column.
Read across the marked cell's row and down its column and cross off each figure that already appears.
Exactly one figure is never crossed off, and that is the one that belongs in the question-mark cell, answer A.
Felix the rabbit has 20 carrots. Every day he eats 2 of them. He has eaten the 12th carrot on a Wednesday. On which day of the week did he start eating the carrots?
Show answer
Answer: E — Friday
Show hints
Hint 1 of 2
Carrots come in pairs each day: day 1 is carrots 1 and 2, day 2 is carrots 3 and 4, and so on.
Still stuck? Show hint 2 →
Hint 2 of 2
Find which day number holds the 12th carrot, then hop backwards on the calendar to day 1.
Show solution
Approach: pair the carrots into days to reach day 6, then count back on the days of the week
Two carrots each day means: day 1 = carrots 1,2; day 2 = 3,4; day 3 = 5,6; day 4 = 7,8; day 5 = 9,10; day 6 = 11,12.
So the 12th carrot is eaten on day 6, which is the Wednesday given.
Count back 5 days from Wednesday: Tue, Mon, Sun, Sat, Friday — that is day 1, answer E.
A rose bush has 8 flowers, with butterflies and dragonflies sitting on them. At most one insect sits on each flower, and more than half of the flowers are taken. The number of butterflies is twice the number of dragonflies. How many butterflies are sitting on the flowers?
Show answer
Answer: C — 4
Show hints
Hint 1 of 2
Twice as many butterflies as dragonflies means the insects come in little teams of 3: two butterflies with one dragonfly.
Still stuck? Show hint 2 →
Hint 2 of 2
More than half of the 8 flowers are taken, so more than 4 insects are sitting, but no more than 8.
Show solution
Approach: group the insects into teams of two butterflies plus one dragonfly and count whole teams
Since there are twice as many butterflies as dragonflies, picture them in teams of 3: two butterflies and one dragonfly each.
More than half of the 8 flowers are filled, so the number of insects is more than 4 but at most 8 — one team is only 3 (too few) and three teams would be 9 (too many).
So there are exactly two teams: 6 insects, which is 4 butterflies and 2 dragonflies, giving 4 butterflies, answer C.
A lion hides in one of three rooms. The note on room 1 reads “The lion is here.” The note on room 2 reads “The lion is not here.” The note on room 3 reads “\(2 + 3 = 2 \times 3\).” Exactly one of the three notes is true. Which room is the lion in?
Show answer
Answer: C — Room 3
Show hints
Hint 1 of 2
The note on room 3 says 2 + 3 = 2 × 3, i.e. 5 = 6 — is that ever true?
Still stuck? Show hint 2 →
Hint 2 of 2
Since that note is false, exactly one of the other two notes must be true; test each room.
Show solution
Approach: use that exactly one note is true
Room 3's note (5 = 6) is false, so the single true note is on room 1 or room 2.
If the lion were in room 1 both notes 1 and 2 would be true; if in room 2 neither would be true.
If the lion is in room 3, only note 2 ('not here') is true — exactly one. So the lion is in Room 3.
A hotel in the Caribbean correctly advertises with the slogan “350 days of sun in the year!” How many days does Mr. Happy have to spend in the hotel in a 365-day year to be guaranteed two consecutive days of sunshine to enjoy?
Show answer
Answer: D — 32
Show hints
Hint 1 of 2
There are 365 − 350 = 15 cloudy days to break up the sunny ones.
Still stuck? Show hint 2 →
Hint 2 of 2
Worst case, each cloudy day separates single sunny days; find the longest stay that could still avoid two sunny days in a row.
Show solution
Approach: worst-case spacing (pigeonhole)
With 15 cloudy days you can separate sunny days into at most 16 single sunny days, giving 15 + 16 = 31 days with no two sunny days adjacent.
So 31 days might not be enough, but on the very next day two sunny days must touch.
He must stay 32 days to be guaranteed two consecutive sunny days.
In the diagram we see 10 islands that are connected by 15 bridges. What is the minimum number of bridges that need to be closed off so that there is no longer any connection from A to B?
Show answer
Answer: C — 3
Show hints
Hint 1 of 2
You want the fewest bridges whose removal leaves no path at all from A to B.
Still stuck? Show hint 2 →
Hint 2 of 2
Look for a bottleneck: the smallest set of bridges that every A-to-B route must use.
Show solution
Approach: find a minimum cut (smallest bottleneck of bridges) separating A from B
Every route from A to B has to cross a narrow set of bridges.
Trace the routes and find the smallest group of bridges that all of them share.
Removing that bottleneck of 3 bridges disconnects A from B, and no smaller set works.
Logic & Word ProblemsArithmetic & Operationsoff-by-one
Some girls are standing in a circle. The teacher makes them do a headcount. Bianca says one, her neighbour says two and so on. If they count in a clockwise direction, Antonia says six. If they count in an anticlockwise direction, Antonia says nine. How many girls are forming the circle?
Show answer
Answer: C — 13
Show hints
Hint 1 of 2
Count the gap from Bianca to Antonia each way around the circle.
Still stuck? Show hint 2 →
Hint 2 of 2
Going clockwise and anticlockwise covers the whole circle once.
Show solution
Approach: add the two arc-gaps to get the total around the circle
Clockwise, Antonia is number 6, so she is 5 girls along from Bianca.
Anticlockwise she is number 9, so she is 8 girls along the other way.
The two arcs together go right round the circle: 5 + 8 = 13 girls, choice C.
Leo and Max are standing in a queue that is made up of 11 people in total. There are 7 people in front of Leo, and Max stands directly behind him in the queue. How many people are behind Max?
Show answer
Answer: B — 2
Show hints
Hint 1 of 2
First work out Leo's place in line from the 7 people ahead of him.
Still stuck? Show hint 2 →
Hint 2 of 2
Max is right behind Leo; the rest of the 11 are behind Max.
Show solution
Approach: locate each person, then count the tail
7 people are in front of Leo, so Leo is 8th.
Max stands directly behind Leo, so Max is 9th.
There are 11 people total, leaving 11 - 9 = 2 behind Max.
Only four players scored goals in a handball game, and each scored a different number of goals. Michael scored the fewest. If the other three players scored 20 goals in total, what is the greatest number of goals Michael could have scored?
Show answer
Answer: C — 4
Show hints
Hint 1 of 2
Michael scored the fewest, and all four totals differ; the other three add to 20.
Still stuck? Show hint 2 →
Hint 2 of 2
To make Michael's count as big as possible, keep the other three just barely above him and distinct.
Show solution
Approach: push the other three players as close to Michael as possible
Michael scored the fewest, so the other three each scored more than him, and all four totals are different.
To let Michael score a lot, the other three should be just barely bigger: the three smallest different scores above Michael are Michael+1, Michael+2 and Michael+3.
If Michael scored 4, the others would be at least 5, 6, 7 = 18, which fits inside 20 (for example 5, 6, 9 add to 20).
If Michael scored 5, the others would be at least 6, 7, 8 = 21, which is already more than 20 — too big.
So Michael could score at most 4 (C).
Same idea with algebraIf Michael scores \(m\), the smallest the other three can total is \((m+1)+(m+2)+(m+3)=3m+6\). We need \(3m+6\le 20\), so \(m\le 4\).
A number is written on each face of a special die. The two numbers on any pair of opposite faces always add up to the same total. Five of the six numbers are 5, 6, 9, 11 and 14. What is the number on the sixth face?
Show answer
Answer: E — 15
Show hints
Hint 1 of 2
Opposite faces share the same total, so the six numbers split into three pairs with equal sums.
Still stuck? Show hint 2 →
Hint 2 of 2
The biggest and smallest known numbers hint at the common pair-sum; find the missing partner.
Show solution
Approach: find the common pair-sum from two known numbers, then complete the last pair
Opposite faces always add to the same total, so the six numbers split into three pairs that all share one sum.
Among the five known numbers, 6 + 14 = 20 and 9 + 11 = 20, so that shared sum must be 20.
The number 5 is left over, so its partner is the sixth face: 20 − 5 = 15.
Boris wants to increase his pocket money. A fairy gives him three magic wands, and he must use every one exactly once: the “+1” wand increases his money by 1 €, the “−1” wand decreases it by 1 €, and the “×2” wand doubles it. In which order should he use the wands to end up with the most money?
Show answer
Answer: D
Show hints
Hint 1 of 2
Doubling multiplies everything that came before, so a +1 done before doubling is worth more than after.
Still stuck? Show hint 2 →
Hint 2 of 2
Add before you double and subtract after you double to end up highest.
Show solution
Approach: order operations so the gain is doubled and the loss is not
The +1 wand is most valuable applied before the ×2 wand, so its euro gets doubled.
The −1 wand should come after the ×2 wand, so only one euro is lost, not two.
Best order: +1, then ×2, then −1, giving (x+1)×2 − 1 = 2x + 1.
David has a stove with two hobs on which he wants to prepare five different dishes. The dishes need 40 min, 15 min, 35 min, 10 min and 45 min until they are fully cooked. He wants to spend as little time in the kitchen as possible but is only allowed to take dishes off the hob when they are fully cooked. How long does he need for the preparation?
Show answer
Answer: C — 75 min
Show hints
Hint 1 of 2
With two hobs, split the five cooking times into two groups; the time needed is the larger group's total.
Still stuck? Show hint 2 →
Hint 2 of 2
Balance the two groups as evenly as you can to make that larger total small.
Show solution
Approach: partition the times into two groups and minimise the larger total
The five times add to 40+15+35+10+45 = 145 minutes across two hobs.
He is free until everything is cooked, so the time needed is the larger group's total.
The most balanced split is 40+35 = 75 against 45+15+10 = 70.
The larger total, and so the time needed, is 75 min.
Diana wants to write whole numbers into each circle in the diagram, so that for all eight small triangles the sum of the three numbers in the corners is always the same. What is the maximum number of different numbers she can use?
Show answer
Answer: C — 3
Show hints
Hint 1 of 2
Equal triangle-sums force many circles to share a value.
Still stuck? Show hint 2 →
Hint 2 of 2
Trace which circles must be equal; only a few can stay distinct.
Show solution
Approach: propagate the equal-sum constraint
The eight small triangles share corners, and equal sums force chains of circles to carry the same number.
Working through the forced equalities leaves at most three independent values.
An explicit labelling achieves three different numbers, so the maximum is 3.
Mona, Asma and Nadja work in the same nursery. On each day from Monday to Friday exactly two of them are working. Mona works three times and Asma works four times per week. How many times does Nadja work per week?
Show answer
Answer: C — 3
Show hints
Hint 1 of 2
Each weekday exactly two of the three work, so add up all the work-days.
Still stuck? Show hint 2 →
Hint 2 of 2
Mona's 3 plus Asma's 4 plus Nadja's count must equal the total work-days.
Show solution
Approach: count the total worker-days
Two people work each of the 5 days, so the week has \(2 \times 5 = 10\) worker-days in total.
Mona fills 3 of them and Asma fills 4, which is 7 worker-days.
Peter wants to guess Paul's password. He already knows the following: the three last characters are digits, and there are at most three capital letters in the password. Which of the following passwords could be Paul's?
Show answer
Answer: C — 1234LLuuaapp4321
Show hints
Hint 1 of 2
Check two rules for each option: the last three characters are digits, and there are at most three capital letters.
Still stuck? Show hint 2 →
Hint 2 of 2
Eliminate any password that breaks either rule.
Show solution
Approach: test each option against both rules
PAUL123 has four capitals (P, A, U, L) — too many.
P0a1u2L3 and 123PAUL do not end in three digits, and Paulin3 also fails the last-three-digits rule.
1234LLuuaapp4321 ends in 321 (three digits) and has only two capitals (L, L), so it satisfies both rules.
Five squirrels A, B, C, D and E are sitting on the marked points. The crosses show 6 nuts that they are collecting. The squirrels start to run at the same time with the same speed to the nearest nut to pick it up. As soon as a squirrel has picked up its first nut it immediately runs on to get another nut. Which squirrel gets a second nut?
Show answer
Answer: C — C
Show hints
Hint 1 of 2
Each squirrel runs to its nearest nut first; mark which nut each one claims.
Still stuck? Show hint 2 →
Hint 2 of 2
After the first round, whichever squirrel is closest to a still-uncollected nut grabs the next one.
Show solution
Approach: assign nearest nuts, then see who reaches the leftover nut soonest
First each squirrel runs to its closest cross (nut); with 6 nuts and 5 squirrels, exactly one nut is still free after this first round.
Whoever finishes its first trip earliest and is closest to that leftover nut grabs it; comparing the distances, squirrel C reaches the leftover nut first.
So the squirrel that gets a second nut is C, choice (C).
Igor writes down all the results of the quarter-finals, semi-finals and final of a tennis tournament. They are listed here in random order: Bert beats Anton, Carl beats Damien, Glen beats Henry, Glen beats Carl, Carl beats Bert, Edon beats Fred, Glen beats Edon. Who plays in the final?
Show answer
Answer: B — Glen and Carl
Show hints
Hint 1 of 2
Use the listed results to rebuild the knockout bracket.
Still stuck? Show hint 2 →
Hint 2 of 2
The final is between the two semifinal winners.
Show solution
Approach: reconstruct the knockout bracket
Quarterfinals from the results: Bert beats Anton, Carl beats Damien, Edon beats Fred, Glen beats Henry.
