The sum of the ages of Tom and Johann is 23. The sum of the ages of Johann and Alex is 24, and the sum of the ages of Alex and Tom is 25. How old is the oldest of them?
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Answer: D — 13
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Hint 1 of 3
Add all three given pair-sums together.
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Hint 2 of 3
The grand total counts every person twice, so half of it is the sum of all three ages.
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Hint 3 of 3
Subtract a known pair from that whole-group total to isolate one person's age.
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Approach: add the equations to get the total, then back out each age
Adding the three pair-sums: 23+24+25 = 72.
Each person is counted twice, so Tom+Johann+Alex = 36.
Alex = 36 - (Tom+Johann) = 36 - 23 = 13; similarly Tom = 12, Johann = 11.
Mike cuts a pizza into four equally big pieces. Then he cuts each piece into three equally big pieces. Into how many equally big pieces did Mike cut the pizza?
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Answer: E — 12
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Hint 1 of 2
Picture the four pieces side by side, then split each one.
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Hint 2 of 2
Each of the four pieces turns into three pieces.
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Approach: count the pieces in equal groups
After the first cuts there are 4 pieces.
Each of those 4 pieces is split into 3, so you get 4 groups of 3.
In a cave there live a starfish, two seahorses and three turtles. They are visited by three starfish, four turtles and five seahorses. How many animals are there now in the cave altogether?
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Answer: E — 18
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Hint 1 of 2
Add up everyone who is now inside the cave: the original animals plus the visitors.
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Hint 2 of 2
Total the residents (1 + 2 + 3) and the visitors (3 + 4 + 5) separately, then combine.
For each sign, count how many mirror lines fold the picture exactly onto itself.
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Hint 2 of 2
A round X-cross folds along four lines, more than a triangle or an arrowed circle.
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Approach: count the mirror lines of each sign
Check each sign for fold lines: the yield triangle (C) has 3, the dead-end sign (E) has 1, and the roundabout arrows (D) and priority sign (B) have none.
The round no-stopping sign (A) is a circle with an X-cross, and an X folds onto itself along 4 lines (two diagonals plus the horizontal and vertical).
A 10 cm long piece of wire is folded so that every part is equally long (see diagram). The wire is then cut through in the two marked positions. How long are the three pieces created in this way?
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Answer: A — 2 cm, 3 cm, 5 cm
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Hint 1 of 3
The wire is folded into equal little segments, so each segment is the same length.
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Hint 2 of 3
Imagine unfolding the wire into one straight 10 cm line and mark where the two cuts land.
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Hint 3 of 3
Count how many equal segments fall in each of the three pieces.
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Approach: unfold the wire and read off the cut positions
Folding 10 cm into equal parts makes a row of equal-length segments.
If you straighten the wire back out, the two marked cuts land on segment boundaries.
Counting the segments in each piece gives lengths of 2 cm, 3 cm and 5 cm (which add back to 10 cm), choice (A).
Ruth takes part in the kangaroo competition where 30 questions have to be answered. She answers every question and each answer is either right or wrong. She has 50% more right than wrong answers. How many of her answers are right?
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Answer: D — 18
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Hint 1 of 2
Let the number of wrong answers be one part.
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Hint 2 of 2
Right is 1.5 parts; right + wrong is 30.
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Approach: parts model
If wrong = 2 parts then right = 3 parts (50% more).
Maria wants to build a bridge across a river. This river has the special feature that from each point along one shore the shortest possible bridge to the other shore always has the same length. Which of the following diagrams is definitely not a sketch of this river?
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Answer: B
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Hint 1 of 3
A constant shortest crossing means the two shores stay a fixed distance apart everywhere.
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Hint 2 of 3
Look for a shore shape where a sharp corner would let you reach the far side by a shorter slanted bridge.
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Hint 3 of 3
At a sharp inside corner of a zig-zag, the nearest point on the far shore is closer than along a straight crossing, so the width can't stay constant.
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Approach: constant-width strip
The condition says the two banks are everywhere the same perpendicular distance apart (a constant-width strip).
Smooth parallel curves can keep a fixed gap.
Banks made of straight segments meeting at sharp angles (the zig-zag) cannot: near an inside vertex the opposite bank is reached by a shorter slanted bridge.
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).
Five points are given in a Cartesian coordinate system: P(−1, 3), Q(0, −4), R(−2, −1), S(1, 1), T(3, −2). Four of these five points are vertices of a square. Which point does not belong there?
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Answer: A — P
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Hint 1 of 2
A square has four equal sides meeting at right angles.
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Hint 2 of 2
Test which four of the five points have all sides equal and perpendicular.
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Approach: check the square
Q(0,−4), R(−2,−1), S(1,1), T(3,−2) give four equal sides (each of squared length 13) with right angles.
Kathi draws a square with side length 10 cm. Then she joins the midpoints of each side to form a smaller square. What is the area of the smaller square?
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Answer: E — 50 cm²
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Hint 1 of 3
Connecting the midpoints leaves a tilted square inside, with a small triangle at each corner.
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Hint 2 of 3
Picture sliding the four corner triangles inward; they exactly fill the tilted square.
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Hint 3 of 3
So the inner square is half of the big square.
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Approach: the midpoint square is half the original
Joining the midpoints cuts off four equal right triangles, one at each corner of the big square.
Those four triangles are the same size as the four triangles that make up the tilted inner square, so the inner square is exactly half of the big square.
The big square has area \(10 \times 10 = 100\), so the inner square is half of that, \(50\text{ cm}^2\), choice (E).
A scatter diagram on the xy-plane gives the picture of a kangaroo as shown on the right. Now the x- and the y-coordinate are swapped for every point. What does the resulting picture look like?
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Answer: A
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Hint 1 of 2
Swapping x and y for every point is a familiar geometric move.
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Hint 2 of 2
It reflects the whole picture across the line y = x (the diagonal).
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Approach: reflection across y = x
Replacing (x,y) by (y,x) reflects each point over the line y = x.
Reflecting the kangaroo across that diagonal gives the picture in A.
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
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Hint 1 of 2
Mark every place where a fork is wrongly on the right or a knife wrongly on the left.
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Hint 2 of 2
One swap fixes one wrong fork together with one wrong knife at the same time.
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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.
Anna has shared her apples fairly between herself and her five girlfriends. Each girl has received half an apple. How many apples did Anna have to start with?
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Answer: B — 3
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Hint 1 of 2
Count how many people get apples, remembering Anna shares with herself too.
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Hint 2 of 2
Six people each get half an apple, so add up six halves.
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Approach: count the equal half-apple shares
Anna and her 5 girlfriends make 6 people in all.
Each gets half an apple, so the total is 6 × ½ = 3 apples.
At Anna’s school 45 teachers come to school by bike, and that is 60% of all the teachers. Only 12% of the teachers come to school by car. How many teachers come to school by car?
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Answer: C — 9
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Hint 1 of 2
First find the total number of teachers from the 60% fact.
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Hint 2 of 2
Then take 12% of that total.
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Approach: find the whole, then a percent of it
45 teachers are 60% of all teachers, so the total is 45 ÷ 0.60 = 75 teachers.
A hen lays white and brown eggs. Lisa takes six of them and puts them in a box as shown. The brown eggs are not allowed to touch each other. What is the largest number of brown eggs Lisa can put in the box?
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Answer: C — 3
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Hint 1 of 2
Two round eggs touch only when their cups are right next to each other (side by side or one above the other).
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Hint 2 of 2
Try putting brown eggs in cups that skip a space, like a checkerboard pattern.
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Approach: spread the brown eggs out so none are next to each other
The box has 6 cups in 2 rows of 3. Eggs touch only when their cups are side by side or one directly above the other.
Put brown eggs in the two top corners and the middle cup of the bottom row — none of these three cups touch.
A fourth brown egg would have to sit next to one of them, so the most Lisa can place is 3.
Lukas invents his own notation for negative numbers. When counting backwards he writes: … 3, 2, 1, 0, 00, 000, 0000, … What is the result of the calculation 000 + 0000 in his notation?
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Answer: C — 000000
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Hint 1 of 2
Match each string of zeros to the negative number it stands for.
Renate combines 555 little piles of 9 stones each into one big pile. Then she splits the big pile into little groups of 5 stones each. How many such groups does she get?
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Answer: A — 999
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Hint 1 of 2
Find the total number of stones first.
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Hint 2 of 2
Then split that total into groups of 5.
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Approach: total then divide
The big pile has 555 × 9 = 4995 stones.
