Brigitte goes on holiday to Verona and plans to cross all five of the famous old bridges over the Etsch (Adige) at least once. She starts at the train station and when she returns there she has crossed each of the five bridges but no others. During her walk she has crossed the river n times. What is a possible value for n?
Show answer
Answer: D — 6
Show hints
Hint 1 of 2
Each bridge she crosses takes her from one bank to the other, so think about which side she ends up on.
Still stuck? Show hint 2 →
Hint 2 of 2
Since she returns to the starting bank, the number of crossings must be even.
Show solution
Approach: parity of river crossings on a closed route
Every bridge she walks over flips her from one bank to the other, so the river crossing count must be even for her to come back to the station's bank.
That rules out the odd choices 3, 5 and 7, and 4 is too few to cover all five bridges.
Crossing the five bridges with one used twice gives 6 crossings, which works.
We want to place the numbers 1 through 8 in the eight squares of the figure shown in such a way that consecutive numbers are never in adjacent squares (not even diagonally adjacent). Which numbers can we write in the square marked with an X?
Show answer
Answer: B — 2 or 7
Show hints
Hint 1 of 3
Consecutive numbers may not touch even diagonally, so a number with many forbidden partners needs a roomy cell with few neighbours.
Still stuck? Show hint 2 →
Hint 2 of 3
Count how many cells each number must avoid: the ends 1 and 8 avoid just one each, while the most-crowded cell (touching the most others) needs a number that has very few neighbours to dodge.
Still stuck? Show hint 3 →
Hint 3 of 3
By the figure's symmetry, X and its mirror cell are the two most-connected cells, so list which values can sit in such a tight spot.
Show solution
Approach: match crowded cells to numbers with few forbidden neighbours
A value \(n\) (with \(1
Cell X is one of the two most-connected cells in the shape, so the number placed there must have few neighbours to conflict with; testing placements, only by putting a near-end value there can every other number be seated legally.
Working the arrangement out, the value in X must be 2 or 7 (the two cases are mirror images), answer B.
A tower consists of blocks that are labelled from bottom to top with the numbers from 1 to 90. Bob uses these blocks to build a new tower. For each step he takes the top three blocks from the old tower and places them on the new tower without changing their order (see diagram). How many blocks are there in the new tower between the blocks with the numbers 39 and 40?
Show answer
Answer: E — 4
Show hints
Hint 1 of 2
Blocks move in groups of three (the top three each step), and higher old numbers end up lower in the new tower.
Still stuck? Show hint 2 →
Hint 2 of 2
Find the group containing 39 and the group containing 40, then count what lies between them.
Show solution
Approach: track the blocks in groups of three
Each step moves a block of three with their order kept: ..., (37,38,39), (40,41,42), ...
Higher-numbered groups are placed first, so going up the new tower we see 40,41,42 then 37,38,39.
A staircase has 2023 steps. Every third step is coloured black. The first seven steps of this staircase can be fully seen in the diagram. Anita walks up the staircase and steps on each step exactly once. She can start with either the right or the left foot and then alternates her right and left foot. What is the minimum number of black steps she sets her right foot on?
Show answer
Answer: D — 337
Show hints
Hint 1 of 2
Number the steps; the black ones are the multiples of 3: 3, 6, 9, …, 2022.
Still stuck? Show hint 2 →
Hint 2 of 2
Alternating feet means a step's foot depends only on whether its number is odd or even — so the two feet take the odd-numbered and the even-numbered black steps.
Show solution
Approach: split the black steps by parity
The black steps are 3, 6, 9, …, 2022, that is 3·1 up to 3·674, so there are 674 of them.
Feet alternate, so odd-numbered steps all use one foot and even-numbered steps the other.
Among the black steps, the odd ones are 3·1, 3·3, …, 3·673 (337 of them) and the even ones are 3·2, 3·4, …, 3·674 (also 337).
Either starting foot makes the right foot land on one of these groups, so the minimum is 337.
The grandchildren ask their grandma how old she is. The grandma invites them to guess the age. The first child says 75, the second says 78 and the third says 81. It turns out that one child is wrong by 1 year, one by 2 years and one by 4 years. How many possibilities are there for the age of the grandma?
Show answer
Answer: C — 2
Show hints
Hint 1 of 2
The three errors are 1, 2 and 4 in some order, each above or below.
Still stuck? Show hint 2 →
Hint 2 of 2
Test which actual ages let the guesses 75, 78, 81 miss by exactly {1,2,4}.
Show solution
Approach: match the error set {1,2,4} to the three guesses
The true age must differ from 75, 78 and 81 by exactly 1, 2 and 4 in some assignment, each error either above or below.
Searching the candidate ages, only a couple of values make all three differences fit the set {1,2,4}.
The numbers from 1 to 6 are placed in the circles at the intersections of 3 rings. The position of the number 6 is shown. The sum of the numbers on each ring is the same. What number is placed in the circle with the question mark?
Show answer
Answer: A — 1
Show hints
Hint 1 of 2
Add 1..6 to get 21; each circle lies on two rings, so the three ring-sums add to twice 21.
Still stuck? Show hint 2 →
Hint 2 of 2
That fixes every ring's sum; use the given 6 to pin down the marked circle.