Semifinals: Carl beats Bert and Glen beats Edon, so Carl and Glen advance.
Chantal has placed numbers in two of the nine cells (see diagram). She wants to place the numbers 1, 2, 3 in every row and every column exactly once. What is the sum of the two numbers in the grey cells?
Show answer
Answer: C — 4
Show hints
Hint 1 of 2
Each row and each column must hold 1, 2, 3 exactly once, like a tiny Sudoku.
Still stuck? Show hint 2 →
Hint 2 of 2
Fill in the forced cells from the given 1 and 2, then read the two grey cells.
Show solution
Approach: fill the 3×3 Latin square from the given clues
The top-left is 1 and the centre is 2; the grey cells are the middle and bottom of the right column.
The top row must finish 1, 3, 2, so the top-right cell is 2.
The right column then needs 1 and 3 in its grey cells, so their sum is 1 + 3 = 4.
10 runners start in a running race. At the finish, there are 3 more runners behind Thomas than there are in front of him. In which position did Thomas finish?
Show answer
Answer: C — 4
Show hints
Hint 1 of 2
Besides Thomas there are 9 other runners, split into a front group and a back group.
Still stuck? Show hint 2 →
Hint 2 of 2
The back group is 3 bigger than the front group; try sharing the 9 runners so the back has 3 more.
Show solution
Approach: share the other 9 runners into front and back groups
Take Thomas out for a moment: the other 9 runners are split into the ones in front and the ones behind.
The back group must be 3 bigger than the front group, so split 9 into 3 in front and 6 behind (6 is 3 more than 3).
With 3 runners in front of him, Thomas is the next one, so he is in 4th place.
Logic & Word Problemssum-constraintcareful-counting
All the boys in a class are born on different days of the week and all the girls are born in different months. If one new girl or boy joins this class, this is definitely no longer true. How many teenagers are in this class?
Show answer
Answer: B — 19
Show hints
Hint 1 of 2
There are only 7 days in a week and 12 months in a year.
Still stuck? Show hint 2 →
Hint 2 of 2
If adding any one new person must break the all-different rule, the class already uses every day and every month.
Show solution
Approach: fill every day and every month exactly
Boys are on different days of the week, so at most 7 boys.
Girls are in different months, so at most 12 girls.
For any new boy OR girl to be forced to repeat, there must already be 7 boys and 12 girls.
Some pirates are climbing onto a ship one after the other using a rope. Their leader is exactly in the middle. He is the eighth pirate to climb onto the ship. How many pirates board the ship?
Show answer
Answer: B — 15
Show hints
Hint 1 of 2
The leader is the 8th to climb and is exactly in the middle of the line.
Still stuck? Show hint 2 →
Hint 2 of 2
If 7 pirates are ahead of him, the same number must be behind him.
Show solution
Approach: balance the pirates on each side of the middle one
The leader is the 8th to climb, so 7 pirates went before him.
Being exactly in the middle, 7 pirates must also come after him.
Each side of each triangle in the diagram is painted either blue (blau), green (grün) or red. Four of the sides are already painted, as shown. Which colour can the line marked x have, if each triangle must have all sides in different colours?
Show answer
Answer: A — only green
Show hints
Hint 1 of 2
Each triangle uses all three colours exactly once, and neighbouring triangles share a side.
Still stuck? Show hint 2 →
Hint 2 of 2
Start from the known blue sides and force each shared side in turn along the strip until you reach x.
Show solution
Approach: propagate the all-different constraint along the strip
Each triangle must use blue, green and red once.
Starting from the two given blue sides and using the shared edges, each triangle's remaining colours are forced one after another.
Following the chain to the side marked x leaves only green possible.
The teacher asks five of her students how many of them had studied the previous day. Azra says: “Nobody.” Berti says: “Only one.” Christa says: “Exactly two.” Doris says: “Exactly three.” Emina says: “Exactly four.” The teacher knows that students always lie if they haven’t studied and are always truthful when they have studied. How many of those students had studied the previous day?
Show answer
Answer: B — 1
Show hints
Hint 1 of 2
The five claims 0, 1, 2, 3, 4 are all different, so at most one of them is true.
Still stuck? Show hint 2 →
Hint 2 of 2
A student tells the truth exactly when they studied, so the number who studied equals the number of true statements.
Show solution
Approach: find the count that makes exactly the truth-tellers truthful
Students who studied tell the truth; the others lie. So the number who studied equals the number of true statements.
The five statements give five different counts, so at most one is true, meaning at most one student studied.
If exactly one studied, that student truthfully said 'Only one' and the other four lie - which is consistent.
Handsome Fritz has a secret e-mail address that is known by only four of his friends. Today he received eight e-mails at this address. Which of the following statements is definitely correct?
Show answer
Answer: D — Fritz received at least two e-mails from one of his friends.
Show hints
Hint 1 of 2
Spread 8 e-mails among only 4 senders — can they all stay at one each?
Still stuck? Show hint 2 →
Hint 2 of 2
This is the pigeonhole principle in disguise.
Show solution
Approach: pigeonhole on senders
Eight e-mails come from just four friends.
If every friend sent at most one, that is at most 4 e-mails — too few.
So at least one friend must have sent two or more.
The statement guaranteed true is (D): at least two e-mails come from one friend.
In a holiday camp, 7 children eat ice cream every day and 9 children eat ice cream every other day. The rest never eat ice cream. Yesterday 13 children ate ice cream. How many children will eat ice cream today?
Show answer
Answer: D — 10
Show hints
Hint 1 of 2
The 7 daily eaters had ice cream yesterday and will have it again today.
Still stuck? Show hint 2 →
Hint 2 of 2
Figure out how many of the every-other-day group ate yesterday; the rest of that group are the ones who eat today.
Show solution
Approach: split the 13 into daily eaters and every-other-day eaters
The 7 daily children ate yesterday, so 13 − 7 = 6 of the every-other-day children also ate yesterday.
Those 6 skip today, so the other 9 − 6 = 3 every-other-day children eat today.
Today's total is the 7 daily children plus those 3, which is 10.
The kangaroos A, B, C, D and E sit in this order, clockwise, around a round table (see picture). After a bell rings, all but one kangaroo swap seats with a neighbour. Afterwards they sit, clockwise, in the order A, E, B, D, C. Which kangaroo did not change places?
Show answer
Answer: B — B
Show hints
Hint 1 of 2
Everyone who moved swapped with a next-door neighbour, so the movers come in pairs.
Still stuck? Show hint 2 →
Hint 2 of 2
Fix the seats and find the one kangaroo whose new neighbours match its old ones — that is the one that stayed put.
Show solution
Approach: anchor one seat and check the rest are neighbour swaps
Before, clockwise: A, B, C, D, E; afterwards, clockwise: A, E, B, D, C.
Suppose B stays in its seat. Reading the after-order clockwise starting at B gives B, D, C, A, E around the circle.
Comparing seat by seat with the before arrangement, C and D have simply swapped (they were neighbours) and A and E have swapped (also neighbours), while B never moved.
Every change is a neighbour swap, so the kangaroo that did not move is B.
The winning team of a football match gets 3 points and the losing team 0 points. In the case of a draw both teams get one point each. Four teams A, B, C and D play a tournament in which each team plays each other team exactly once. At the end of the tournament team A has 7 points, and teams B and C have 4 points each. How many points does team D have?
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Answer: B — 1
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Hint 1 of 2
Six games are played; figure out the total number of points handed out, depending on draws.
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Hint 2 of 2
A's 7 points pin down how A's three games went.
Show solution
Approach: reconstruct results from the point totals
Each of the 6 games gives out 3 points (a win) or 2 points (a draw).
A has 7 = 3+3+1, so A won twice and drew once (A played 3 games).
B and C each have 4 points; working through the remaining games consistently forces D to total just 1 point.
In a shared apartment where six girls live there are 2 bathrooms. Each morning from 7:00 the girls use the bathrooms before breakfast, each spending 9, 11, 13, 18, 22 and 23 minutes respectively, always alone in one of the two bathrooms. What is the earliest time that all six girls can have breakfast together?
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Answer: B — 7:49
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Hint 1 of 2
Two bathrooms run in parallel; minimise the longer of the two totals.
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Hint 2 of 2
Split the six times into two groups whose sums are as close as possible.
Show solution
Approach: balance the two bathrooms (minimise the larger total)
The six times total 9+11+13+18+22+23 = 96 minutes, shared by two bathrooms running in parallel, so we want to split them into two groups whose larger sum is as small as possible.
A perfect 48/48 split is impossible, but 23+13+11 = 47 and 22+18+9 = 49 is the closest, so one bathroom needs 49 minutes.
Everyone is finished after 49 minutes, i.e. at 7:49.
Each of six lone heroes has captured some wanted people. In total they captured 20 wanted people: the first hero caught one, the second two, the third three. The fourth hero caught more than any other hero. What is the smallest number the fourth hero could have caught so that this is possible?
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Answer: B — 6
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Hint 1 of 2
The first three heroes account for 1+2+3 = 6, leaving the rest of the 20.
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Hint 2 of 2
Make the fifth and sixth heroes as large as possible while staying below the fourth.
Show solution
Approach: minimise the largest under a sum constraint
30 children took part in a sports competition, playing football and handball. 15 took part in football and 20 took part in handball. How many children took part in both sports?
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Answer: E — 5
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Hint 1 of 3
If you add the football players and the handball players, the children who played both get counted twice.
Still stuck? Show hint 2 →
Hint 2 of 3
Add 15 and 20 and compare that total to the real number of children, 30.
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Hint 3 of 3
The extra amount above 30 is exactly the children who were counted twice.
Show solution
Approach: the double-counted children are the ones who played both
Add the football players and handball players: 15 + 20 = 35.
But there are only 30 children, so 35 is 5 too many.
Those extra 5 are the children counted in both lists, so 5 played both, which is answer E.
Andi, Berti, Christa, Doris and Edi were born on these days: 20.02.2000, 12.03.2000, 20.03.2000, 12.04.2000 and 23.04.2000 (in some order). Andi and Edi have their birthday in the same month. Berti and Christa also share a birthday month. Andi and Christa were born on the same day of different months. Doris and Edi were also born on the same day of different months. Which of these children is the youngest?
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Answer: B — Berti
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Hint 1 of 2
Match the four clues to the five dates; same-month and same-day pairs are very restrictive.
Still stuck? Show hint 2 →
Hint 2 of 2
Once everyone has a date, the youngest is simply the latest one.
Show solution
Approach: deduce the date assignment
Same-month pairs force Andi&Edi into one of the doubled months and Berti&Christa into the other.
Same-day clues then fix Andi = 12.03, Christa = 12.04, Edi = 20.03, Doris = 20.02, Berti = 23.04.
A travel agency organises four different trips for a certain group. Each trip has a participation rate of 80%. What is the minimum percentage of the group which has taken part in all four roundtrips?
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Answer: D — 20 %
Show hints
Hint 1 of 2
Count who could MISS a trip rather than who takes it.
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Hint 2 of 2
Spread the 20% who skip each trip over different people to make the all-four group as small as possible.
Show solution
Approach: bound the overlap using the missers
Each trip is skipped by 20% of the group.
Across four trips at most 4×20% = 80% skip at least one trip.
So at least 100% − 80% = 20% took part in all four.
A flea stands on the floor and wants to climb the 10 steps. Each time it can either jump 3 steps up or jump 4 steps down. What is the smallest number of jumps it must make?
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Answer: E — 8
Show hints
Hint 1 of 2
Each up-jump moves him 3 steps, so just adding up-jumps lands on 3, 6, 9, 12, ... — never exactly 10.
Still stuck? Show hint 2 →
Hint 2 of 2
To fix that miss, mix in a down-jump of 4 and count how many jumps that takes.
Show solution
Approach: skip-count the up-jumps, then patch with a down-jump
Only going up, the flea lands on 3, 6, 9, 12, 15, 18, ... — he hops right over 10 and never lands on it.
So he must overshoot and then drop down 4: from 12 a down-jump lands on 8, and from 18 a down-jump lands on 14, still not 10.
Try going higher: 6 up-jumps reach step 18, then 2 down-jumps of 4 take him 18 → 14 → 10, landing exactly on 10.
That is 6 + 2 = 8 jumps, and nothing shorter ever lands on 10, so the fewest jumps is 8.
Take four cards and on each one write one of the numbers 2, 5, 7, 12. On the back of each card write one of the following properties: “divisible by 7”, “prime number”, “odd”, “greater than 100” so that the number on the other side does not have this property. Every number and every property is used exactly once. Which number is on the card with the property “greater than 100”?
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Answer: C — 7
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Hint 1 of 3
The property written on a card must be FALSE for the number on that card.
Still stuck? Show hint 2 →
Hint 2 of 3
Pin down the forced pairings first: which number can sit with 'prime'?
Still stuck? Show hint 3 →
Hint 3 of 3
Match the most restrictive properties one at a time and see which number is left for 'greater than 100'.
Show solution
Approach: forced matching
Each card's stated property must be false for its own number; the numbers are 2, 5, 7, 12.
'Prime' must go on the only non-prime, 12; 'odd' must go on an even number, and with 12 used that leaves 2.
'Divisible by 7' must go on a non-multiple of 7, so not 7; only 5 remains for it, leaving 'greater than 100' for the last number 7.