Splitting into groups of 5 gives 4995 ÷ 5 = 999 groups.
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?
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Answer: C — 3
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Hint 1 of 2
Equal triangle-sums force many circles to share a value.
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Hint 2 of 2
Trace which circles must be equal; only a few can stay distinct.
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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.
Four girls are sleeping in a room with their heads on the grey pillows. Bea and Pia are sleeping on the left-hand side of the room with their faces towards each other; Mary and Karen are on the right-hand side with their backs towards each other. How many girls sleep with their right ear on the pillow?
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Answer: C — 2
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Hint 1 of 3
For each girl, picture which cheek is pressed into the pillow and so which ear is underneath.
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Hint 2 of 3
When two girls lie side by side facing opposite ways, they rest on opposite ears.
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Hint 3 of 3
So in a face-to-face pair (and in a back-to-back pair) exactly one girl is on her right ear.
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Approach: pair up the girls and use mirror directions
Bea and Pia lie facing each other: since they point opposite ways, one rests on her left ear and the other on her right ear, so that pair gives 1 right-ear girl.
Mary and Karen lie back to back, again pointing opposite ways, so that pair also gives exactly 1 right-ear girl.
Adding the two pairs, \(1 + 1 = 2\) girls sleep on their right ear, choice (C).
Grandma stands in the courtyard calling her cat and all her chickens. After a little while, 20 legs come running towards her. How many chickens does Grandma have?
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Answer: C — 8
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Hint 1 of 2
The cat has 4 legs; every chicken has 2.
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Hint 2 of 2
Take away the cat's legs first, then split the rest into pairs.
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Approach: remove the cat's legs, then divide by two
The cat accounts for 4 of the 20 legs, leaving 20 − 4 = 16 chicken legs.
Each chicken has 2 legs, so there are 16 ÷ 2 = 8 chickens.
I have some unusual dice. On their faces are the digits 1 to 6 as usual, however the odd numbers are negative (so −1, −3, −5 instead of 1, 3, 5). I throw two such dice at the same time. Which of the following sums can I definitely not achieve with one such throw?
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Answer: D — 7
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Hint 1 of 2
List the six face values: −1, 2, −3, 4, −5, 6.
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Hint 2 of 2
Try to write each target as a sum of two of those values.
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Approach: reachable sums
Each die shows one of −5, −3, −1, 2, 4, 6.
Sums 3, 4, 5, 8 are all possible (e.g. −1+4, 2+2, −1+6, 2+6).
The given net is folded along the dotted lines to form an open box. The box is placed on the table so that the opening is on top. Which side is facing the table?
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Answer: B — B
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Hint 1 of 2
Fold the net up in your head into an open box (one face missing for the opening).
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Hint 2 of 2
With the opening on top, the face opposite the opening is the one on the table.
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Approach: fold the net up and find the bottom face
Fold the net up so the four sides stand and one face is missing; that missing face is the opening on top.
The face that lies flat at the bottom, opposite the opening, is the one touching the table, which is B, choice (B).
If you add up the digits of the year 2016 (2 + 0 + 1 + 6), the result is 9. What is the next year after 2016 for which the sum of the digits is 9 again?
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Answer: B — 2025
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Hint 1 of 2
The next year must be after 2016 and have digits adding to 9.
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Hint 2 of 2
Try years just after 2016 and add their digits.
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Approach: check years after 2016 for digit sum 9
Add the digits of each year right after 2016: 2017 gives 10, 2018 gives 11, and they keep climbing, so none of 2017–2024 lands back on 9.
A house has 12 rooms. Each room has two windows and one light. Only when the light is on in a room are both of its windows lit. Yesterday evening, 18 windows were lit. In how many of the rooms was the light off?
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Answer: B — 3
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Hint 1 of 2
A room shows light in both its windows only when its light is on.
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Hint 2 of 2
Find how many rooms were lit, then compare with the 12 rooms total.
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Approach: windows come in pairs per lit room
Each lit room shows 2 illuminated windows, so 18 windows means 18 ÷ 2 = 9 rooms were lit.
There are 12 rooms in all, so 12 − 9 = 3 rooms had the light off.
Step by step the word VELO is changed into the word LOVE. In every step two adjacent letters are allowed to be swapped around. What is the minimum amount of steps needed?
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Answer: B — 4
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Hint 1 of 2
Number the target positions and look at how out of order the start is.
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Hint 2 of 2
The fewest adjacent swaps equals the number of inversions.
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Approach: count inversions
Target LOVE means positions L,O,V,E; the start VELO reads as 3,4,1,2 in that order.
The out-of-order pairs are (3,1),(3,2),(4,1),(4,2) — four of them.
Alex has one rope 1 m long and another 2 m long. He cuts up both ropes so that all the pieces are of equal length. Which of the following numbers of pieces can he not obtain this way?
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Answer: B — 8
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Hint 1 of 2
Every piece must fit a whole number of times into the 1 m rope and into the 2 m rope.
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Hint 2 of 2
If each piece is 1/n of a metre, the short rope gives n pieces and the long rope gives 2n, so the total is always a multiple of 3.
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Approach: the total number of pieces is always a multiple of 3
Equal pieces must divide both ropes exactly, so each piece is 1/n of a metre for some whole number n.
Then the 1 m rope makes n pieces and the 2 m rope makes 2n pieces, for 3n pieces in all — always a multiple of 3.
Among the options 6, 9, 12 and 15 are multiples of 3, but 8 is not, so 8 pieces cannot be obtained.
A mouse wants to escape a labyrinth (see picture). On her way out she is only allowed to go through each opening once at most. How many different ways can the mouse choose to get outside?
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Answer: D — 6
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Hint 1 of 2
Trace each route from the mouse to the outside, never reusing an opening.
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Hint 2 of 2
Count the separate paths one by one and don't repeat any.
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Approach: trace and count the distinct escape routes
Start at the mouse and follow the openings outward, using each opening at most once.
List every different route that reaches the outside.
Sven writes five different single-digit positive whole numbers on a board. He realises that no sum of two of these numbers is equal to 10. Which of the following numbers has Sven definitely written on the board?
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Answer: E — 5
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Hint 1 of 2
Group the digits 1–9 into pairs that add to 10.
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Hint 2 of 2
You can use at most one number from each pair; count how many that allows.
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Approach: pigeonhole on pairs
The pairs summing to 10 are {1,9},{2,8},{3,7},{4,6}; the digit 5 has no partner.
From four pairs you may pick only one each — at most 4 numbers.
To reach five numbers you must also use 5, so 5 is definitely written.
During a cycle race starting at D and finishing at B, every connecting road between the towns A, B, C and D shown in the diagram is ridden along exactly once. How many possible routes are there for the race?
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Answer: C — 6
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Hint 1 of 2
A valid race rides every drawn road exactly once, starting at D and ending at B (an Euler trail).
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Hint 2 of 2
List the routes by the first road taken out of D, then trace each to the end at B.
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Approach: count Euler trails from D to B
The five roads are A-B, A-D, B-D, B-C and D-C; the race must use each exactly once, leaving D and arriving at B.
Organise by the first move from D: starting D-A leads to 2 finishing routes, starting D-B leads to 2, and starting D-C leads to 2.
The diagram shows a circle with centre O as well as a tangent that touches the circle at point P. The arc AP has length 20 and the arc BP has length 16. What is the size of the angle ∠AXP?
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Answer: E — 10°
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Hint 1 of 3
Notice A, O, B are collinear, so AB is a diameter and the arc from A through P to B is a semicircle.
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Hint 2 of 3
The line through X is tangent at P and a secant cutting A and B, so its angle equals half the difference of the two intercepted arcs.
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Hint 3 of 3
Turn the arc lengths into degrees first, using that the two arcs add to 180.
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Approach: convert arc lengths to degrees, then use the tangent-secant angle
Since A, O, B lie on one line, AB is a diameter, so arc AP + arc PB is a semicircle: \(20 + 16 = 36\) units of length equal \(180^\circ\), i.e. 1 unit is \(5^\circ\).
Thus arc AP \(= 100^\circ\) and arc PB \(= 80^\circ\).
The tangent at P and the secant through B and A meet at X, so \(\angle AXP = \tfrac12(\text{arc }AP - \text{arc }PB) = \tfrac12(100^\circ - 80^\circ) = 10^\circ\).
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?
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Answer: C — 3
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Hint 1 of 2
Each weekday exactly two of the three work, so add up all the work-days.