Frank laid out his dominoes as shown in the picture. (Dominoes that touch must always show the same number of points.) Before his brother George removed two dominoes, there were 33 points altogether. How many points is the question mark worth?
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Answer: C — 4
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Hint 1 of 2
Touching halves must match, so equal numbers run along where dominoes meet.
Still stuck? Show hint 2 →
Hint 2 of 2
Use that the whole chain held 33 points to pin down the hidden half.
Show solution
Approach: use matching ends and the total of 33
Where two dominoes touch the pips are equal, which fixes most of the hidden values along the chain.
Adding all the visible pips and using that the full set held 33 points leaves the question-mark half determined.
Carrying out that bookkeeping gives the missing value.
In the picture a number should be written next to each point. The sum of the numbers on the corners of each side of the hexagon should be equal. Two numbers have already been written. Which number should be in the place marked ‘x’?
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Answer: A — 1
Show hints
Hint 1 of 3
Every side of the big hexagon must have the same corner-sum, so use the two known numbers to find what that common sum is.
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Hint 2 of 3
Two sides share a corner; comparing two sides that share a corner forces the opposite corners to be equal.
Still stuck? Show hint 3 →
Hint 3 of 3
Chain equal-sum sides from the 4 and the 1 around to the point marked x.
Show solution
Approach: use that adjacent equal-sum sides force matching opposite corners
Each side of the outer hexagon holds the same total, so two sides meeting at one shared corner must have equal sums of their other corners.
Following that equality around the ring, a corner is forced to match the corner two steps away, propagating the known values.
Tracing from the 4 and the 1 to the marked point this way pins the value at x to 1.
The 17 houses in my street are numbered consecutively, on one side with the odd numbers 1, 3, 5,… and on the other side with the numbers 2, 4, 6,…. My house is the last one on the even side and has the number 12. Yours is the last one on the odd side. Which number does your house have?
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Answer: E — 21
Show hints
Hint 1 of 2
First figure out how many houses are on the even side, given the largest even number is 12.
Still stuck? Show hint 2 →
Hint 2 of 2
The rest are on the odd side; find the last (largest) odd number there.
Show solution
Approach: split the 17 houses across the two sides
My house is the last even number, 12, so the even side has 2,4,6,8,10,12 — that is 6 houses.
The other 17−6 = 11 houses are on the odd side.
The 11th odd number is 21, so your house is number 21.
Felix the Tomcat catches 12 fish in 3 days. On the second day he catches more than on the first. On the third day he catches more than on the second but less than on the first two days together. How many fish does he catch on day three?
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Answer: A — 5
Show hints
Hint 1 of 2
The three numbers add to 12 and increase, with day 3 less than the first two days combined.
Still stuck? Show hint 2 →
Hint 2 of 2
Try small increasing whole numbers that sum to 12 and check the day-3 condition.
Show solution
Approach: use the inequalities to pin the three day totals
Let the catches be d1 < d2 on days 1 and 2, with total 12.
Day 3 is more than day 2 but less than d1+d2.
Testing d1=3, d2=4 gives d3=5: indeed 5>4 and 5<7, and 3+4+5=12.
All the children in a class have at least one pet, and at most two pets. They write down how many pets they have together. Two children each have a dog and a fish. Three each have a cat and a dog. No child has two cats. Altogether they have eight cats, six dogs, and two fish. How many children are in the class?
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Answer: A — 11
Show hints
Hint 1 of 2
Sort the children by which pets they own, using the given groups first.
Still stuck? Show hint 2 →
Hint 2 of 2
Account for all 8 cats, 6 dogs and 2 fish without giving anyone two cats.
Show solution
Approach: tally each pet type
2 children have dog+fish (using both fish and 2 dogs); 3 children have cat+dog (using 3 cats and 3 dogs).
Dogs left: 6 − 5 = 1, so one child has just a dog. Cats left: 8 − 3 = 5, so five children have just a cat (no one has two cats).
The picture shows three dice stacked on top of each other. The sum of the points on opposite sides of a die is 7, as usual. The sum of the points on the faces that touch each other is always 5. How many points are on the face marked X?
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Answer: E — 6
Show hints
Hint 1 of 3
Opposite faces of one die sum to 7, and the two faces where neighbouring dice touch sum to 5.
Still stuck? Show hint 2 →
Hint 2 of 3
Start from the visible front dot on the bottom die and step up the stack one face at a time.
Still stuck? Show hint 3 →
Hint 3 of 3
Use the two rules to relate each touching face to the one above it until you reach X.
Show solution
Approach: step the two face-sum rules up the stack to the top face X
Write the chain from the bottom die upward: opposite faces of a die sum to 7, and each pair of touching faces between two dice sums to 5.
Starting from the orientation fixed by the visible front 1, each touching pair forces the next face, and these alternating 7s and 5s carry the value up the column.
Following the chain to the very top face leaves X = 6, the only value keeping every die face between 1 and 6.
Marie has 9 pearls which weigh, in order, 1 g, 2 g, 3 g, 4 g, 5 g, 6 g, 7 g, 8 g and 9 g. She makes four rings, each with two pearls. The pearls on those rings weigh, in order, 17 g, 13 g, 7 g and 5 g. How much does the pearl which has not been used weigh?
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Answer: C — 3 g
Show hints
Hint 1 of 2
You do not need to figure out the actual pairs — think about totals.
Still stuck? Show hint 2 →
Hint 2 of 2
Subtract the combined ring weight from the weight of all nine pearls.
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Approach: the unused pearl is the total minus the four ring sums
The pearls 1..9 weigh 1+2+...+9 = 45 g in total.
The four rings weigh 17+13+7+5 = 42 g and use 8 pearls.
Anna, Bob, Cleo, Dido, Eva, and Ferdl each roll a die. Each person rolls a different number. Anna’s number is twice as big as Bob’s. Anna’s number is three times as big as Cleo’s. Dido’s number is four times as big as Eva’s. Which number did Ferdl roll?
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Answer: D — 5
Show hints
Hint 1 of 2
Anna's number is both double and triple of someone's — what must it be?
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Hint 2 of 2
Once Anna is fixed, the other multiples force every name, leaving Ferdl's.
Show solution
Approach: pin down Anna first
Anna is 2×Bob and 3×Cleo, so Anna is divisible by 2 and 3 — on a die that means Anna = 6, Bob = 3, Cleo = 2.
Dido = 4×Eva forces Eva = 1, Dido = 4. The five used numbers are 6, 3, 2, 1, 4, so Ferdl rolled 5, answer D.
A quizshow has the following rules: each contestant begins with 10 points. They must answer 10 questions. For each correct answer they get a point and for each incorrect answer they lose a point. Mrs Blandorfer finished the show with 14 points. How many questions had she answered incorrectly?
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Answer: D — 3
Show hints
Hint 1 of 3
First imagine she got every single question right and see how many points that would give.
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Hint 2 of 3
Compare that perfect score with her real score of 14 points.
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Hint 3 of 3
Each wrong answer instead of a right one costs her 2 points: one she does not gain, plus one she loses.
Show solution
Approach: compare with a perfect score
If all 10 answers were right she would start with 10 and add 10, reaching 20 points.
She actually finished with 14, which is 20 − 14 = 6 points short of perfect.
Each wrong answer costs 2 points (she misses the +1 and also loses 1), so 6 ÷ 2 = 3 questions were wrong, answer D.
The managing director of a company claims “Every one of our employees is at least 25 years old.” It turns out, he is wrong. Which of the following statements is correct?
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Answer: D — One of the employees of the company is less than 25 years old.
Show hints
Hint 1 of 2
The boss's 'every employee is at least 25' is false.
Still stuck? Show hint 2 →
Hint 2 of 2
The exact opposite of 'all are at least 25' is just one counterexample.
Show solution
Approach: negate a universal statement
'Every employee is at least 25' being wrong means the claim fails for someone.
The negation of 'all ≥ 25' is 'at least one is below 25'.
On the playground some children measure the length of the playground in strides. Anni takes 15 strides, Betty 17, Denis 12 and Ivo 14. Who has the longest stride?
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Answer: C — Denis
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Hint 1 of 2
They all cross the same length, so fewer strides means each stride is longer.
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Hint 2 of 2
Find who used the fewest strides.
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Approach: fewest strides means the longest stride
The playground length is fixed, so the longest stride belongs to whoever takes the fewest steps.
A certain film lasts 90 minutes. It begins at 17:10. During the film there are two advert breaks, one lasting eight minutes and the other five minutes. At what time will the film end?
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Answer: D — 18:53
Show hints
Hint 1 of 2
Add the film length and both break lengths to the start time.
Still stuck? Show hint 2 →
Hint 2 of 2
Work in minutes from 17:10.
Show solution
Approach: add all the durations to the start time
The film runs 90 minutes and the breaks add 8 + 5 = 13 minutes.
Total time on screen and breaks: 90 + 13 = 103 minutes.
Starting at 17:10, add 103 minutes: 17:10 + 1 h 43 min = 18:53.
In the diagram we want to colour the fields with the colours A, B, C and D so that adjacent fields always have different colours. (Even fields that share only one corner count as adjacent.) Some fields have already been coloured in. In which colour can the grey field be coloured?
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Answer: D — either C or D
Show hints
Hint 1 of 2
Start from the filled 2×2 block and propagate the ‘different even diagonally’ rule outward.
Still stuck? Show hint 2 →
Hint 2 of 2
Track exactly which colours are forbidden for the grey field by its coloured neighbours and their forced neighbours.
Show solution
Approach: propagate the king-adjacency constraints
Filling cells one at a time, each new cell must differ from all eight neighbours (corners included).
Carrying the forced colours across the grid leaves the grey field free to be coloured C or D, but not A or B.
A farmer has 30 cows, some chickens and no other animals. The total number of chicken legs is equal to the total number of cow legs. How many animals does the farmer have?
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Answer: B — 90
Show hints
Hint 1 of 2
Count the cow legs first.
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Hint 2 of 2
Chicken legs match cow legs, so work out how many chickens that is.
Show solution
Approach: match leg counts, then total the animals
30 cows have 30 × 4 = 120 legs.
The chickens have the same number of legs: 120.
Each chicken has 2 legs, so there are 120 ÷ 2 = 60 chickens.
In a group of 2009 kangaroos each one is either light or dark. The smallest of the light kangaroos is bigger than exactly 8 dark kangaroos. One light one is bigger than exactly 9 dark ones, another light one is bigger than exactly 10 dark ones, and so on. Exactly one light kangaroo is bigger than all dark kangaroos. How many light kangaroos are there?
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Answer: B — 1001
Show hints
Hint 1 of 2
Line up the light kangaroos by size and read off how many darks each one beats.
Still stuck? Show hint 2 →
Hint 2 of 2
The counts 8, 9, 10, … rise by one per light kangaroo until one beats every dark.
Show solution
Approach: match each light kangaroo to its ‘beats-this-many-darks’ count
The k-th smallest light kangaroo beats exactly 7 + k dark kangaroos.
The largest light beats all D darks, so 7 + L = D, where L lights and D darks total 2009.
On the island of the truth-tellers and the liars, there are 25 people standing in a line. The person at the front claims that everybody standing behind him is a liar. Everybody else claims that the person standing in front of them is a liar. How many liars are standing in the line? (Truth-tellers always tell the truth and liars always lie.)
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Answer: C — 13
Show hints
Hint 1 of 2
Suppose the front person tells the truth and check whether the chain stays consistent.
Still stuck? Show hint 2 →
Hint 2 of 2
Once the front person must be a liar, the labels alternate down the line.
Show solution
Approach: test the front person, then alternate
If person 1 were truthful, everyone behind would be a liar; but then liar person 3 calling liar person 2 a liar would be true - contradiction. So person 1 lies.
Person 2 truthfully calls person 1 a liar, so person 2 tells the truth; the labels then alternate, putting liars at the odd positions.
Among 25 people the odd positions 1, 3, ..., 25 give 13 liars.
Fabio never tells the truth on Tuesdays, Thursdays and Saturdays, while he always tells the truth on the other days of the week. One day Mateo had the following conversation with Fabio:
Mateo: “What day is today?” Fabio: “Saturday” Mateo: “What day will tomorrow be?” Fabio: “Wednesday”
On which day of the week did the conversation take place?
Show answer
Answer: D — Thursday
Show hints
Hint 1 of 2
On a truth day both of Fabio’s answers are true; on a lying day both are false.
Still stuck? Show hint 2 →
Hint 2 of 2
Test each weekday against his two statements until one fits.
Show solution
Approach: check each day for consistency
Fabio lies on Tue/Thu/Sat and tells the truth otherwise; both his answers share that day’s truth status.
He says ‘today is Saturday’ and ‘tomorrow is Wednesday’.
Only Thursday works: it is a lying day, and both statements are indeed false (it isn’t Saturday, and tomorrow is Friday not Wednesday).
The calendar shows the days of a month, with the columns running Monday, Tuesday, …, Sunday, but the dates are missing. The two dark grey boxes are a Thursday and a Wednesday. If you add the two dates in the dark grey boxes, you get 29. What day of the week is the 1st of the month?
Show answer
Answer: D — Thursday
Show hints
Hint 1 of 3
Count the days from the grey Thursday box down to the grey Wednesday box - it is exactly 13 days later.