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Hint 2 of 2
Mona's 3 plus Asma's 4 plus Nadja's count must equal the total work-days.
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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?
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Answer: C — 1234LLuuaapp4321
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Hint 1 of 2
Check two rules for each option: the last three characters are digits, and there are at most three capital letters.
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Hint 2 of 2
Eliminate any password that breaks either rule.
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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.
Gerda walks along the road and writes down the letters she can see on her right-hand side. Which word is formed while Gerda walks from point 1 to point 2?
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Answer: A — KNAO
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Hint 1 of 2
Walk from point 1 toward point 2 and only write a letter when it is on Gerda's right.
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Hint 2 of 2
Read off the right-hand letters in the order she passes them.
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Approach: record only the right-side letters along the route
Walk Gerda's path from point 1 to point 2 and look at each sign — write it down only when it is on her right-hand side.
In the order she passes them, the right-side letters are K, N, A, O.
Inside the square ABCD there are four identical rectangles (see diagram). The perimeter of each rectangle is 16 cm. What is the perimeter of the square?
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Answer: E — 32 cm
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Hint 1 of 2
Let a rectangle have long side a and short side b; the square's side equals a + b.
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Hint 2 of 2
Each rectangle's perimeter 2(a+b) = 16 gives a + b directly.
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Approach: relate the square side to a rectangle half-perimeter
From the arrangement, the square's side equals one long side plus one short side, a + b.
Each rectangle has perimeter 2(a + b) = 16 cm, so a + b = 8 cm, which is the square's side.
a, b, c, d are positive whole numbers for which \(a + 2 = b - 2 = c \times 2 = d \div 2\) holds true. Which of the four numbers a, b, c and d is biggest?
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Answer: D — d
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Hint 1 of 2
Set the common value equal to k and write each of a, b, c, d in terms of k.
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Hint 2 of 2
Compare a = k-2, b = k+2, c = k/2, d = 2k.
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Approach: express all four through the common value
Let a+2 = b-2 = 2c = d/2 = k.
Then a = k-2, b = k+2, c = k/2, d = 2k.
For positive whole numbers 2k exceeds the others, so d is biggest.
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?
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Answer: C — C
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Hint 1 of 2
Each squirrel runs to its nearest nut first; mark which nut each one claims.
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Hint 2 of 2
After the first round, whichever squirrel is closest to a still-uncollected nut grabs the next one.
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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).
A 3×3 field is made up of 9 unit squares. In two of these squares, circles are inscribed as shown in the diagram. How big is the shortest distance between these circles?
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Answer: A — 2√2 − 1
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Hint 1 of 2
The two circles sit in opposite corner squares; find the distance between their centres first.
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Hint 2 of 2
Subtract one radius from each circle from the centre-to-centre distance.
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Approach: distance of centres minus radii
Each circle has radius 1/2 and its centre at the middle of a corner unit square.
The centres are 2 right and 2 up apart, a distance of 2√2.
Removing the two radii gives the gap 2√2 − 1, so 2√2 − 1.
In this number pyramid each number in a higher cell is equal to the product of the two numbers in the cells immediately underneath it. Which of the following numbers cannot appear in the topmost cell, if the cells on the bottom row hold only natural numbers greater than 1?
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Answer: D — 105
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Hint 1 of 2
Write the top cell as a product of the three bottom entries.
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Hint 2 of 2
The top equals a*b^2*c, so it must contain a perfect-square factor bigger than 1.
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Approach: express the apex as a*b^2*c
With bottom cells a, b, c, the middle cells are ab and bc, and the top is ab*bc = a*b^2*c.
So the top number must be divisible by some square b^2 with b greater than 1.
Among the options, 105 = 3*5*7 is square-free, so it cannot appear: answer 105 (D).
There are 30 girls and boys in a class. Two students always share a desk. Every boy shares a desk with a girl. Exactly half the girls share a desk with a boy. How many boys are in the class?
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Answer: D — 10
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Hint 1 of 3
A boy-girl desk has one boy and one girl, so counting those desks counts the boys.
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Hint 2 of 3
Those same boy-girl desks use up exactly half the girls, so there are as many boys as half-the-girls.
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Hint 3 of 3
That means there are twice as many girls as boys.
Show solution
Approach: match boys to half the girls
Every boy sits at a boy-girl desk, so the number of boy-girl desks equals the number of boys.
Those desks contain exactly half the girls, so the number of boys equals half the girls; in other words there are twice as many girls as boys.
Split 30 into 1 part boys and 2 parts girls: 3 equal parts make 30, so each part is 10, and the boys are 1 part, giving 10 boys, choice (D).
With algebraIf \(b\) boys and \(g\) girls, then \(b+g=30\) and \(g=2b\), so \(3b=30\) and \(b=10\).
Five children each have a black square, a grey triangle and a white circle made of paper. The children place the three shapes on top of each other as shown in the pictures. In how many pictures was the triangle placed after the square?
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Answer: D — 3
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Hint 1 of 2
If the triangle was placed after the square, the triangle must cover the square.
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Hint 2 of 2
Check each picture: is the square hidden by the triangle, or does it show on top?
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Approach: decide the stacking order from what covers what
The triangle is placed after the square when the triangle sits on top of (hides part of) the square.
Go through the five pictures and mark the ones where the triangle covers the square.
Three of the pictures show the triangle on top of the square.
Five sparrows are sitting on a rope (see picture). Some of them are looking to the left, some of them are looking to the right. Every sparrow whistles as many times as the number of sparrows it can see sitting in front of it. For example, the third sparrow whistles exactly twice. How many times do all the sparrows whistle altogether?
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Answer: D — 10
Show hints
Hint 1 of 2
Each sparrow only sees the birds in the direction its beak is pointing.
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Hint 2 of 2
Count how many birds are in front of each sparrow, then add all five counts.
Show solution
Approach: count each sparrow's forward view and add them up
Look at which way each beak points, then count the sparrows in front of it.
From left to right the sparrows look right, left, right, left, right, so they see 4, 1, 2, 3, and 0 birds in front.
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?
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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.
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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.
Hansi writes the number 2581953764 on a strip of paper. Twice he cuts through the strip between two digits, getting three numbers which he adds. What is the smallest sum he can obtain in this way?
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Answer: B — 2975
Show hints
Hint 1 of 3
A long piece is worth a lot (thousands or more), so very long pieces make the sum big.
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Hint 2 of 3
The ten digits split into three pieces whose lengths add to 10, so keep every piece short — at most four digits.
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Hint 3 of 3
Among the short splits, choose the one whose biggest piece has the smallest leading digits.
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Approach: keep the pieces short, then shrink the leading digits
Two cuts make three pieces, and a piece with 5 or more digits already passes every answer, so each piece should have at most 4 digits.
That means the lengths are 3, 4, 3 in some order; the 4-digit piece dominates the sum, so we want it to start with the smallest digits.
Cutting as \(258 + 1953 + 764\) makes the 4-digit piece start with 1, and the total is \(258 + 1953 + 764 = 2975\).
No split beats this, so the smallest sum is 2975, choice (B).
Konrad dries mushrooms. From 4 kg of fresh mushrooms he gets 1 kg of dried mushrooms. How many kilograms of mushrooms does he have to pick in order to receive 4 kg of dried mushrooms?
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Answer: B — 16 kg
Show hints
Hint 1 of 2
Every 4 kg of fresh mushrooms shrinks to 1 kg dried.
Still stuck? Show hint 2 →
Hint 2 of 2
To get 4 kg dried, scale that 4-to-1 relationship up four times.
Show solution
Approach: scale the fresh-to-dried ratio
4 kg fresh gives 1 kg dried.
For 4 kg dried, he needs 4 times as much fresh: 4 × 4 kg = 16 kg.
What percentage of the area of the triangle is coloured in grey in the adjacent diagram?
Show answer
Answer: C — 88%
Show hints
Hint 1 of 3
Each side of the big triangle is split into 1 + 3 + 1 = 5 equal parts.
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Hint 2 of 3
Each small white corner triangle has the same angles as the big one, so it is a scaled copy.
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Hint 3 of 3
A shape scaled by a length factor of 1/5 has area (1/5)² of the original.
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Approach: the white corners are scale-1/5 copies
Every side of the big triangle reads 1 + 3 + 1 = 5, so each side has length 5 in these units.
Each white corner triangle shares the big triangle's angles and has its two cut-off sides equal to 1, so it is the big triangle shrunk by a factor of 1/5.
Its area is therefore (1/5)² = 1/25 = 4% of the whole, and the three corners remove 3×4% = 12%.
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?
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Answer: B — Glen and Carl
Show hints
Hint 1 of 2
Use the listed results to rebuild the knockout bracket.
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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.
In rectangle ABCD the side BC is exactly half as long as the diagonal AC. Let X be the point on CD for which |AX| = |XC| holds true. How big is the angle ∠CAX?
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Answer: E — another angle
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Hint 1 of 2
Since BC is half the diagonal AC, find the angles the diagonal makes first.
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Hint 2 of 2
Then use the isosceles condition AX = XC; the resulting angle is not one of the first four options.
Show solution
Approach: angle chase, then check against the listed values
BC = AC/2 makes the diagonal split the corner so that angle BAC = 30 degrees and angle ACD = 30 degrees.
With X on CD and AX = XC, triangle AXC is isosceles, giving angle CAX = 30 degrees.
That value is not 12.5, 15, 27.5 or 42.5 degrees, so the answer is another angle (E).
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?
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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.
Jilly makes up a multiplication magic square using the numbers 1, 2, 4, 5, 10, 20, 25, 50 and 100. The products of the numbers in each row, column and diagonal should be equal. In the diagram it can be seen how she has started. Which number goes into the cell with the question mark?
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Answer: B — 4
Show hints
Hint 1 of 3
Multiply all nine numbers to find the constant product, then take its cube root for the centre.
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Hint 2 of 3
Use the constant product along a row, column and the diagonal to fill the cells.
Still stuck? Show hint 3 →
Hint 3 of 3
The product of all nine values is 10^9, so each line multiplies to 1000.
Show solution
Approach: multiplicative magic square
All nine numbers multiply to 109, so each line's product is the cube root, 1000, and the centre is 10.
Top row 20×1×? = 1000 gives 50; the diagonal 20×10×? = 1000 gives 5 at the bottom-right.
The right column 50×?×5 = 1000 forces the marked cell to be 4.
Diana cuts a rectangle of area 2016 into 56 identical squares. The side lengths of the rectangle and the squares are all whole numbers. For how many different rectangles can she do this? (Two rectangles are said to be different if they are not congruent.)
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Answer: B — 4
Show hints
Hint 1 of 2
Each square has the same area, so find that area first.
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Hint 2 of 2
2016/56 = 36, so squares are 6x6; count the rectangle shapes from the factor pairs of 56.
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Approach: fix the square size, count factor pairs
Each square has area 2016/56 = 36, so side 6 (a whole number, good).
The 56 squares form an m-by-n grid with m*n = 56.
Non-congruent factor pairs: (1,56),(2,28),(4,14),(7,8), so 4 rectangles.
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.
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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.
Jack wants to keep six tubes each of diameter 2 cm together using a rubber band. He chooses between the two possible variations shown. How are the lengths of the rubber bands related to each other?
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Answer: E — Both bands are equally long.
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Hint 1 of 2
Each band is straight segments plus curved arcs that wrap the outer tubes.
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Hint 2 of 2
The arcs always join to one full circle; compare only the straight parts.
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Approach: straight parts plus one circle
In both arrangements the curved pieces add up to exactly one full circle (2π).
The straight pieces trace the outline of the centres; both layouts give the same total straight length of 12.
Tim, Tom and Jim are triplets. Their twin brothers John and James are 3 years younger. All five have their birthday today. Which of the following numbers could be the sum of the ages of the five brothers?
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Answer: B — 89
Show hints
Hint 1 of 2
Let the triplets be age t; the twins are t − 3 each. Write the total.
Still stuck? Show hint 2 →
Hint 2 of 2
The sum is 5t − 6, so the answer plus 6 must be a multiple of 5.
Show solution
Approach: express the sum and test the form
Three triplets are each age t and two twins are each t − 3.
Total = 3t + 2(t − 3) = 5t − 6, so (sum + 6) must be a multiple of 5.
89 + 6 = 95 = 5 × 19 works (t = 19); the other options fail, so the sum can be 89.
The square shown in the diagram has a perimeter of 4. The perimeter of the equilateral triangle is
Show answer
Answer: B — \(3 + \sqrt{3}\)
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Hint 1 of 2
The square has side 1; work out how far the triangle's side must reach past the square.
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Hint 2 of 2
The triangle's base extends beyond the square's foot by a segment set by the \(60^\circ\) slope, which has length \(\tfrac{1}{\sqrt3}\).
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Approach: read the triangle's side from the unit square
The square has perimeter 4, so its side is 1.
Where a slanted \(60^\circ\) side rises a height of 1 (the square's height), it runs sideways by \(\tfrac{1}{\sqrt3}\), the extra base length beyond the square.
This makes the triangle's side \(1 + \tfrac{1}{\sqrt3}\), so its perimeter is \(3\left(1 + \tfrac{1}{\sqrt3}\right) = 3 + \sqrt3\).
Tim, Tom and Jim are triplets. Their brother Carl is exactly 3 years younger. All four are having their birthdays today. How old can the four brothers be altogether?
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Answer: A — 53
Show hints
Hint 1 of 3
The three triplets are all the same age, and Carl is 3 years younger than that age.
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Hint 2 of 3
Imagine giving Carl 3 extra birthdays so all four are the same age: then the total would be 4 equal ages.
Still stuck? Show hint 3 →
Hint 3 of 3
So the real total is 3 less than some multiple of 4.
Show solution
Approach: make all four ages equal, then adjust by 3
If Carl were the same age as the triplets, all four ages would be equal, so the total would be 4 times one age, a multiple of 4.
Carl is actually 3 years younger, so the real total is 3 less than a multiple of 4.
Checking the choices, only \(53 + 3 = 56 = 4 \times 14\) is a multiple of 4, so the total is 53, choice (A).
With algebraIf each triplet is \(t\), Carl is \(t-3\) and the total is \(3t+(t-3)=4t-3\); setting \(4t-3=53\) gives \(t=14\).
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?
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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.
A 3 cm wide strip of paper is dark on one side and light on the other. The folded strip lies exactly inside a rectangle 27 cm long and 9 cm wide (see diagram). How long is the strip of paper?
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Answer: D — 57 cm
Show hints
Hint 1 of 3
The strip is 3 cm wide and fills a 27 cm by 9 cm rectangle, which is three strip-widths tall.
Still stuck? Show hint 2 →
Hint 2 of 3
Unroll the zig-zag: each slanted fold spans the full 9 cm height, so it is longer than a flat 3 cm-wide piece would be.
Still stuck? Show hint 3 →
Hint 3 of 3
Add the flat horizontal runs to the longer slanted fold pieces.
Show solution
Approach: unroll the folded strip and add the pieces
The 3 cm wide strip lies inside the 27 by 9 rectangle (three strip-widths tall), folding up and down as a zig-zag.
Unrolling it, the flat horizontal stretches plus the slanted fold pieces (each crossing the full 9 cm height) recombine into one straight strip.
Summing the straight runs and the longer slanted pieces gives a total length of 57 cm.
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?
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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.
Richard writes down all numbers with the following properties: the first digit is 1; each of the following digits is at least as big as the previous one; the sum of the digits is 5. How many such numbers can Richard write down?
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Answer: B — 5
Show hints
Hint 1 of 3
The first digit is 1, so the digits after it must add up to the remaining 4.
Still stuck? Show hint 2 →
Hint 2 of 3
Those later digits can never go down, so list them from smallest going up.
Still stuck? Show hint 3 →
Hint 3 of 3
Just write out every non-decreasing way to make 4 (using digits at least 1).
Show solution
Approach: list every non-decreasing way the leftover digits add to 4
The number starts with 1, so the digits that follow are non-decreasing and add up to \(5 - 1 = 4\).
Listing the ways to make 4 without ever decreasing: 4, then 1+3, then 2+2, then 1+1+2, then 1+1+1+1.
These give the numbers 14, 113, 122, 1112, 11111 — that is 5 numbers, choice (B).
Tick, Trick and Track are triplets. Their brother Franz is exactly 3 years older. All four children are having their birthdays today. How old can the four brothers be altogether?
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Answer: B — 27
Show hints
Hint 1 of 2
The three triplets are all the same age, and Franz is just 3 more than that.
Still stuck? Show hint 2 →
Hint 2 of 2
If you take Franz's extra 3 years off the total, what is left should split into four equal ages.
Show solution
Approach: take off Franz's extra 3, then split the rest into four equal ages
All four boys would be the same age except Franz, who has 3 extra years, so take those 3 away from the total first.