Still stuck? Show hint 2 →
Hint 2 of 3
So the two grey dates are 13 apart and add up to 29; find two such numbers.
Still stuck? Show hint 3 →
Hint 3 of 3
Once you know the grey Thursday's date, count back by 7s to reach the 1st.
Show solution
Approach: use the 13-day gap and the sum of 29, then walk back to the 1st
Going down two rows and one box to the left, the grey Wednesday lands 13 days after the grey Thursday.
Take away those extra 13 days from the sum 29, and 16 is left for two equal Thursday dates, so each is 16 ÷ 2 = 8 - the grey Thursday is the 8th.
Counting back a week, the 1st of the month is also a Thursday (8, then 1).
To compare the weights of a red square, a star and a green circle, Mona uses a beam balance (see picture). The lower pan holds the heavier side. Each shape always has the same weight, different shapes have different weights, and every weight is 1, 2, 3, 4 or 5 kg. How many kilograms does the red square weigh?
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Answer: C — 3 kg
Show hints
Hint 1 of 2
The lower pan is heavier: one square outweighs two stars, and two circles outweigh three squares.
Still stuck? Show hint 2 →
Hint 2 of 2
Try weights 1 to 5 for the square: it must be more than two stars yet light enough that two circles beat three of it.
Show solution
Approach: read both balances, then test the square's weight
Left balance: the square pan is lower, so one square is heavier than two stars.
Right balance: the two-circle pan is lower, so two circles are heavier than three squares.
One square beating two stars means the square is at least 3 (two different weights of 1 and 2 already make 3).
If the square is 4, three squares weigh 12, but two circles can be at most 2×5 = 10 - too light. So the square is 3, with stars 1 and circles 5: 3 > 2 and 10 > 9 both hold, giving 3 kg, option C.
Patricia has written a number in each box of a \(7 \times 10\) table. The sum of the numbers in each rectangle of size \(3 \times 4\) or \(4 \times 3\) is zero. Patricia reveals two of the numbers, as shown in the diagram. What is the sum of all the numbers in the table?
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Answer: D — -45
Show hints
Hint 1 of 3
Subtracting two overlapping zero-sum rectangles that differ by one row (or column) shows two far-apart cells must be equal.
Still stuck? Show hint 2 →
Hint 2 of 3
This forces the entries to repeat with a small period, so most of the \(7\times10\) grid can be tiled by zero-sum blocks.
Still stuck? Show hint 3 →
Hint 3 of 3
Whatever cells are left over after tiling are pinned to the two revealed numbers 20 and 25.
Show solution
Approach: tile with zero-sum blocks, then add the leftover cells
Two \(3\times4\) blocks sharing three columns but shifted one column differ only in their end columns, so those columns have equal sums; the same holds vertically, forcing a repeating pattern.
Most of the \(7\times10\) board splits into \(3\times4\)/\(4\times3\) rectangles that each sum to 0, leaving only a few cells whose values equal \(-20\) and \(-25\) by the periodicity.
Adding those leftover cells gives total \(= -(20+25) = \textbf{-45}\), choice (D).
Mike has three bags. Each bag contains three balls. On one bag there is a sign saying “1 white, 2 black”, on the second a sign saying “2 white, 1 black” and on the third a sign saying “3 white”. However, the signs have been swapped so that none of them is correct now. On each turn, Mike chooses a bag that still contains balls, draws one blindly and places it visibly next to the bag. What is the minimum number of balls that he has to draw to know for sure which sign should have been on which bag?
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Answer: C — 2
Show hints
Hint 1 of 3
Since every label is wrong, the labels form a derangement of three—so naming one bag's true contents forces the other two.
Still stuck? Show hint 2 →
Hint 2 of 3
Pick the bag that gives the most information per draw, and ask what a single draw can fail to settle.
Still stuck? Show hint 3 →
Hint 3 of 3
Test whether one draw can ever be ambiguous, then whether two draws always resolve it.
Show solution
Approach: use that the labels are a derangement, then bound the draws
Because no label is correct, the three labels are a derangement of three, so identifying any one bag's true contents determines all three.
Draw from the bag labelled "2 white, 1 black": its real contents are either "1 white, 2 black" or "3 white," so one black ball settles it—but a first white ball is still ambiguous, so one draw is not enough.
A second draw from that same bag always distinguishes the two cases, so 2 draws suffice and are necessary, choice (C).
Every day, penguin Paula catches 12 fish for her two children. Every day she gives one of her children 7 fish and the other 5 fish. After several days, one of her children has received 44 fish. How many fish did the other child receive?
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Answer: D — 52
Show hints
Hint 1 of 2
One child's daily share is sometimes 7 and sometimes 5; over all days the two children together get 12 each day.
Still stuck? Show hint 2 →
Hint 2 of 2
If you can find the number of days, the other child's total is just the grand total minus 44.
Show solution
Approach: find the number of days, then subtract
Over d days the children share 12d fish total.
One child's total 5d + 2·(number of 7-days) = 44 forces d = 8 (with 2 such days).
Total fish = 12 × 8 = 96, so the other child got 96 − 44 = 52.
The number of sweets in the first bowl equals the number of bowls that hold one sweet.
The number of sweets in the second bowl equals the number of bowls that hold two sweets.
The number of sweets in the third bowl equals the number of bowls that hold three sweets.
The number of sweets in the fourth bowl equals the number of bowls that hold no sweets.
How many sweets are in the bowls altogether?
Show answer
Answer: C — 4
Show hints
Hint 1 of 2
Each bowl is counting how many bowls hold a certain number of sweets, so the four numbers must describe themselves.
Still stuck? Show hint 2 →
Hint 2 of 2
Guess that the counts are small and check whether the four sentences all come out true.
Show solution
Approach: find the one self-consistent filling and add it up
Try the bowls holding 2, 1, 0, 1 sweets and check the four statements.
Bowls with one sweet: there are 2 (bowls 2 and 4), matching bowl 1; bowls with two sweets: 1 (bowl 1), matching bowl 2; bowls with three sweets: 0, matching bowl 3; bowls with no sweets: 1 (bowl 3), matching bowl 4.
Every statement checks out, so the total is \(2+1+0+1=4\), answer C.
The picture shows a honeycomb with 16 cells. Some cells (but not all) are filled with honey. The number in a cell tells how many of its neighbouring cells are filled with honey. How many cells of the honeycomb are filled with honey?
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Answer: C — 9
Show hints
Hint 1 of 3
Start at the easiest clues: a cell saying 0 means none of its neighbours have honey, so colour all of them empty.
Still stuck? Show hint 2 →
Hint 2 of 3
A cell whose number equals how many neighbours it has must have every neighbour filled, so fill all of them.
Still stuck? Show hint 3 →
Hint 3 of 3
Keep going back and forth — each thing you mark empty or full gives new clues — until every cell is decided, then count the filled ones.
Show solution
Approach: start from the strongest clues and fill in the rest step by step
A number in a cell just counts its honey-filled neighbours, so a 0 makes all its neighbours empty, and a number as big as the cell's neighbour-count makes them all filled.
Begin with those certain cells, then each cell you settle tells you more about the cells touching it, like a chain of dominoes.
Spreading these forced choices around the comb leaves exactly one pattern that fits every number.
Counting the filled cells in that pattern gives 9.
We have 6 cards and there is one number written on each side of each card. The pairs of numbers on the cards are (5, 12), (3, 11), (0, 16), (7, 8), (4, 14) and (9, 10). The cards can be placed in the empty squares of the calculation \(\square+\square+\square-\square-\square-\square\) in any order with any side up. What is the smallest possible result of the calculation?
Show answer
Answer: D — −26
Show hints
Hint 1 of 3
The calculation adds three cards and subtracts three, and for any card you may show either side.
Still stuck? Show hint 2 →
Hint 2 of 3
A subtracted card should show its big number and an added card its small number, so decide which three cards to subtract.
Still stuck? Show hint 3 →
Hint 3 of 3
Moving a card from the added group to the subtracted group lowers the total by (its small side + its big side), so subtract the three cards with the largest two-number totals.
Show solution
Approach: subtract the three cards with the biggest totals, showing their large sides
Three cards are added and three are subtracted; clearly each added card should show its small number and each subtracted card its big number.
Switching a card from 'added' to 'subtracted' changes the total by \(-(\text{small}+\text{big})\), so the three subtracted cards should be the ones with the largest pair-totals: \((9,10)=19\), \((4,14)=18\), \((5,12)=17\).
Subtracting their big sides gives \(10+14+12=36\); the remaining cards add their small sides \(0+3+7=10\).
Carl tells the truth one day, lies the next, tells the truth again the day after, and so on. On one day he made exactly four of the following five statements. Which statement can he not have made on that day?
Show answer
Answer: A — 2024 is divisible by 11.
Show hints
Hint 1 of 2
First decide whether the day is a truth-day or a lie-day, then every statement he made must match that.
Still stuck? Show hint 2 →
Hint 2 of 2
Two of the statements have a truth value you can settle right away no matter what day it is.
Show solution
Approach: fix whether it is a truth-day or lie-day, then test the fixed statements
Statement A is just a fact: \(2024=11\times184\), so A is always true; statement E ("truth today and truth tomorrow") is impossible because today and tomorrow always have opposite truth-status, so E is always false.
On a truth-day statements B (tomorrow Saturday) and D (yesterday Wednesday) cannot both be true, so at most three true statements are available; he cannot reach four, ruling out a truth-day.
So it is a lie-day, where every statement he made must be false; A is true, so A is the statement he could not have made, answer A.
Captain Flint asks four of his pirates to note down how many gold, silver and bronze coins there were in the treasure chest. Their answers are shown in the diagram. Unfortunately, part of the paper was damaged. Only one of the four pirates told the truth; the other three lied on every single one of their answers. The total number of coins was 30. Who told the truth?
Show answer
Answer: B — Al
Show hints
Hint 1 of 3
The visible numbers are: Tom 9 silver and 11 bronze; Al 7 gold and 12 bronze; Pit 10 gold and 10 bronze; Jim 9 gold and 10 silver.
Still stuck? Show hint 2 →
Hint 2 of 3
Whoever is honest gives the real counts, so every other pirate's visible numbers must each be wrong, and the real gold, silver and bronze add to 30.
Still stuck? Show hint 3 →
Hint 3 of 3
Try each pirate as the honest one and check that the three real counts (totalling 30) make every other pirate wrong on each entry.
Show solution
Approach: assume each pirate honest in turn and force the real counts to total 30
Suppose Al is honest: then gold = 7 and bronze = 12, so silver = 30 − 7 − 12 = 11.
Check the liars against (gold, silver, bronze) = (7, 11, 12): Tom's 9 silver and 11 bronze are both wrong, Pit's 10 gold and 10 bronze are both wrong, and Jim's 9 gold and 10 silver are both wrong.
So every other pirate lied on every answer, exactly as required, and trying any other pirate as honest forces a clash, so this is the only possibility.
A game board is composed of 8 squares on which we want to stack coins. Initially, all squares are empty. On each turn we choose four adjacent squares and place one coin on each of those squares. The numbers show how high the stacks are. Unfortunately, the table wobbled and five of the stacks fell over (shown by ★). How many coins were on the field indicated with a question mark before the stack fell?
★
30
42
★
★
36
?
★
Show answer
Answer: A — 24
Show hints
Hint 1 of 3
Each move adds 1 to a block of four neighbouring squares, so differences between nearby stacks reveal hidden move counts.
Still stuck? Show hint 2 →
Hint 2 of 3
Let \(m_i\) be how many moves start at square \(i\) (covering squares \(i\) to \(i+3\)); write each known stack as a sum of the \(m_i\).
Still stuck? Show hint 3 →
Hint 3 of 3
The \(?\) stack at square 7 equals \(m_4+m_5\), which you can pull out of the known stacks 2, 3 and 6.
Show solution
Approach: express each stack as a sum of move-counts and subtract
Let \(m_i\) be how many moves start at square \(i\), so square \(p\)'s height is the sum of all windows covering it.
Square 2 gives \(m_1+m_2=30\) and square 3 gives \(m_1+m_2+m_3=42\), so \(m_3=12\).
Five clocks are hanging on the wall. One clock is one hour ahead. Another one is one hour late and one is correct. Two clocks have stopped working. Which clock shows the correct time?
Show answer
Answer: D — D
Show hints
Hint 1 of 2
The fast, slow and correct clocks differ by exactly one hour from each other in a chain.
Still stuck? Show hint 2 →
Hint 2 of 2
Find three clocks whose times are consecutive hours; the middle one is correct.
Show solution
Approach: locate the +1, correct and -1 trio of consecutive-hour clocks
The accurate clock, the one an hour fast, and the one an hour slow show three times spaced one hour apart.
Only one set of three clock faces forms such a consecutive-hour chain.
The middle time of that chain is the correct clock, shown in D.
A teacher wants to write the numbers from 1 to 7 into the circles, exactly one number in each circle. When he adds up the two numbers of circles that are next to each other, he gets the number written between those two circles. Which number does he write in the circle with the question mark?
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Answer: D — 4
Show hints
Hint 1 of 2
The number sitting between two circles is just those two circles added together.
Still stuck? Show hint 2 →
Hint 2 of 2
Start with a gap whose two circles you can guess, then fill the circles one at a time like a chain.
Show solution
Approach: treat each in-between number as the sum of its two circles and fill the chain step by step
Each number written between two circles is the sum of the two circles next to it, and the seven circles use 1 through 7 once each.
Begin at a gap where only one pair of circle numbers can add up to it, write those in, then move along filling each next circle so its new gap matches.
When the chain is complete, the circle with the question mark holds 4.
Maria colours exactly 5 cells of this grid in grey. Then she has her 5 friends guess which cells she has coloured in, and their answers are the five patterns A, B, C, D and E. Maria looks at the patterns and says: „One of you is right. The others have each guessed exactly four cells correctly.“ Which pattern did Maria paint? (Choose from pictures A–E.)
Show answer
Answer: E
Show hints
Hint 1 of 2
Each friend's guess is a set of 5 cells; one matches all 5, the others match exactly 4.
Still stuck? Show hint 2 →
Hint 2 of 2
Find the pattern that differs from each of the four near-miss guesses in just one cell.
Show solution
Approach: find the true pattern consistent with one exact and four near-misses
The correct answer shares all 5 cells with one guess and exactly 4 cells with each of the other four.
So Maria's pattern must overlap four of the diagrams in 4 cells and one in all 5.
The numbers from 1 to 11 are written in the empty hexagons. The sums of the three numbers in three hexagons with a common bold point are always equal. Three of the eleven numbers are already written in (see diagram). Which number is written in the hexagon with the question mark?
Show answer
Answer: E — 9
Show hints
Hint 1 of 2
Each bold point ties three hexagons to a common sum.
Still stuck? Show hint 2 →
Hint 2 of 2
Chain those equal-sum conditions across the honeycomb, using the three given numbers.
Show solution
Approach: propagate the equal-triple-sum conditions
Every bold point forces its three surrounding hexagons to share one common sum value.
Linking these conditions with the placed numbers 6, 4 and 11 and the requirement that 1–11 each appear once determines the value at the marked hexagon.
The vertices of a 20-gon are labelled using the numbers 1 to 20 so that adjacent vertices always differ by 1 or 2. The sides of the 20-gon whose vertices are labelled with numbers that only differ by 1 are drawn in red. How many red sides does the 20-gon have?
Show answer
Answer: B — 2
Show hints
Hint 1 of 2
Going around, the labels almost always jump by 2; jumps of 1 are forced only at the extremes.
Still stuck? Show hint 2 →
Hint 2 of 2
Try odds rising then evens falling and count the steps of size 1.
Each animal in the picture stands for a whole number greater than zero, and different animals stand for different numbers. The two animals in each column add up to the number written beneath that column. What is the largest possible value of the sum of the four numbers in the top row?
Show answer
Answer: C — 20
Show hints
Hint 1 of 3
Add the four column sums together: that total equals the top row plus the bottom row.
Still stuck? Show hint 2 →
Hint 2 of 3
The top row is biggest when the bottom row is as small as possible.
Still stuck? Show hint 3 →
Hint 3 of 3
All eight animal numbers must be different, so the bottom row cannot just be all 1s - find the smallest different numbers that still fit.
Show solution
Approach: make the bottom row as small as the all-different rule allows
The four column sums add up to 36, and that 36 splits into the top row plus the bottom row, so a smaller bottom row means a bigger top row.
Every animal stands for a different number, so the four bottom animals must be four different numbers; the smallest four different positive whole numbers are 1, 2, 3, 4 - but each top animal must beat its own bottom partner and also stay different from all the rest.
Choosing bottom values 1, 3, 5, 7 (top partners 2, 4, 6, 8) keeps all eight numbers different and gives the smallest workable bottom row of 16.
Then the top row is 36 − 16 = 20, and no allowed arrangement does better, so the answer is C.
30 people are sitting around a round table. Some of them are wearing a hat. People without a hat must tell the truth; people with a hat may either tell the truth or lie. Each person claims: “At least one of my two neighbours is wearing a hat.” What is the largest number of people who can be without a hat?
Show answer
Answer: D — 20
Show hints
Hint 1 of 2
A truth-teller (no hat) says truly that a neighbour wears a hat, so a no-hat person cannot sit between two no-hat people.
Still stuck? Show hint 2 →
Hint 2 of 2
That means no three no-hat people in a row - fit as many no-hat people as that allows around 30 seats.
Show solution
Approach: forbid three hatless people in a row, then pack the most
A hatless person always tells the truth, so their claim 'a neighbour wears a hat' must be true; thus no hatless person sits between two hatless people.
So at most two hatless people can sit consecutively, in a pattern like (no-hat, no-hat, hat) repeated.
Around 30 seats that gives 2 out of every 3, i.e. 20 hatless people.
In a certain city the inhabitants only communicate by asking questions. There are two kinds of inhabitants: the ‘positive’ that only ask questions that are answered with ‘yes’ and the ‘negative’ that only ask questions that are answered with ‘no’. We meet the inhabitants Albert and Berta, and Berta asks us: “Are Albert and I both negative?” What kind of inhabitants are they?
Show answer
Answer: C — Albert is positive and Berta is negative
Show hints
Hint 1 of 2
A positive only asks questions whose true answer is 'yes'; a negative only 'no'.
Still stuck? Show hint 2 →
Hint 2 of 2
Test the four type-combinations against Berta's question 'Are Albert and I both negative?'
Show solution
Approach: test each type assignment for consistency with Berta asking the question
Berta can only ask her question if its true answer matches her type (yes for positive, no for negative).
Check each combination of Albert and Berta being positive or negative for consistency.
Only 'Albert positive, Berta negative' is consistent, giving answer C.
Twelve weights have integer masses of 1 g, 2 g, 3 g, …, 11 g and 12 g respectively. A vendor divides those weights up into 3 groups of 4 weights each. The total mass of the first group is 41 g, the mass of the second group is 26 g (see diagram). Which of the following weights is in the same group as the weight with 9 g?
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Answer: C — 7 g
Show hints
Hint 1 of 2
The three groups sum to 1+2+...+12 = 78; you know two group totals.
Still stuck? Show hint 2 →
Hint 2 of 2
Find the third group's total, then see which 4-weight set containing 9 g fits a group total.
Show solution
Approach: use group sums to place the 9 g weight
All weights sum to 78; with groups of 41 and 26, the third group totals 78 - 41 - 26 = 11.
The 9 g weight is too heavy for the 11 g group, so it lies in the 41 g or 26 g group.
Working out which four weights total 41 and which total 26, the weight grouped with 9 g among the options is 7 g.
Four villages A, B, C and D lie (not necessarily in this order) along a straight road. A and C are 75 km apart, B and D are 45 km apart, and B and C are 20 km apart. Which of these distances cannot be the distance from A to D?
Show answer
Answer: C — 80 km
Show hints
Hint 1 of 2
Place the points on a line using the known gaps AC = 75, BD = 45, BC = 20; AD depends on the left-right order.
Still stuck? Show hint 2 →
Hint 2 of 2
Try the possible orderings of A, B, C, D consistent with the gaps and list the AD distances that can occur.
Show solution
Approach: place the points with signed positions on the line
Put C at 0. Then B is at +20 or -20 (since BC = 20), A is at +75 or -75 (since AC = 75), and D sits 45 from B.
Trying all the consistent combinations, A to D comes out as 10, 50, 100 or 140 km.
AD = 80 km never appears, so the distance that cannot occur is C.
Mowgli asks a bear and a panther what day of the week it is. The bear always lies on Monday, Tuesday and Wednesday; the panther always lies on Thursday, Friday and Saturday. On every other day both tell the truth. The bear says, “Yesterday was one of my lying days.” The panther says, “Yesterday was also one of my lying days.” On which day did this conversation take place?
Show answer
Answer: A — Thursday
Show hints
Hint 1 of 2
Each animal claims 'yesterday was one of MY lying days'; an animal telling the truth today is honest, while a liar inverts its claim.
Still stuck? Show hint 2 →
Hint 2 of 2
Check each weekday: decide for the bear and panther whether today is truth or lie, then test if both statements can hold.
Show solution
Approach: test each weekday against both animals
The bear lies Mon-Tue-Wed; the panther lies Thu-Fri-Sat; both tell the truth on Sunday.
On Thursday the bear tells the truth and yesterday (Wed) really was its lying day; the panther lies today, and since yesterday (Wed) was not its lying day, its statement is false as required of a liar.
Both statements are consistent only on Thursday, so the answer is A.
Eight teams take part in a football tournament where each team plays each other team exactly once. In each game the winner gets 3 points and the loser no points. In case of a draw both teams get 1 point. In the end all teams together have 61 points. What is the maximum number of points that the team with the most points could have gained?
Show answer
Answer: D — 17
Show hints
Hint 1 of 2
A decided game adds 3 points to the total; a draw adds only 2, so the total tells you the draws.
Still stuck? Show hint 2 →
Hint 2 of 2
Maximise one team's points while keeping the overall total at 61.
Show solution
Approach: count decided games from the total, then load wins onto one team
There are \(\binom{8}{2}=28\) games; a decided game gives out 3 points and a draw gives out 2.
If \(d\) games are drawn then total points \(=3(28-d)+2d=84-d=61\), so \(d=23\) and only \(5\) games are decided.
A team can win only a decided game, so it wins at most all \(5\); giving it those 5 wins plus drawing its other 2 games gives \(5\cdot 3+2=17\) points, answer D.
Stan has five toys: a ball, a set of blocks, a game, a puzzle and a car. He puts each toy on a different shelf of the bookcase (shelf 1 at the bottom up to shelf 5 at the top). The ball is higher than the blocks and lower than the car. The game is directly above the ball. On which shelf can the puzzle not be placed?
Show answer
Answer: C — 3
Show hints
Hint 1 of 3
The ball is the key: many clues are about where the ball sits.
Still stuck? Show hint 2 →
Hint 2 of 3
The game sits right on top of the ball, with the blocks below the ball and the car above it.
Still stuck? Show hint 3 →
Hint 3 of 3
Try each possible shelf for the ball and see where the puzzle is allowed to land.
Show solution
Approach: test the ball's shelf and rule one out
If the ball is on shelf 2: blocks on 1, game on 3, and the car and puzzle take shelves 4 and 5 in some order — puzzle is on 4 or 5.
If the ball is on shelf 3: game on 4, car on 5, and the blocks and puzzle take shelves 1 and 2 — puzzle is on 1 or 2.
(The ball can not be on 1, with nothing below it, or on 4, leaving no room for the car above the game.)
So the puzzle can land on 1, 2, 4 or 5, but never on shelf 3.
Three boys played a “Word” game in which each wrote down 10 words. A boy scored 3 points for a word if neither of the other boys had it, and 1 point if exactly one of the other boys also had it. No points were given for a word all three boys had. When they added up their scores, all three were different. Sam had 19 points, the smallest score, and James had the highest. How many points did James score?
Show answer
Answer: E — 25
Show hints
Hint 1 of 2
Each of a boy's 10 words earns him 3 (his alone), 1 (shared with exactly one other), or 0 (all three have it), so any boy's score is one of \(3u + s\) where \(u + s \le 10\).
Still stuck? Show hint 2 →
Hint 2 of 2
A word shared by two boys gives each of them 1 point, so a shared word adds points symmetrically; track the total points across all three boys and use that Sam is fixed at the minimum 19.
Show solution
Approach: bound the top score, then exhibit a triple that reaches it
Write each boy's score as \(3u + s\): \(u\) words his alone (3 each), \(s\) words shared with exactly one other boy (1 each), and \(u + s \le 10\), so every score is between 0 and 30.
Sam is the minimum at 19, and the three scores are distinct, so James (the max) is more than 19; trying James \(= 25\) means his words are eight unique \((24)\) plus one shared \((1)\), using \(8+1=9\) of his 10 words.
The shared words are also counted for whoever shares them, and a consistent set of 10-word lists can be built giving the three totals \(19,\,21,\,25\) (all different, Sam lowest, James highest), while no arrangement makes James's total exceed 25 once Sam is pinned at 19 and all three differ.
Five kangaroos named A, B, C, D and E have one child each, named a, b, c, d and e, not necessarily in that order. In the first group photo shown exactly 2 of the children are standing next to their mothers. In the second group photo exactly 3 of the children are standing next to their mothers. Whose child is a?
Show answer
Answer: D — \(D\)
Show hints
Hint 1 of 2
Each child stands by a different kangaroo in the two photos, so a child can be correctly placed in at most one photo.
Still stuck? Show hint 2 →
Hint 2 of 2
Since 2 + 3 = 5 children, every child is correct in exactly one photo; sort out which assignment makes both counts work.
Show solution
Approach: match children to mothers by an exactly-one-photo argument
In photo 1 the pairings are A-d, B-a, C-b, D-c, E-e (2 correct); in photo 2 they are A-b, B-e, C-d, D-a, E-c (3 correct).
No child keeps the same partner across photos, so each child is correct in just one photo and all five split as 2 + 3.
The only consistent bijection takes photo-1 pairs b–C, d–A and photo-2 pairs a–D, c–E, e–B, so a's mother is D.
Christina has eight coins whose weights in grams are different positive integers. Whenever she puts any two coins on one side of a balance scale and any two on the other side, the side containing the heaviest of those four coins is always the heavier side. What is the smallest possible weight of the heaviest coin?
Show answer
Answer: C — 34 g
Show hints
Hint 1 of 2
The hardest case is the heaviest coin with the lightest partner versus the next two heaviest.