What is left must split evenly into four equal ages (one for each boy).
Take 3 off each choice and see which splits into 4 equal whole numbers: 27 − 3 = 24, and 24 shared by 4 is 6 each, so the ages are 6, 6, 6 and 9.
So the four brothers can be 27 years old altogether, choice B.
Eight cards with the numbers 1, 2, 4, 8, 16, 32, 64, 128 are each in an unmarked envelope. Eva randomly chooses some of these eight envelopes. Ali takes the remaining ones. Both add their numbers together. They find out that Eva’s sum is 31 bigger than Ali’s sum. How many envelopes has Eva chosen?
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Answer: D — 5
Show hints
Hint 1 of 2
The eight cards are the powers of two from 1 to 128; their total is 255.
Still stuck? Show hint 2 →
Hint 2 of 2
If Eva's sum beats Ali's by 31 out of 255, find Eva's exact total, then read off its binary digits.
Show solution
Approach: binary split
The cards total 1+2+…+128 = 255.
Eva − Ali = 31 and Eva + Ali = 255, so Eva = (255+31)/2 = 143.
The two kangaroos Jump and Hop start at the same time from the same line in the same direction. Each jumps exactly once per second. Jump always jumps 6 m. Hop jumps 1 m, then 2 m, then 3 m, and so on. After how many jumps does Hop catch up with Jump?
Show answer
Answer: B — 11
Show hints
Hint 1 of 2
After n jumps Jump has gone 6n m; Hop has gone 1 + 2 + ... + n m.
Still stuck? Show hint 2 →
Hint 2 of 2
Set n(n+1)/2 = 6n and solve.
Show solution
Approach: equate the distances after n jumps
After n jumps Jump is at 6n m and Hop is at 1 + 2 + ... + n = n(n+1)/2 m.
Hop catches up when n(n+1)/2 = 6n, i.e. n + 1 = 12, so n = 11.
Three three-digit numbers are built using the digits 1 to 9 so that each of the nine digits is used exactly once. Which of the following numbers cannot be the sum of the three numbers?
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Answer: A — 1500
Show hints
Hint 1 of 2
The sum of all nine digits is fixed.
Still stuck? Show hint 2 →
Hint 2 of 2
1+2+...+9 = 45, so the total of the three numbers must be a multiple of 9.
Show solution
Approach: digit-sum divisibility by 9
The nine digits 1..9 sum to 45, a multiple of 9, so the total of the three numbers is divisible by 9.
Among the options only 1500 is not a multiple of 9 (1+5 = 6), so it cannot be the sum (A).
The perimeter of rectangle ABCD is 30 cm. Three more rectangles are added so that their centres are at the corners A, B and D and their sides are parallel to the rectangle (see diagram). The sum of the perimeters of these three rectangles is 20 cm. What is the length of the border of the whole shape (the thick black line)?
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Answer: C — 40 cm
Show hints
Hint 1 of 3
Look at one corner rectangle: its centre is on the corner, so exactly half of it sits outside the big rectangle and half sits inside.
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Hint 2 of 3
Trace the thick line around that corner and compare it to the plain corner it replaced.
Still stuck? Show hint 3 →
Hint 3 of 3
Each corner rectangle adds only half of its own perimeter to the outline.
Show solution
Approach: see how much each corner rectangle adds to the outline
Because each small rectangle is centred on a corner, the big rectangle's two edges cut it into four equal quarters, so half of the small rectangle pokes outside.
Tracing the thick line, the bits that poke out add length while the bits tucked inside hide the same length, so each corner rectangle adds exactly half of its own perimeter to the outline.
The three corner rectangles have perimeters adding to 20 cm, so together they add \(\tfrac{1}{2}\times 20 = 10\) cm.
The outline is the big rectangle's perimeter plus this, \(30 + 10 = 40\) cm, choice (C).
In a magic garden there are magic trees. On each tree there are either 6 pears and 3 apples, or 8 pears and 4 apples. In total there are 25 apples on the magic trees. How many pears in total are hanging on the magic trees altogether?
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Answer: D — 50
Show hints
Hint 1 of 2
Use the apple counts (3 per first kind of tree, 4 per second) to total 25 apples.
Still stuck? Show hint 2 →
Hint 2 of 2
Notice each apple comes with a fixed bundle of pears, so the pear total may not depend on the exact mix.
Show solution
Approach: find the tree counts from apples, then total the pears
A '6-pears/3-apples' tree carries 2 pears per apple, and an '8-pears/4-apples' tree also carries 2 pears per apple.
Since every apple is matched by exactly 2 pears on either kind of tree, the 25 apples come with 2 × 25 = 50 pears.
Seven identical dice (each with 1, 2, 3, 4, 5 and 6 points on its faces) are glued together to form the solid shown. Faces that are glued together always show the same number of points. How many points can be seen on the surface of the solid?
Show answer
Answer: D — 105
Show hints
Hint 1 of 3
Each die has 1 + 2 + ... + 6 = 21 points; seven dice hold 7 × 21.
Still stuck? Show hint 2 →
Hint 2 of 3
Subtract the hidden glued faces, which come in equal-number pairs.
Still stuck? Show hint 3 →
Hint 3 of 3
The six gluings hide pairs that total a fixed amount; remove it.
Show solution
Approach: total points minus the hidden glued faces
Seven dice have 7 × 21 = 147 points in all.
The central die is glued to 6 others; each gluing hides two equal faces, and over the 6 contacts the hidden faces sum to 42.
Each of the ten points in the diagram is labelled with one of the numbers 0, 1 or 2. It is known that the sum of the numbers in the corner points of each white triangle is divisible by 3, while the sum of the numbers in the corner points of each black triangle is not divisible by 3. Three of the points are already labelled as shown in the diagram. With which numbers can the inner point be labelled?
Show answer
Answer: A — only 0
Show hints
Hint 1 of 3
Read every condition modulo 3: a white triangle's three corners sum to 0, a black triangle's do not.
Still stuck? Show hint 2 →
Hint 2 of 3
Two triangles that share an edge differ only in their third corner, so compare the apex labels of triangles sitting on the same base.
Still stuck? Show hint 3 →
Hint 3 of 3
Chase the conditions outward from the given 0 and 2 to pin the inner point's residue.
Show solution
Approach: residues mod 3 on the triangular grid
Work mod 3: each white (upward) triangle's corners sum to 0, each black (downward) triangle's corners do not.
An upward and the downward triangle resting on the same edge share two corners, so their third corners must have different residues; this forces neighbouring apex labels apart.
Starting from the labelled corners 0 and 2 and propagating these forced differences leaves the inner point only one possible residue.
That residue is 0, so the inner point can be only 0 (A).
Luigi owns a few square tables and some chairs for his little restaurant. If he sets out his tables individually with 4 chairs each, he is 6 chairs short. If he always puts two tables together to make a bigger table with 6 chairs, he has 4 chairs left over. How many tables does Luigi have?
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Answer: B — 10
Show hints
Hint 1 of 3
Pairing tables uses 6 chairs for every 2 tables, which is 3 chairs per table; singling uses 4 chairs per table.
Still stuck? Show hint 2 →
Hint 2 of 3
So switching from paired to single needs 1 more chair for each table.
Still stuck? Show hint 3 →
Hint 3 of 3
Compare how the chair situation swings from '4 left over' to '6 short'.
Show solution
Approach: see how 1 extra chair per table swings the count
Paired up, the tables use 6 chairs per 2 tables, which is 3 chairs per table; set out singly they use 4 chairs per table.
So going from paired to single costs 1 extra chair for each table.
The situation swings from 4 chairs left over to 6 chairs short, a change of \(4 + 6 = 10\) chairs, and that swing is exactly 1 chair per table, so there are 10 tables, choice (B).
With algebraChairs \(=4T-6\) (singly) and \(=3T+4\) (paired); setting them equal, \(4T-6=3T+4\) gives \(T=10\).
In an enclosure there are 2016 kangaroos. Each of them is either red or grey, and there is at least one red and at least one grey kangaroo amongst them. For each kangaroo K we calculate the fraction obtained, if you take the number of kangaroos of the other colour divided by the kangaroos of the own colour (including K itself). Determine the sum of these 2016 fractions.
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Answer: A — 2016
Show hints
Hint 1 of 2
Group the kangaroos by colour: say r red and g grey, with r + g = 2016.
Still stuck? Show hint 2 →
Hint 2 of 2
Add the fractions colour by colour and watch the counts cancel.
Show solution
Approach: sum by colour
Each red kangaroo contributes g/r and there are r of them, totalling g.
Each grey kangaroo contributes r/g and there are g of them, totalling r.
There are 20 students in total, girls and boys, in a class. Two students always share a desk. One third of the boys share a desk with a girl, and half of the girls share a desk with a boy. How many boys are in the class?
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Answer: B — 12
Show hints
Hint 1 of 3
Let b boys and g girls with b + g = 20; count the mixed desks two ways.
Still stuck? Show hint 2 →
Hint 2 of 3
One third of the boys equals half the girls: b/3 = g/2.
Still stuck? Show hint 3 →
Hint 3 of 3
Solve b/3 = g/2 together with b + g = 20.
Show solution
Approach: count the mixed desks two ways
Let b boys and g girls, b + g = 20.
Mixed desks from the boys' side = b/3; from the girls' side = g/2; these are equal, so b/3 = g/2.
Then 2b = 3g; with b + g = 20 this gives b = 12, g = 8, so there are 12 boys.
Bettina chooses five points A, B, C, D and E on a circle and draws the tangent to the circle at point A. She realizes that the five angles marked x are all equally big. (Note that the diagram is not drawn to scale!) How big is the angle ∠ABD?
Show answer
Answer: C — 72°
Show hints
Hint 1 of 3
The tangent at A and the chords AB, AC, AD, AE split the straight tangent line into the five equal angles x.
Still stuck? Show hint 2 →
Hint 2 of 3
A tangent-chord angle equals half its intercepted arc, so each angle x cuts off an equal arc.
Still stuck? Show hint 3 →
Hint 3 of 3
Add up the equal arc pieces that the inscribed angle ABD intercepts.
Clara is forming one big triangle made up of identical little triangles. She has already put some triangles together (see diagram). What is the minimum number of little triangles she still has to add?
Show answer
Answer: B — 9
Show hints
Hint 1 of 3
A big triangle with little triangles has a square-number count: side 2 holds 4, side 3 holds 9, side 4 holds 16.
Still stuck? Show hint 2 →
Hint 2 of 3
Find the smallest such big triangle that still fits around the pieces already placed.
Still stuck? Show hint 3 →
Hint 3 of 3
Then subtract the pieces already there from that total.
Show solution
Approach: complete to the smallest big triangle that fits
The widest row already placed forces the big triangle to be 4 little triangles along each side, and a side-4 triangle holds \(4 \times 4 = 16\) little triangles.
Counting what is already placed and taking it away from 16, Clara must add 9 more little triangles, choice (B).
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.
A creeping plant twists exactly 5 times around a post with circumference 15 cm (as shown in the diagram) and thus reaches a height of 1 m. While the plant grows the height of the plant also grows with constant speed. How long is the creeping plant?
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Answer: C — 1.25 m
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Hint 1 of 2
Unroll one full twist into a flat right triangle.
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Hint 2 of 2
Its base is the post's circumference and its height is the rise per twist.
Show solution
Approach: unroll the spiral
Each of the 5 twists rises 100/5 = 20 cm while going 15 cm around.
In a square of area 36 there are grey parts as shown in the diagram. The sum of the areas of all the grey parts is 27. How long are the distances a, b, c and d together?
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Answer: D — 9
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Hint 1 of 2
Every grey triangle shares the same corner of the square and has its base (one of a, b, c, d) on an edge.
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Hint 2 of 2
From that corner each base is the full side away, so every triangle has height 6; the grey total is 3(a + b + c + d).
Show solution
Approach: grey area in terms of the four base segments
The square has side 6 (area 36). Every grey triangle has its apex at the same corner, with base a, b, c or d lying on an edge a full side (6) away.
So the total grey area is (1/2)(6)(a + b + c + d) = 3(a + b + c + d) = 27.
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?
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Answer: D — 13
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Hint 1 of 3
Every side of the pentagon uses three circles and they all add to the same total.
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Hint 2 of 3
Find a side that already has two numbers filled in to pin down that common total.
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Hint 3 of 3
Then walk around the pentagon, filling each missing circle from the side total until you reach X.
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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).
Eva writes seven numbers on a piece of paper, one of which is 201. She adds up these seven numbers and gets 2016. Now she replaces the 201 with the number 102 and again adds up the seven numbers. Which result does she get now?
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Answer: C — 1917
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Hint 1 of 2
Only one number changes, so only the change in that number changes the total.
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Hint 2 of 2
Find how much smaller 102 is than 201, and subtract that from 2016.
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Approach: adjust the sum by the change in the single number
Replacing 201 with 102 lowers that number by 201 − 102 = 99.
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?
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Answer: D — 12:30
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Hint 1 of 2
Turn Theo's belief into the real time, then read Leo's watch and Leo's belief.
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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.
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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.
A quadrilateral has an inner circle (i.e. all four sides of the quadrilateral are tangents to the circle). The ratio of the perimeter of the quadrilateral to the circumference of the circle is 4:3. The ratio of the area of the quadrilateral to that of the circle is therefore
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Answer: E — 4:3
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Hint 1 of 2
A tangential polygon's area is the inradius times its semiperimeter.
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Hint 2 of 2
Combine that with the given perimeter-to-circumference ratio.
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Approach: area = r * semiperimeter for a tangential quadrilateral
For a quadrilateral with an inscribed circle of radius r, area = r * (perimeter/2).
Given perimeter:circumference = 4:3, perimeter = (4/3)(2*pi*r) = 8*pi*r/3.
Area = r * (4*pi*r/3) = 4*pi*r^2/3; circle area = pi*r^2; ratio = 4:3 (E).
The symbols ◯, □ and ◇ stand for three different digits. If the digits of the number ◯□◯ are added, you get the two-digit number □◇. If the digits of the two-digit number □◇ are added, you get the single-digit number □. Which digit does ◯ stand for?
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Answer: E — 9
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Hint 1 of 3
Start from the last clue: adding the two digits of □◇ gives □ again.
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Hint 2 of 3
If □ + ◇ = □, the diamond ◇ must be 0.
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Hint 3 of 3
Now use that the digits of ◯□◯ add to the two-digit number □0.
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Approach: chase the digit clues from the smallest one up
The last clue says □ + ◇ = □, and the only way adding ◇ leaves □ unchanged is ◇ = 0, so the two-digit number □◇ ends in 0.
The first clue says the digits of ◯□◯ add to that number, so ◯ + □ + ◯ = (a multiple of 10), i.e. 2◯ + □ is 10, 20, ...
Trying □0 = 20 gives 2◯ + 2 = 20, so ◯ = 9, and the digits are all different (9, 2, 0).
Check: 929 has digit sum \(9+2+9 = 20\), and \(2+0 = 2\) — it works, so ◯ = 9, choice (E).
Leo has built a stick made up of 27 building blocks (see picture). He splits the stick into two pieces so that one part is twice as long as the other. He keeps repeating this: each time he takes one of the two pieces and splits it so that one piece is twice as long as the other. Which of the following pieces can never result in this way? (The choices are pieces of length 2, 4, 6, 8 and 10 blocks.)
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Answer: E — 10
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Hint 1 of 2
Splitting a piece into a 2:1 ratio only works when its length divides into three equal parts.
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Hint 2 of 2
List every length you can reach starting from 27 and see which option never appears.
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Approach: track which lengths a 2:1 split can produce
27 splits into 18 and 9; 18 splits into 12 and 6; 9 splits into 6 and 3; 12 splits into 8 and 4; 6 splits into 4 and 2.
The reachable lengths are 2, 3, 4, 6, 8, 9, 12, 18, 27.
A 10-block piece never appears, so 10 can never result.
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
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Hint 1 of 3
Only the grey target cells must change colour (an odd number of times); count them.
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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.
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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.
Twelve girls met up in a pastry shop. On average they each ate 1.5 muffins. None of them ate more than two muffins, and two of them ate nothing. How many girls ate two muffins?
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Answer: E — 8
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Hint 1 of 2
Twelve girls averaging 1.5 muffins ate 18 muffins total.
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Hint 2 of 2
Two ate 0; of the other 10 let x eat 2 and the rest 1, then use the total 18.
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Approach: set up the total-muffin equation
Total muffins = 12 × 1.5 = 18.
Two girls ate 0; of the remaining 10 let x eat 2 and (10 − x) eat 1: 2x + (10 − x) = 18.