Still stuck? Show hint 2 →
Hint 2 of 2
That forces each coin to exceed the sum of the previous two minus the lightest — a Fibonacci-style growth from 1, 2.
Show solution
Approach: push the worst pairing to a Fibonacci-type bound
The binding condition is: heaviest + lightest > (2nd heaviest) + (3rd heaviest), with the analogous rule for every coin acting as the maximum.
Taking the lightest as 1 forces each weight to be at least the sum of the two preceding ones.
2021 balls are arranged in a row and numbered from 1 to 2021. Each ball is coloured green, red, yellow or blue. Among any five consecutive balls there is exactly one red, one yellow and one blue ball. After any red ball, the next ball is yellow. The balls numbered 2, 20 and 202 are green. What colour is the ball numbered 2021?
Show answer
Answer: D — blue
Show hints
Hint 1 of 2
Among any 5 in a row there is one each of red, yellow, blue, so two of every five are green — the colouring repeats every 5.
Still stuck? Show hint 2 →
Hint 2 of 2
Use the green positions (2, 20, 202) to find which residues mod 5 are green, then place red-then-yellow.
Show solution
Approach: show period 5, then locate residues by colour
Any five consecutive balls contain one red, one yellow, one blue and thus two greens, which forces a period-5 colouring.
Balls 2, 20, 202 are green, so positions with remainder 2 and remainder 0 (mod 5) are green.
The red, yellow, blue sit at remainders 1, 3, 4; 'red then yellow' forces red at 3, yellow at 4, leaving blue at remainder 1.
Ball 2021 has remainder 1, so it is blue, choice (D).
Five friends decided to spend their vacation together. In a conversation, Adam said, “Yesterday was Wednesday.” Beto said, “Tomorrow will be Friday.” Carlos said, “The day before yesterday was Tuesday.” David said, “The day after tomorrow is Saturday.” Finally, Eli said, “Today is Monday.” One of them was wrong. Who was wrong?
Show answer
Answer: E — Eli
Show hints
Hint 1 of 2
Translate each statement into 'today is ___' and see if they agree.
Still stuck? Show hint 2 →
Hint 2 of 2
Four friends point to the same day; the odd one out is wrong.
Show solution
Approach: convert each clue to today's weekday
Adam (yesterday Wed), Beto (tomorrow Fri), Carlos (day-before-yesterday Tue) and David (day-after-tomorrow Sat) all mean today is Thursday.
Eli says today is Monday, disagreeing with the other four, so Eli was wrong.
Geometry & MeasurementLogic & Word Problemsperimeterspatial-reasoning
In each of the four corners of a swimming pool, 10 m wide by 25 m long, there is a child. The swim instructor is sitting almost in the middle of one of the long edges of the pool. When he calls the children, they all choose the longest path along the edges to reach him. What was the sum of the distances covered by the four children?
Show answer
Answer: E — 210 m
Show hints
Hint 1 of 2
The pool's perimeter is 2×(10+25) = 70 m; the instructor sits near the middle of a long edge.
Still stuck? Show hint 2 →
Hint 2 of 2
Each child walks the longer way round, which is 70 minus the short way; add the four long routes.
Show solution
Approach: use the perimeter and take the long way each time
Perimeter = 2 × (10 + 25) = 70 m, with the instructor about 12.5 m from each end of a long edge.
The two near corners take the long route 70 − 12.5 = 57.5 m each; the two far corners take 70 − 22.5 = 47.5 m each.
Counting & ProbabilityLogic & Word Problemscaseworkcareful-counting
Twelve colored cubes are lined up side by side: three blue, two yellow, three red and four green, but not in that order. There is a red cube at one end and a yellow one at the other end. The red cubes are all together, and the green cubes are all together. The tenth cube from the left is blue. In how many ways can the cubes be lined up?
Show answer
Answer: D — 9
Show hints
Hint 1 of 2
The red block of 3 sits at the red end, the green block of 4 stays together, and position 10 is blue.
Still stuck? Show hint 2 →
Hint 2 of 2
Place the two big blocks and the end yellow first, then count where the loose blues and the other yellow can go.
Show solution
Approach: place the blocks, then count the rest
Red (3 together) occupies the red end and yellow occupies the far end; green stays as a block of 4.
With position 10 forced blue, the green block and the remaining blues and yellow have only a few valid placements.
Counting all consistent arrangements gives 9 ways.
A panel has 4 circles. When Lucy touches a circle, that circle and every circle touching it switch colour (white ↔ black), as shown. Starting with all circles white, at least how many circles must Lucy touch, one after another, to make all four black?
Show answer
Answer: C — 4
Show hints
Hint 1 of 2
Touching a circle flips it and its neighbours; each circle must end flipped an odd number of times.
Still stuck? Show hint 2 →
Hint 2 of 2
Find the smallest set of touches making every circle's flip-count odd.
Show solution
Approach: make every circle flip an odd number of times
Touching a circle toggles it and its neighbours; all four start white and must end black (odd flips each).
One or two touches cannot make all four flip an odd number of times.
The fewest touches achieving this is 4 (touching each circle once works).
Carlos always tells the truth on alternate days. On the other days, he tells only lies. Today he made exactly four of the five statements that follow. Which one was not made today by him?
Show answer
Answer: C — My name is Carlos.
Show hints
Hint 1 of 2
On a truth day every statement he makes is true; on a lie day every one is false.
Still stuck? Show hint 2 →
Hint 2 of 2
Check whether all of A, B and D could be true together — if not, today must be a lie day.
Show solution
Approach: rule out a truth day, forcing a lie day
If today were a truth day, A (prime total) and B (equal counts) force exactly 1 male and 1 female friend, contradicting D (three older male friends).
So today is a lie day: every statement he makes is false.
'My name is Carlos' is true, so on a lie day he cannot make it.
Counting & ProbabilityLogic & Word Problemscareful-countingcasework
Ten people each order an ice cream: four vanilla, three chocolate, two lemon and one mango. As toppings they order four umbrellas, three cherries, two wafers and one chocolate gum — one topping per ice cream. Since no two of the ten ice creams may be exactly alike, which of the following combinations is possible?
Show answer
Answer: E — Lemon and cherry.
Show hints
Hint 1 of 2
Each flavour gets a different topping, with limited supply (4 umbrellas, 3 cherries, 2 wafers, 1 gum).
Still stuck? Show hint 2 →
Hint 2 of 2
A pairing is impossible if it forces two identical ice creams or runs a topping short.
Show solution
Approach: test each pairing against the flavour and topping counts
Ten ice creams (4 vanilla, 3 chocolate, 2 lemon, 1 mango) each get a distinct topping (4 umbrellas, 3 cherries, 2 wafers, 1 gum), no two identical.
Each option claims a specific flavour-topping pair; check it can fit a full valid assignment.
Only option E (lemon with cherry) is consistent with completing the whole table.
Ana wants to write positive whole numbers, one in each of the squares shown, so that the sums of the four numbers in each row and the four numbers in each column are equal. She has already written some numbers, as shown. She wants to write the missing numbers so that the sum of these six numbers is as small as possible. What is this sum?
Show answer
Answer: B — 24
Show hints
Hint 1 of 2
All four row sums and all four column sums must equal one common value S.
Still stuck? Show hint 2 →
Hint 2 of 2
Each already-filled row or column caps how small its blanks can be, so push S down to the smallest value that still leaves every blank a positive whole number.
Show solution
Approach: minimise the common row/column sum, then add the forced blanks
Every row and every column must add to the same total S, so the given numbers in any line bound S from below.
Choosing the smallest S that keeps all six blanks positive whole numbers and fits every line, the blanks are forced.
Adding those six forced entries gives the minimum sum 24, choice B.
The map shows some islands connected by bridges. A navigator wants to visit each island exactly once. He started at Cang Island and wants to finish at Uru Island, and he has just reached the black island in the centre. In which direction must he go now to be able to complete his route?
Show answer
Answer: C — South.
Show hints
Hint 1 of 2
He has to visit every island exactly once, so the move he makes now must not strand any island he still needs to reach.
Still stuck? Show hint 2 →
Hint 2 of 2
If he picks a direction that walks him into a dead-end corner before the rest are visited, he can never get back; only one direction keeps a path open all the way to Uru.
Show solution
Approach: choose the only move that leaves a single-visit path to Uru
He must pass through each island once and finish at Uru, so from the centre he cannot step toward any group of islands he would later be unable to leave.
Going North, East or West leads him into a part of the map he would have to enter or leave twice, leaving some island unvisited.
Heading South is the one move that still lets him reach every remaining island exactly once and end at Uru.
Adam and Bruna try to find out which is Carla's favourite figure, among the figures shown. Carla told Bruna the shape of the figure, and told Adam the colour of the figure. Then this conversation takes place. Adam: “I don't know what Carla's favourite figure is, and I know that Bruna doesn't know either.” Bruna: “At first I didn't know what Carla's favourite figure was, but now I know.” Adam: “Now I know too.” What is Carla's favourite figure?
Show answer
Answer: E
Show hints
Hint 1 of 2
Adam knows the colour, Bruna the shape; each statement eliminates possibilities.
Still stuck? Show hint 2 →
Hint 2 of 2
Adam knowing Bruna cannot know rules out any colour containing a one-of-a-kind shape.
Show solution
Approach: step through the knowledge statements
Adam (knows colour) is sure Bruna cannot know yet, so the colour has no unique-shape figure — this rules out white (the lone hexagon), leaving green and pink.
Bruna (knows shape) now can tell: among green/pink only square, star and triangle are unique, so the figure is green square, green star, or pink triangle.
Adam now knows from the colour, which is unique only for pink, so it is the pink triangle (E).
Peter colours each of the eight circles either red, yellow or blue. Two circles that are directly joined by a line are not allowed to be the same colour. Which two circles must Peter definitely colour the same?
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Answer: A — 5 and 8
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Hint 1 of 2
With only three colours, look for two circles forced into the same colour by their shared neighbours.
Still stuck? Show hint 2 →
Hint 2 of 2
Find two circles that are both adjacent to the same two differently-coloured circles.
Show solution
Approach: forced colour from a 3-colouring constraint
Each circle differs in colour from every circle it is joined to.
Two circles each connected to the same pair of other circles (which take the two remaining colours) are forced to share the one leftover colour.
Tracing the connections, circles 5 and 8 are forced to be the same colour.
Logic & Word Problemscareful-countingspatial-reasoning
A graph consists of 16 points and several connecting lines, as shown in the diagram. An ant is at point A. With every move the ant can move from the point where it currently is, along one of the connecting lines, to an adjacent point. At which of the points P, Q, R, S and T can the ant be after 2019 moves?
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Answer: B — only at P, R, S or T, not at Q
Show hints
Hint 1 of 3
Two-colour the 16 points so every line joins different colours.
Still stuck? Show hint 2 →
Hint 2 of 3
After an odd number of moves the ant must sit on the colour opposite to A's.
Still stuck? Show hint 3 →
Hint 3 of 3
2019 is odd, so the ant ends on the opposite colour class from its start.
Show solution
Approach: bipartite two-colouring and parity
The graph is bipartite: colour the points so each edge joins the two colours.
A starts on one colour; after 2019 (odd) moves it must be on the other colour.
Of P, Q, R, S, T only Q shares A's colour, so the ant can be at P, R, S or T but not at Q.
Logic & Word Problemssum-constraintcaseworkmagic-square
Numbers are placed in the square grid shown so that each of 1, 2, 3, 4 and 5 appears exactly once in every row and in every column. In addition, the sum of all the numbers in each of the three black-bordered sections must be the same. Which number must be written in the top right cell?
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Answer: C — 3
Show hints
Hint 1 of 2
It is a 5×5 Latin square (1–5 once per row and column) split into three black-bordered regions of equal sum.
Still stuck? Show hint 2 →
Hint 2 of 2
All 25 cells sum to 75, so each region sums to 25; combine that with the given 2 and the row/column rules to force the top-right cell.
Show solution
Approach: use the equal region sums of 25 with the Latin-square rules
Each row contains 1–5 and sums to 15, so the whole grid sums to 75.
The three black-bordered regions have equal sums, so each region sums to \(75 \div 3 = 25\).
Tracking the cells in each region together with the placed 2 and the once-per-row/column constraint forces the entries step by step.
Logic & Word ProblemsNumber Theorycaseworkcareful-counting
Paul wants to write a positive whole number onto every tile in the number wall shown, so that every number is equal to the sum of the two numbers on the tiles that are directly below. What is the maximum number of odd numbers he can write on the tiles?
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Answer: B — 14
Show hints
Hint 1 of 2
Work in parity (odd/even): a tile is odd exactly when the two below it differ in parity.
Still stuck? Show hint 2 →
Hint 2 of 2
Search the bottom row patterns of the six-row wall to maximise odd tiles.
Show solution
Approach: reduce to parity and optimise the bottom row of the wall
Only odd/even matters: a tile is odd exactly when the two tiles below it have different parities, so the whole wall is fixed once the bottom row's pattern of odds and evens is chosen.
For the six-row (21-tile) wall, testing the bottom-row patterns, the best choice (such as even, odd, odd, even, odd, odd) makes 14 of the 21 tiles odd.
So the maximum number of odd tiles is 14, choice B.