Two three-digit numbers are made up of six different digits. The first digit of the second number is twice as big as the last digit of the first number. (Note: 0 is also a digit, but cannot be the first digit of a number.) What is the smallest possible sum of the two numbers?
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Answer: C — 537
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Hint 1 of 3
The hundreds digits matter most, so make those two as small as you can.
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Hint 2 of 3
The second number's first digit is twice the first number's last digit, so test small even leading digits like 2 or 4.
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Hint 3 of 3
Once the hundreds and that linked pair are fixed, fill the leftover spots with the smallest unused digits.
Show solution
Approach: make the hundreds digits smallest, respecting the doubling rule
The two hundreds digits drive the sum, so we want them tiny; the smallest nonzero hundreds digit is 1 for the first number.
The second number's hundreds digit must be double the first number's units digit, and the smallest workable pair is units 2 with hundreds 4, so the numbers look like 1?2 and 4?? .
Fill the remaining slots with the smallest unused digits 0, 3, 5: that gives 102 and 435, all six digits different.
Their sum \(102 + 435 = 537\) is the smallest possible, choice (C).
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?
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Answer: B — Bernhard
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Hint 1 of 3
Each sparrow whistles once for every sparrow it can see in the direction its beak points.
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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.
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Hint 3 of 3
Look for the sparrow who can see only a few birds now but would see many more after turning.
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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.
A motorboat drives in the middle of a stream. Downstream it needs four hours to get from X to Y. In order to drive back from Y to X it needs six hours. Tree trunks are also floating on the stream. How many hours does it take for a tree trunk to float in the middle of the stream from X to Y?
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Answer: E — 24
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Hint 1 of 2
Let the boat speed and current speed combine downstream and oppose upstream.
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Hint 2 of 2
The tree floats at the current's speed alone; find that from the two trip times.
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Approach: relative speeds
Downstream speed is D/4 and upstream is D/6 (D the distance).
Subtracting, twice the current is D/4 − D/6 = D/12, so the current is D/24.
Little Red Riding Hood is taking waffles to three grandmothers. Her basket starts completely full. Just before she reaches each grandmother’s house, the wolf eats half of the waffles in the basket. When she leaves the third grandmother, the basket is empty. Each grandmother gets the same number of waffles. The original number of waffles can definitely be divided by which of the following numbers?
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Answer: D — 7
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Hint 1 of 2
Work backward: before each grandmother the basket held twice what it held after.
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Hint 2 of 2
Each grandmother gets the same amount; reconstruct the start as a multiple, then see what must divide it.
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Approach: work backward through the halvings
The wolf halves the basket before each of the 3 grandmothers, and each grandmother takes the same amount g.
Working backward through the three halvings and equal gifts, the original amount comes out as a multiple of 7.
So the original number of waffles is always divisible by 7.
In the right-angled triangle ABC (with the right angle at A) the angle bisectors of the acute angles intersect at point P. The distance of P to the hypotenuse is \(\sqrt{8}\). What is the distance of P to A?
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Answer: E — 4
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Hint 1 of 3
Two angle bisectors meeting is the incentre, so its distance to every side is the same inradius.
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Hint 2 of 3
At the right angle A, the incentre sits on the bisector of a \(90^\circ\) angle, a \(45^\circ\) line from each leg.
Still stuck? Show hint 3 →
Hint 3 of 3
Relate AP to the inradius using that \(45^\circ\) geometry.
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Approach: incentre geometry at the right angle
The two acute-angle bisectors meet at the incentre P, so its distance to the hypotenuse is the inradius \(r = \sqrt{8}\).
P is also distance \(r\) from each leg, so from the right-angle vertex A it lies along the \(45^\circ\) bisector at distance \(r\sqrt{2}\).
\(AP = \sqrt{8}\cdot\sqrt{2} = \sqrt{16} = 4\), answer E.
In the Kangaroo Republic, every month has 40 days, which are numbered through from 1 to 40. Every day with a number that is divisible by 6 is a public holiday, and likewise every day with a prime number. How often per month does it occur that there is exactly one working day between two public holidays?
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Answer: A — 1
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Hint 1 of 2
Mark every day that is a multiple of 6 or a prime as a holiday.
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Hint 2 of 2
Look for a holiday, then a single working day, then another holiday.
Show solution
Approach: list and scan
Holidays are the primes and the multiples of 6 up to 40.
Scanning for the pattern holiday–working–holiday, only day 4 sits alone between holidays 3 and 5.
A big cube is made of 64 small cubes. Exactly one of them is grey (see diagram). Two cubes are neighbours if they share a common face. On day one the grey cube colours all of its neighbours grey. On day two all grey cubes again colour all of their neighbours grey. How many of the 64 little cubes are grey at the end of the second day?
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Answer: E — 17
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Hint 1 of 3
After day one the grey cube and all cubes sharing a face with it are grey.
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Hint 2 of 3
After day two add every cube sharing a face with those, so grey reaches anything within 2 face-steps of the start.
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Hint 3 of 3
Count the cubes you can reach in at most 2 face-steps, remembering the block is only 4 by 4 by 4 so some directions run off the edge.
Show solution
Approach: count cubes reachable within two face-steps
A cube ends up grey exactly when it can be reached from the start in at most two face-to-face steps (one step on day one, one on day two).
The grey cube sits on the top face just in from the back, so day one greys its 5 face-neighbours, and day two greys the new cubes one more step out.
Counting every cube within two face-steps (stopping at the outer faces of the 4 by 4 by 4 block) gives 17 grey cubes.
The equations \(x^2 + ax + b = 0\) and \(x^2 + bx + a = 0\) both have real solutions. It is known that the sum of the squares of the solutions of the first equation is equal to the sum of the squares of the solutions of the second equation, and that \(a \ne b\). Then \(a + b\) equals
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Answer: B — -2
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Hint 1 of 3
By Vieta, the sum of the squares of the roots is \((\text{sum})^2 - 2(\text{product})\).
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Hint 2 of 3
Set the two sums-of-squares equal, then factor the resulting symmetric equation.
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Hint 3 of 3
The condition \(a \ne b\) lets you cancel one factor.
Show solution
Approach: Vieta plus factoring
For \(x^2+ax+b\) the roots have sum \(-a\) and product \(b\), so their squares sum to \(a^2 - 2b\); for \(x^2+bx+a\) it is \(b^2 - 2a\).
Natural numbers, no two the same, are written on a board. The product of the two smallest is 16, and the product of the two largest is 225. What is the sum of all the numbers on the board?
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Answer: C — 44
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Hint 1 of 3
The two smallest numbers multiply to 16; the two largest to 225. Factor each.
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Hint 2 of 3
Pick distinct factor pairs that can be the smallest two and largest two, with no room for numbers in between.
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Hint 3 of 3
Then add all the numbers.
Show solution
Approach: identify the extreme pairs by factoring
The two smallest distinct naturals with product 16 are 2 and 8; the two largest distinct with product 225 are 9 and 25.
Since 8 and 9 are consecutive, no other number can fit between the small and large pairs, so the board holds exactly 2, 8, 9, 25.
In a solid cube P is a point on the inside. We cut the cube into 6 (sloping) pyramids. Each pyramid has one face of the cube as its base and point P as its top. The volumes of five of these pyramids are 2, 5, 10, 11 and 14. What is the volume of the sixth pyramid?
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Answer: C — 6
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Hint 1 of 2
Pair up pyramids on opposite faces of the cube.
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Hint 2 of 2
Two opposite pyramids have heights adding to the cube's edge, so each opposite pair has the same total volume.
Jakob writes down four consecutive positive whole numbers. He calculates all possible sums of three of those numbers and realises that none of those sums is a prime number. What is the smallest number that Jakob could have written down?
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Answer: C — 7
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Hint 1 of 2
Adding three of four consecutive numbers leaves out one; write the four sums.
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Hint 2 of 2
Two of the sums are automatically multiples of 3; make the other two composite too.
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Approach: search the smallest start
For a start n the four sums are 3n+3, 3n+4, 3n+5, 3n+6; the first and last are multiples of 3.
Need 3n+4 and 3n+5 composite as well; the first n that works is n = 7 (25 and 26).
The diagram shows a pentagon with the length of each side marked. Five circles are drawn with centres A, B, C, D and E. On each side of the pentagon, the two circles centred at the ends of that side touch each other. Which point is the centre of the biggest circle?
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Answer: A — A
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Hint 1 of 3
Touching circles mean neighbouring radii add to the side between their centres.