Logic & Word ProblemsGeometry & Measurementsymmetrycasework
30 dancers are standing in a circle facing the centre. The dance instructor shouts “Left” and many of them turn 90° to the left. Unfortunately, some are confused and turn right, so that some dancers are now directly facing each other. All of the ones that are facing each other are shaking their head. It turns out that 10 dancers shake their head. Then the dance instructor says “Turn around” and all of them turn 180° to look in the opposite direction. Again, all of the ones that are directly facing each other shake their head. How many dancers are shaking their head second time round?
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Answer: A — 10
Show hints
Hint 1 of 2
Two dancers facing each other still form a special pair after both turn 180°.
Still stuck? Show hint 2 →
Hint 2 of 2
Turning everyone around swaps who faces whom, but the count is preserved by symmetry.
Show solution
Approach: track neighbouring pairs facing each other versus back-to-back before and after the turn-around
After turning, each dancer looks clockwise or anticlockwise; a neighbouring pair faces each other when both look toward the gap between them, and is back-to-back when both look away from it.
Going once around the circle, every switch from clockwise-runs to anticlockwise-runs is matched by a switch back, so the number of facing gaps always equals the number of back-to-back gaps.
The first round has 10 head-shakers, i.e. 5 facing gaps, hence also 5 back-to-back gaps; turning everyone 180° reverses all directions, so those 5 back-to-back gaps become the new facing gaps.
That gives 5 facing pairs again, so 10 dancers shake their heads the second time, choice A.
Counting & ProbabilityLogic & Word Problemscaseworkcareful-counting
Three weights are randomly placed on each tray of a beam balance. The balance dips to the right hand side as shown on the picture. The masses of the weights are 101, 102, 103, 104, 105 and 106 grams. For how many percent of the possible distributions is the 106-grams-weight on the right (heavier) side?
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Answer: B — 80 %
Show hints
Hint 1 of 2
List the splits of the six weights into two trays of three with the right side heavier.
Still stuck? Show hint 2 →
Hint 2 of 2
Among those, count how often the heaviest weight (106) is on the right.
Show solution
Approach: enumerate the heavier-right splits and check where 106 sits
Of the C(6,3)=20 ways to fill the right tray, exactly half (10) make the right side heavier.
Counting those, the 106-gram weight is on the right in 8 of the 10 cases.
2017 people live on an island. Each person is either a liar (who always lies) or a nobleman (who always tells the truth). Over a thousand of them attend a banquet where they all sit together around one big round table. Everyone says, “Of my two neighbours, one is a liar and one is a nobleman.” What is the maximum number of noblemen on the island?
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Answer: A — 1683
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Hint 1 of 2
The banquet statement constrains only the people seated at the round table; the others on the island are free.
Still stuck? Show hint 2 →
Hint 2 of 2
Work out the densest valid liar/nobleman pattern around the table, then add the unseated islanders.
Show solution
Approach: maximise noblemen given the round-table constraint plus free islanders
A seated nobleman truly has one liar neighbour, so two noblemen can sit together but never three in a row; a seated liar lies, so its two neighbours match (both noblemen or both liars).
The densest legal seating repeats the block nobleman-nobleman-liar, making at most \(\frac{2}{3}\) of the seated people noblemen, so a table of \(n\) (a multiple of 3) seats up to \(\frac{2n}{3}\) noblemen.
Everyone not at the banquet can be a nobleman, so the island total is \((2017-n) + \frac{2n}{3} = 2017 - \frac{n}{3}\), maximised by the smallest legal \(n\) over a thousand, namely \(n = 1002\).
That gives \(2017 - 334 = 1683\) noblemen, answer A.
We consider a 5×5 square that is split up into 25 fields. Initially all fields are white. In each move it is allowed to change the colour of two fields that are horizontally or vertically adjacent (i.e. white fields turn black and black ones turn white). What is the smallest number of moves needed to obtain the chessboard colouring shown in the diagram?
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Answer: B — 12
Show hints
Hint 1 of 3
Only the grey target cells must change colour (an odd number of times); count them.
Still stuck? Show hint 2 →
Hint 2 of 3
Each move flips exactly two cells, so think about how few moves can deliver an odd flip to every grey cell.
Still stuck? Show hint 3 →
Hint 3 of 3
Try pairing up grey cells, or grey cells with a shared white neighbour, to use each move well.
Show solution
Approach: count the cells to flip, then pair them
In the target, 12 grey cells must each be flipped an odd number of times while the 13 white cells stay flipped an even number of times.
Each move flips exactly two adjacent cells, so to give all 12 grey cells an odd flip you need at least 12 moves (no single move can settle two grey cells, since grey cells are never adjacent).
Twelve moves are enough: for each grey cell, flip it together with one chosen white neighbour, arranging the choices so every white cell is touched an even number of times.
Four sportswomen and sportsmen are sitting around a round table for dinner. They do four different sports: ice skating, skiing, hockey and sledging. The person who skies sits to the left of Sandra. The person who ice skates sits opposite Benjamin. Eva and Philipp sit next to each other. A woman sits next to the person who plays hockey. Which sport does Eva do?
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Answer: A — Ice skating
Show hints
Hint 1 of 3
Note the women are Sandra and Eva; the men are Benjamin and Philipp.
Still stuck? Show hint 2 →
Hint 2 of 3
Seat Eva and Philipp side by side, then use 'ice skater opposite Benjamin' to test who skates.
Still stuck? Show hint 3 →
Hint 3 of 3
Finish with 'a woman sits next to the hockey player' to rule out the wrong seating.
Show solution
Approach: place the two friends, then test the opposite clue
Put Eva and Philipp next to each other; the remaining two seats hold Sandra and Benjamin, opposite Eva and Philipp respectively.
If Benjamin sits opposite Philipp, then the ice skater (opposite Benjamin) is Eva; if Benjamin sits opposite Eva, the skater is Philipp.
In the second seating the hockey player ends up with only men as neighbours, breaking 'a woman sits next to the hockey player', so that seating is impossible.
Only the first seating survives, and there the ice skater opposite Benjamin is Eva, so Eva does ice skating.
On a board there are blue and red rectangles. Exactly 7 of the rectangles are squares. There are 3 more red rectangles than blue squares. There are also two more red squares than blue rectangles. How many blue rectangles are there on the board?
Show answer
Answer: B — 3
Show hints
Hint 1 of 2
Treat squares as a special kind of rectangle and name four counts: blue squares, blue non-square rectangles, red squares, red non-square rectangles.
Still stuck? Show hint 2 →
Hint 2 of 2
Translate each sentence into an equation, then require every count to be a non-negative whole number.
Show solution
Approach: set up four counts and solve with non-negativity
Let TB be the total blue rectangles (squares included) and use: blue squares + red squares = 7; red rectangles = blue squares + 3; red squares = TB + 2.
These give blue squares = 5−TB, blue non-square rectangles = 2·TB−5, red non-square rectangles = 6−2·TB.
All must be ≥ 0, which forces TB = 3, so there are 3 blue rectangles (B).
Logic & Word Problemswork-backwardcareful-counting
The 96 members of a counting club are standing in a circle counting. They start with 1, 2, 3, etc., each person in the circle saying the next number in turn. If a member of the club says an even number, he steps out of the circle. The remaining members continue, starting the second round with 97. They continue in this way until only one member of the club is left. Which number did this person say in round one?
Show answer
Answer: D — 65
Show hints
Hint 1 of 2
Saying an even number = being eliminated, so this is the every-second-person Josephus elimination.
Still stuck? Show hint 2 →
Hint 2 of 2
Use the 2·L+1 rule after writing 96 as a power of two plus a remainder.
Show solution
Approach: Josephus elimination with step 2
Saying an even number removes that person, so persons are knocked out two-at-a-time in the classic every-second elimination.
Writing 96 = 64 + 32, the survivor's position is 2·32 + 1 = 65.
In round one the person in position 65 says the number 65 (D).
Andy fills a \(3\times 3\) table with the digits 1 to 9 so that each cell contains exactly one digit. He has already placed the digits 1, 2, 3 and 4 as shown in the diagram. Two numbers are ‘neighbouring’ when the cells they are in share one side. After finishing the table he noticed that the sum of the numbers neighbouring 9 equals 15. What is the sum of the numbers neighbouring 8?
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Answer: E — 27
Show hints
Hint 1 of 2
The corners already hold 1, 2, 3, 4, so 5, 6, 7, 8, 9 fill the four edge cells and the centre.
Still stuck? Show hint 2 →
Hint 2 of 2
An edge cell touches its two corner digits plus the centre, so test where 9 can sit to make its neighbours add to 15.
Show solution
Approach: place 9 from its neighbour-sum, then total the neighbours of 8
The corners are 1, 3 (top) and 2, 4 (bottom); 5–9 fill the four edge cells and the centre.
An edge cell's neighbours are its two adjacent corners plus the centre. For the neighbours of 9 to total 15, the only fit is 9 on the right edge (3 + 4 + centre = 15), forcing the centre to be 8.
The neighbours of the centre 8 are all four edge cells, which hold 9 and the three remaining digits 5, 6, 7.
A set of scales does not always show the correct mass. If something weighs less than 1000 g it shows the exact mass; when something weighs 1000 g or more it shows some mass over 1000 g. You have 5 balls with masses A g, B g, C g, D g and E g, each less than 1000 g. Weighing them in pairs, the scales show: \(B+D=1200\), \(C+E=2100\), \(B+E=800\), \(B+C=900\), \(A+E=700\). Which ball is the heaviest?
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Answer: D — D
Show hints
Hint 1 of 2
A reading is trustworthy only when the true pair-sum is under 1000g; otherwise it just signals 'over 1000'.
Still stuck? Show hint 2 →
Hint 2 of 2
Pick out the readings below 1000 as exact and see which ball that leaves as the heaviest.
Show solution
Approach: keep only the readings under 1000g as exact
Readings under 1000g are exact; readings of 1200 and 2100 only mean 'the true sum is over 1000g'.
The trustworthy exact sums are B+E = 800, B+C = 900 and A+E = 700, so B, C, E, A are all fairly light.
Since B+D exceeds 1000g while B is small, D must be large; combined with the small reliable sums, D comes out heaviest.
A group of 25 people is made up of knights, rascals and shilly-shalliers. The knights always tell the truth, the rascals are always untruthful, and the shilly-shalliers answer alternately truthfully and falsely (in either order). After the first question to everybody, “Are you a knight?”, 17 answered “Yes!”. After the second question, “Are you a shilly-shallier?”, 12 answered “Yes!”. After the third question, “Are you a rascal?”, 8 answered “Yes!”. How many knights are in this group?
Show answer
Answer: B — 5
Show hints
Hint 1 of 2
Work out how each type answers each question; note that knights and rascals both say 'yes' to 'are you a knight?'.
Still stuck? Show hint 2 →
Hint 2 of 2
Track the shilly-shalliers by their two alternating patterns and turn the three 'yes' counts into equations.
Show solution
Approach: translate each yes-count into an equation
Split shilly-shalliers into those answering true-false-true and those answering false-true-false across the three questions.
Question 3 ('are you a rascal?'): only the false-true-false shillies say yes, so that group has 8 people.
Question 2 ('are you a shilly?'): rascals plus those same shillies say yes: r + 8 = 12, so r = 4.
Question 1 ('are you a knight?'): knights, rascals and those shillies say yes: k + 4 + 8 = 17, so k = 5 knights.
In the forests of a magical island kingdom there are three kinds of animals: lions, wolves and goats. Wolves can eat goats, and lions can eat both wolves and goats. Since it is a magical island kingdom, a wolf that eats a goat changes into a lion, a lion that eats a goat changes into a wolf, and a lion that eats a wolf changes into a goat. To begin with there were 17 goats, 55 wolves and 6 lions on the island. After some time no more eating is possible. What is the maximum number of animals that can still be on the island?
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Answer: D — 23
Show hints
Hint 1 of 2
Every meal removes exactly one animal, so the total only goes down.
Still stuck? Show hint 2 →
Hint 2 of 2
Eating stops only when a single species remains — which species can grow largest?
Show solution
Approach: track the invariant; eating ends with one species
Each eating event removes one animal, so the herd can only shrink, and it stops only when just one kind of animal is left (any two different kinds can still eat).
Following the allowed transformations from 17 goats, 55 wolves, 6 lions, the largest single-species end state reachable is all lions.
Two buttons showing smiling faces and two showing sad faces are in a row, as shown. Pressing a button changes its face and also the faces of its neighbours. What is the least number of button presses needed so that only smiling faces are showing?
Show answer
Answer: B — 3
Show hints
Hint 1 of 3
Pressing a button flips that face and the faces right next to it (sad becomes happy and happy becomes sad).
Still stuck? Show hint 2 →
Hint 2 of 3
Try pressing one button and watch which faces change, then plan the next press from the new picture.
Still stuck? Show hint 3 →
Hint 3 of 3
Check whether 1 or 2 presses can ever work before settling on a small number.
Show solution
Approach: try short sequences and watch the faces flip
Each press flips the button you push and its direct neighbours.
One or two presses always leave at least one sad face showing, so they are not enough.
Pressing the right three buttons in turn flips all the sad faces happy, so the fewest presses is 3, choice B.
2013 people live on an island; some are truth-tellers (who always tell the truth) and the rest are liars (who always lie). Each day one person says ‘When I have left the island, the number of truth-tellers will equal the number of liars,’ and then leaves. After 2013 days no one is left on the island. How many liars were living there to begin with?
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Answer: B — 1006
Show hints
Hint 1 of 3
Remember the rule: whatever a truth-teller says is really true, and whatever a liar says is really false.
Still stuck? Show hint 2 →
Hint 2 of 3
Look at the very first speaker: after he leaves there are 2012 people, and an even number can split into two equal halves.
Still stuck? Show hint 3 →
Hint 3 of 3
Decide whether that first speaker must be a truth-teller or a liar, and what that forces about the rest.
Show solution
Approach: reason about the first speaker, then balance the counts
After the first person leaves, 2012 remain. The claim 'truth-tellers equal liars' would mean \(1006 = 1006\), which is possible, so the first speaker can be a truth-teller telling the truth.
That makes the remaining 1006 truth-tellers and 1006 liars, and the very first speaker an extra truth-teller, giving \(1006 + 1 = 1007\) truth-tellers at the start.
The rest are liars: \(2013 - 1007 = 1006\) liars to begin with, which is choice B.
40 boys and 28 girls hold hands in a big circle. Exactly 18 boys give their right hand to a girl. How many boys give their left hand to a girl?
Show answer
Answer: A — 18
Show hints
Hint 1 of 3
Each boy-and-girl neighbour pair is one boy's hand joined to one girl's hand.
Still stuck? Show hint 2 →
Hint 2 of 3
Walk around the circle one way and mark every spot where a boy is followed by a girl, and every spot where a girl is followed by a boy.
Still stuck? Show hint 3 →
Hint 3 of 3
Since the circle closes up, those two kinds of spots must come in equal numbers.
Show solution
Approach: match boy→girl and girl→boy spots around the circle
Walk around the circle one way. Every place where a boy is followed by a girl is a boy giving that hand to a girl; every place a girl is followed by a boy is a girl giving her hand to a boy.
Going around a loop, the number of boy→girl spots equals the number of girl→boy spots, because the two kinds must alternate.
So just as many boys give their left hand to a girl as give their right hand to a girl: 18.
The runners Kann, Gu and Ru are favourites to win the marathon. Before the race three experts gave their predictions for the outcome of the race.
Expert 1: “Either Kann or Gu will win.” Expert 2: “If Gu is second Ru will win.” Expert 3: “If Gu is third Kan will not win.” Expert 4: “Either Gu or Ru will come second.”
After the race all four predictions were proven correct. In which order, did the three runners finish the race?
Show answer
Answer: D — Gu, Ru, Kan
Show hints
Hint 1 of 2
List the possible finishing orders and test each against all four statements.
Still stuck? Show hint 2 →
Hint 2 of 2
All four predictions are true at once, so eliminate any order that breaks even one.
Show solution
Approach: test each finishing order against all four statements
Try the finishing orders of the three runners; require all four experts correct at once.
The order Gu first, Ru second, Kan third works: Expert 1 (Kann or Gu wins) holds since Gu wins; Expert 4 (Gu or Ru second) holds since Ru is second; Experts 2 and 3 are 'if Gu is 2nd/3rd...' which never happen, so they hold automatically.
A goldsmith has 12 double-links of chain. Out of these he wants to make a single closed chain with 24 links. What is the minimum number of links that he must open (and close again)?
Show answer
Answer: A — 8
Show hints
Hint 1 of 2
Opening both links of one whole double-link frees two connectors.
Still stuck? Show hint 2 →
Hint 2 of 2
How many pieces must you sacrifice so the freed links join all the rest into one loop?
Show solution
Approach: sacrifice whole pieces to use their links as connectors
There are 12 double-links (pieces). Opening both links of one piece gives two open links that can join other pieces, while using up that whole piece.
If you open p whole pieces you get 2p connector links; the remaining 12 − p pieces need 12 − p joins to close into one loop.
Require 2p ≥ 12 − p, i.e. 3p ≥ 12, so p ≥ 4; opening 4 pieces means opening 2 × 4 = 8 links (A).
Of 5 lamps each one can be set to “ON” or “OFF”. Each time when the switch of one lamp is changed, not only does the status of that particular lamp change but also that of one other lamp chosen at random. If the same switch is changed several times not always the same other lamp changes. Initially all lamps are set to “OFF”. Then 10 switching operations are carried out. After that one can say that
Show answer
Answer: C — definitely not all lamps are switched to “ON”;
Show hints
Hint 1 of 2
Each switch flips its own lamp AND one other — so two lamps flip at once.
Still stuck? Show hint 2 →
Hint 2 of 2
Track the parity (even/odd) of how many lamps are ON.
Show solution
Approach: parity of the number of ON lamps
Every operation flips exactly two lamps, so the number of lamps that are ON changes by an even amount each time.
Starting from 0 (all OFF, even), the count of ON lamps stays even after any number of operations.
All five ON would be 5, an odd number, which is impossible — so we can definitely say not all lamps are switched to ON (C).
Each one of the three birds Isaak, Max and Oskar has its own nest. Isaak says: “I am more than twice as far away from Max as I am from Oskar.” Max says: “I am more than twice as far away from Oskar as I am from Isaak.” Oskar says: “I am more than twice as far away from Max as I am from Isaak.” At least two of them speak the truth. Who is lying?
Show answer
Answer: B — Max
Show hints
Hint 1 of 2
The statements form a cycle, so they cannot all be true at once.
Still stuck? Show hint 2 →
Hint 2 of 2
Try making each bird the single liar and see which case lets the other two be truthful.
Show solution
Approach: test which single liar makes at least two statements consistent
The three 'more than twice as far' claims cannot all hold together for any placement of nests.
Assume each bird in turn is the liar and check whether the other two statements can both be true.
Only if Max is lying can Isaak's and Oskar's statements both hold, so Max is lying.
Three big boxes P, Q and R are stored in a warehouse. The upper picture on the right shows their placement seen from above. The boxes are so heavy that they can only be rotated 90° around a vertical edge, as indicated in the pictures below. Now the boxes should be rotated to stand against the wall in a certain order. Which arrangement is possible? (Choice E: all four arrangements are possible.)
Show answer
Answer: B
Show hints
Hint 1 of 3
Each allowed move is a 90° turn about a vertical edge, which both slides a box and rotates its top label a quarter turn.
Still stuck? Show hint 2 →
Hint 2 of 3
So a box's final orientation is tied to how far it travelled: track position and label-rotation together.
Still stuck? Show hint 3 →
Hint 3 of 3
Test each pictured line-up and reject any whose letters point the wrong way for the moves needed to reach it.
Show solution
Approach: couple each box's final position to its forced label orientation
Tipping a box 90° about a vertical edge moves it one step and turns its label a quarter turn, so position and orientation change together — they cannot be set independently.
Walking the boxes to the wall one move at a time, each box's letter must end pointing the way that number of quarter-turns dictates.
Comparing the four pictured line-ups, only one has every letter oriented consistently with the moves that put the boxes there.
Logic & Word Problemswork-backwardcareful-counting
Mark plays a computer game on a 4×4 board. The cells each have a colour which is initially hidden. If he clicks on a cell it turns red or blue. He knows that there are exactly two blue cells and that they share one side. What is the smallest number of clicks with which he can definitely find the blue cells?
Show answer
Answer: B — 10
Show hints
Hint 1 of 2
Think of the two blue cells as a domino placed somewhere on the board.
Still stuck? Show hint 2 →
Hint 2 of 2
You need a clicking pattern that cannot miss the domino in any position — count the minimum needed.
Show solution
Approach: find the fewest clicks that always reveal the two adjacent blue cells
The two blue cells share an edge, so they form a domino somewhere in the 4×4 grid.
A clever set of clicks must hit at least one blue cell no matter where that domino lies, then confirm its partner.
The smallest number of clicks that guarantees finding both is 10.
Logic & Word Problemswork-backwardcareful-counting
To decide who gets the last piece of Leni’s birthday cake, five children use a counting rhyme. Leni, Sara, Hannes, Petra and Arno stand in this order, clockwise in a circle. They count clockwise: KAN – GA – ROO – OUT – ARE – YOU. One child is counted for each syllable, and whoever is counted on YOU is out. They repeat this until only one child is left. Leni may choose who starts the count. Whom must she choose if she wants Arno to get the piece of cake?
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Answer: B — Sara
Show hints
Hint 1 of 2
The rhyme has 6 syllables, so count 6 children around the circle and the 6th one is out.
Still stuck? Show hint 2 →
Hint 2 of 2
Try each possible starting child and act it out around the circle until one child is left — you want that to be Arno.
Show solution
Approach: act it out starting with Sara
Start the count on Sara: counting 6 (Sara, Hannes, Petra, Arno, Leni, Sara) puts Sara out.
Keep going from the next child each time: the next two rounds put Petra out, then Hannes out.
That leaves Leni and Arno, and the final count of 6 puts Leni out, so Arno survives — Leni must choose Sara (answer B).
100 people take part in a race where no one can tie. Everybody is questioned after the race as to which place they have achieved and all answer with a number between 1 and 100. The sum of all answers is 4000. What is the minimum number of people who have lied about their result?
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Answer: D — 12
Show hints
Hint 1 of 2
If everyone told the truth the answers would sum to 1+2+...+100.
Still stuck? Show hint 2 →
Hint 2 of 2
Each liar can only pull the total down by so much; find the most one liar removes.
Show solution
Approach: compare the true sum with the stated sum
Truthful answers sum to 5050, but the stated total is 4000, short by 1050.
A truthful person states their real rank; a liar can drop their claim, removing at most (rank − 1).
The 11 highest ranks lying remove at most 99+98+...+89 = 1034, not enough.
Six-legged, seven-legged and eight-legged octopuses serve Neptune, king of the sea. The seven-legged ones always lie, while the six-legged and eight-legged ones always tell the truth. One day four octopuses meet. The blue one says: “Together we have 28 legs.” The green one says: “Together we have 27 legs.” The yellow one says: “Together we have 26 legs.” The red one says: “Together we have 25 legs.” Which colour octopus is telling the truth?
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Answer: C — green
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Hint 1 of 2
Truth-tellers all know the real total, so if there were two of them they would say the same number.
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Hint 2 of 2
Since all four said different numbers, only one can be telling the truth, so the other three are 7-legged liars.
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Approach: find the total that makes exactly one honest
Only one octopus can be honest (the others would otherwise repeat the same total), so the other three are 7-legged liars with \(3 \times 7 = 21\) legs.
The honest one has 6 or 8 legs; \(21 + 6 = 27\) matches a claim, while \(21 + 8 = 29\) matches nobody, so the real total is 27.
The octopus that truthfully says 27 is the green one — the answer is C.
In Tautostadt there are only nobles and liars. Every sentence spoken by a noble is true, and every sentence spoken by a liar is false. One day some of them meet in a room, and three of them speak as follows:
The first one says: “There are no more than three in this room. We are all liars.”
The second one says: “There are no more than four in this room. We are not all liars.”
The third one says: “In this room we are five. Three of us are liars.”
How many people are in the room, and how many of them are liars?
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Answer: C — four people, two of which are liars
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Hint 1 of 2
Nobles speak only truths and liars only falsehoods — a liar's whole sentence is false.
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Hint 2 of 2
Notice that anyone claiming everyone is a liar cannot be telling the truth; start there.
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Approach: test the speakers' truth values for consistency
The first speaker's claim includes that everyone is a liar, which a noble could not say, so he is a liar and his whole sentence is false.
With four people in the room, the second speaker's claim (at most four, not all liars) is true, so he is a noble, while the third speaker's claim of five people is false, so he is a liar.
That gives a room of four people, two of whom are liars (option C).
In a haunted house the house ghost suddenly disappears. At that moment in time all clocks show 6:15. However, there is also one strange clock in the house that showed the correct time before that event — starting from the disappearance it begins to count backwards. At 19:30 in real time the house ghost reappears. What time does the odd clock show at that moment?
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Answer: A — 17:00
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Hint 1 of 2
The vanishing happens at 6:15 in the evening, that is 18:15.
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Hint 2 of 2
Find how much real time passes, then subtract it because the odd clock runs backwards.
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Approach: run the time backwards by the elapsed amount
When the ghost vanishes the clocks read 6:15 p.m. = 18:15.
The ghost returns at 19:30, so 1 hour 15 minutes of real time pass.
The odd clock counts backwards, so it shows 18:15 − 1:15 = 17:00.
In Funny-Foot-Land men and women wear the same sort of shoes. Each man has a left foot that is two sizes bigger than his right foot. Each woman has a left foot that is one size bigger than her right foot. However, shoes are only sold in pairs of the same size. To save money some friends decide to buy shoes together. After putting on their new shoes, two shoes are left over — one of size 36 and one of size 45. What is the minimum number of people in that group?
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Answer: A — 5
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Hint 1 of 2
Each person uses two different shoe sizes; shoes come only in same-size pairs.
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Hint 2 of 2
Think of going from size 36 up to size 45 in steps of 1 (a woman) or 2 (a man) — how few steps reach 45?
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Approach: link size 36 to size 45 with the fewest people
A man uses sizes that differ by 2; a woman uses sizes that differ by 1.
Because exactly the sizes 36 and 45 are each left with one spare shoe, the people must form a chain of shared sizes from 36 to 45.
From 36 to 45 is a gap of 9; using mostly steps of 2, the fewest people needed is 5.