A rectangular piece of paper ABCD is 5 cm wide and 50 cm long. The paper is white on one side and grey on the other. Christina folds the strip as shown so that the vertex B coincides with M, the midpoint of the edge CD. Then she folds it so that the vertex D coincides with N, the midpoint of the edge AB. How big is the area of the visible white part in the diagram?
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Answer: B — 60 cm²
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Hint 1 of 3
Each fold flips a corner flap, so its grey back shows and it hides the white strip underneath it.
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Hint 2 of 3
Find the visible white as the leftover middle strip minus the white that the two folded grey flaps cover.
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Hint 3 of 3
Each flap, once folded inward, lands as a triangle whose base is 13 and height 5 over the white middle.
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Approach: subtract the white hidden by the two folded-in grey flaps
The strip has area \(5 \times 50 = 250\); each fold turns over an end flap of area 62.5, leaving a white middle band of area \(250 - 2 \times 62.5 = 125\).
Folding B onto M (and D onto N) lays each grey flap back onto that middle band, where it covers a triangle of base 13 and height 5, area \(\tfrac12 \times 13 \times 5 = 32.5\).
The two covered triangles sit in opposite halves and do not overlap, so they hide \(2 \times 32.5 = 65\) of white.
Visible white \(= 125 - 65 = 60\,\text{cm}^2\), answer B.
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
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Hint 1 of 3
Note the women are Sandra and Eva; the men are Benjamin and Philipp.
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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.
Susi writes a different positive whole number on each of the 14 cubes of the pyramid (see diagram). The sum of the numbers on the nine cubes on the bottom is 50. The number on every other cube equals the sum of the numbers on the four cubes directly underneath it. What is the biggest number that can be written on the topmost cube?
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Answer: E — 118
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Hint 1 of 3
Each cube above is the sum of the four directly under it; the top is a weighted sum of the nine bottom numbers.
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Hint 2 of 3
The centre bottom cube counts 4 times, edge-centres twice, corners once; the nine distinct numbers total 50.
Still stuck? Show hint 3 →
Hint 3 of 3
Put the largest values where the weight is biggest to maximise the top.
Show solution
Approach: weighted sum of the bottom layer
Adding up the pyramid, the top equals a weighted sum of the nine bottom numbers with weights: centre 4, the four edge-middles 2 each, the four corners 1 each, so top = (sum of all nine) + 3·(centre) + (sum of the four edges) = 50 + 3·centre + (edge sum).
To make this biggest, push value into the centre: give the four corners the smallest distinct values 1, 2, 3, 4 (sum 10), leaving 40 for the centre plus four edges.
Make the four edges the next-smallest distinct values 5, 6, 7, 8 (sum 26), so the centre is 40 − 26 = 14; then top = 50 + 3·14 + 26 = 118.
Anna chooses a positive whole number n and writes down the sum of all positive whole numbers from 1 to n. A prime number p divides this sum but none of the summands. Which of the following numbers is a possible value of n + p?
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Answer: A — 217
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Hint 1 of 3
The sum \(1+2+\cdots+n\) equals \(\tfrac{n(n+1)}{2}\).
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Hint 2 of 3
A prime that divides the sum but none of the summands \(1,\dots,n\) must be larger than \(n\).
Still stuck? Show hint 3 →
Hint 3 of 3
The only way such a prime appears in \(\tfrac{n(n+1)}{2}\) is as \(n+1\) itself.
Show solution
Approach: the special prime must equal n+1
The sum is \(\tfrac{n(n+1)}{2}\); if prime \(p\) divides it but no summand \(1,\dots,n\), then \(p > n\).
A prime bigger than \(n\) can only come from the factor \(n+1\), so \(p = n+1\) and \(n+p = 2n+1\) is odd.
Test the odd options: \(217 = 2(108)+1\) needs \(n=108\), \(p=109\), and 109 is prime, while the others give composite \(n+1\).
For \(n=108\), \(p=109\) divides \(\tfrac{108\cdot109}{2}\) but no summand, so 217 works (A).
A date can be written in the form DD.MM.YYYY; e.g. today’s date is 17.03.2016. We call a date “surprising” if all 8 digits used in this notation are different. In which month does the next surprising date occur?
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Answer: B — June
Show hints
Hint 1 of 3
All eight digits must differ, so the year YYYY itself must already use four distinct digits.
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Hint 2 of 3
A day's first digit is 0-3 and a month's first digit is 0 or 1, which sharply limits which years can work.
Still stuck? Show hint 3 →
Hint 3 of 3
Step forward from 2016 to the first year whose digits leave a legal day and month with no repeats.
Show solution
Approach: find the first year that leaves room for a valid day and month
Any year from 2017 onward that starts 20.. reuses the 0 (months and small days also need a 0 or repeat), so no surprising date appears in the 2000s.
Checking the 2100s, 2200s and early 2300s, the leading digits keep colliding with the only small digits a valid month (01-12) and day (01-31) can use, so none works.
The first year that frees up enough distinct small digits is 2345, and its earliest surprising date is 17.06.2345 (digits 1,7,0,6,2,3,4,5 all different).
In each of the five carriages of a train there is at least one passenger. Two passengers are neighbours if they are in the same carriage or in two successive carriages. Each passenger has either exactly 5 or exactly 10 neighbours. How many passengers are on the train?
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Answer: C — 17
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Hint 1 of 3
A passenger's neighbour count = (own carriage size) + (adjacent carriage sizes) − 1, and it must be 5 or 10 for everyone.
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Hint 2 of 3
So for each carriage the sum of it and its neighbours is fixed at 6 or 11; the two end carriages have only one neighbour-carriage.
Still stuck? Show hint 3 →
Hint 3 of 3
Pin down the middle carriage first, then the two pairs at the ends.
Show solution
Approach: fix the carriage sizes from the neighbour counts
Each passenger's neighbours = (own carriage) + (touching carriages) − 1, so for every carriage the block-sum of it and its neighbours is 6 or 11.
The two end pairs must sum to 6 (an end carriage plus its single neighbour), and the middle three must sum to 11, which forces the middle carriage to hold 5.
So the train is (end-pair sum 6) + 5 + (end-pair sum 6) = 6 + 5 + 6 = 17 passengers, e.g. 3, 3, 5, 3, 3.
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 three fields that are adjacent in a horizontal or vertical line (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: A — less than 10
Show hints
Hint 1 of 3
Each move flips exactly three in-line cells, so a cell ends black only if it is flipped an odd number of times.
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Hint 2 of 3
Try to cover several needed black cells with each single move instead of one at a time.
Still stuck? Show hint 3 →
Hint 3 of 3
Look for a clever overlap of horizontal and vertical triples that hits the target with very few moves.
Show solution
Approach: find an efficient flip sequence
A move toggles three adjacent cells in a line, so the goal is to flip exactly the target-black cells an odd number of times and the rest an even number.
Because moves overlap, a single well-placed triple can settle several target cells at once.
A careful set of fewer than ten moves produces the whole chessboard pattern, so the answer is less than 10 (A).
Exactly 2016 people are taking part in a conference. They are registered as P1 to P2016 in the system. Each person from P1 to P2015 has shaken exactly the amount of other hands that his/her own system number indicates. How many people did P2016 shake hands with?
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Answer: D — 1008
Show hints
Hint 1 of 2
P2015 shook everyone, so each person shares a handshake with the busiest people first.
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Hint 2 of 2
Pair the busiest with the least busy and peel inward; track who is left for P2016.
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Approach: pair from the extremes
P2015 shakes everyone (so P1's single handshake is with P2015); P2014 shakes all but P1 (matching P2's two), and so on.
Peeling these matched pairs inward, P2016 ends up shaking exactly half of the others.
The positive whole number N has exactly six different (positive) factors, including 1 and N. The product of five of these factors is 648. Which of these numbers is the sixth factor of N?
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Answer: C — 9
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Hint 1 of 3
Divisors pair up to multiply to N, so the product of all of them is a power of N.
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Hint 2 of 3
With six divisors, all six multiply to \(N^3\).
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Hint 3 of 3
Then the sixth factor is \(N^3\) divided by the given product 648.
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Approach: product of all divisors = N^(d/2)
For \(N\) with 6 divisors, the product of all six is \(N^3\) (divisors pair up to give \(N\)).
So the sixth factor is \(N^3 / 648\), meaning \(648\) must equal \(N^3\) divided by one divisor.
Trying \(N = 18\): its factors \(1,2,3,6,9,18\) multiply to \(5832 = 18^3\), and \(5832 / 648 = 9\).