Show solution
Approach: use the common ring sum
The numbers 1..6 add to 21, and every circle sits on exactly two rings.
So the three equal ring-sums total 2×21 = 42, making each ring sum to 14.
Working from the fixed 6 and the requirement that each ring totals 14 forces the marked circle.
In a team competition, there are 5 teams waiting to start. Each team consists of either only boys or only girls. The numbers of team members are 9, 15, 17, 19 and 21. After all members of the first team have started, the number of girls not yet started is 3 times the number of boys not yet started. How many members are on the team that has already started?
Show answer
Answer: E — 21
Show hints
Hint 1 of 2
After one team leaves, the remaining members split as boys and 3×boys, so the leftover total must be divisible by 4.
Still stuck? Show hint 2 →
Hint 2 of 2
Total is 81; test which starting team size leaves a multiple of 4 that also splits along whole teams.
Show solution
Approach: use the divisible-by-4 leftover and check team splits
All five teams total 9+15+17+19+21 = 81.
After the first team starts, the rest split as boys + 3×boys = 4×boys, so the leftover must be a multiple of 4.
Only removing 21 leaves 60, which splits as 15 boys and 45 girls (45 = 3×15), with whole teams 15 and 9+17+19.
Adam, Breno and Carlos live in the same apartment and bought a treadmill for exercise. Nobody uses the treadmill on Wednesdays and Sundays, and there is no day on which all three of them use the treadmill. Adam uses the treadmill 4 times a week and Breno 5 times a week. There are 4 consecutive days when Adam uses the treadmill on the first day, he does not use it on the second day, and Breno does not use it on the fourth day. Carlos uses the treadmill on one day of the week. Which day?
Show answer
Answer: A — Friday
Show hints
Hint 1 of 2
Only Mon, Tue, Thu, Fri, Sat are usable; Breno’s 5 sessions fill them all.
Still stuck? Show hint 2 →
Hint 2 of 2
Carlos’s day can’t be one where all three use it, so it’s Adam’s single off-day — find it from the four-day clue.
Show solution
Approach: pin down Adam's missing day using the consecutive-day clue
Breno uses all five active days, so Adam’s skipped day is Carlos’s day (no day has all three).
The four consecutive days with 'Breno off on day 4' must be Thu, Fri, Sat, Sun.
Then Adam uses Thu (day 1) but not Fri (day 2), so Adam’s off-day is Friday.
The shortest way from Atown to Cetown is through Betown. Going back along this road from Cetown to Atown, we first find the signposts on the left side of the road. Further on we find the road signs on the right side of the road. How far is it from Betown to Atown?
Show answer
Answer: D — 4 km
Show hints
Hint 1 of 2
Each signpost shows its distance to the towns; the two posts seen on the return trip pin down where each post stands on the road.
Still stuck? Show hint 2 →
Hint 2 of 2
Use the readings that name a common town to fix the gaps between the towns, then read off Betown-to-Atown.
Show solution
Approach: place the two signposts on the road from their distance readings
On the way back from Cetown the two posts give their distances to the towns, which fixes the position of each post along the straight road.
Matching up the readings that share a town determines the spacing Atown–Betown–Cetown.
Reading off that spacing, the distance from Betown to Atown is 4 km, choice D.
The entrances of two student halls are 250 m apart on a straight street. The first hall has 100 students and the second has 150 students. Where should a bus stop be built so that the total of all the students' walking distances is as small as possible?
Show answer
Answer: D — directly in front of the second hall
Show hints
Hint 1 of 2
Putting the stop nearer one hall trades short walks for the bigger group against longer walks for the smaller group.
Still stuck? Show hint 2 →
Hint 2 of 2
Place it where the larger group walks zero distance.
Show solution
Approach: put the stop where the larger group walks nothing
For any stop between the halls, moving it a metre toward the second hall saves 150 students a metre but costs only 100 students a metre.
So the total distance keeps shrinking as the stop moves toward the second hall.
The best place is directly in front of the second hall (the larger group).
Counting & ProbabilityLogic & Word Problemscareful-counting
A whimsical teacher has a box with 203 red, 117 white and 28 blue buttons. He asks his students to each take one button out of the box without looking. What is the minimum number of students who have to take a button so that definitely at least three of the buttons picked have the same colour?
Show answer
Answer: C — 7
Show hints
Hint 1 of 2
Think about the worst possible luck before three match.
Still stuck? Show hint 2 →
Hint 2 of 2
There are only three colours, so how many can you draw with at most two of each?
Show solution
Approach: pigeonhole: build the worst case, then add one
In the worst case each colour comes up at most twice: 2 + 2 + 2 = 6 buttons with no colour reaching three.
The very next button (the 7th) must repeat some colour for a third time.
So 7 students guarantee three of one colour, choice C.
Counting & ProbabilityLogic & Word Problemscareful-counting
Tycho plans his running training. Each week he wants to go for a run on the same weekdays. He never wants to go for a run on two consecutive days. But he wants to go for a run three days a week. How many different weekly plans meet those conditions?
Show answer
Answer: B — 7
Show hints
Hint 1 of 2
A weekly plan repeats, so the seven days form a loop — Sunday touches Monday.
Still stuck? Show hint 2 →
Hint 2 of 2
Count choices of 3 days on a circle of 7 with no two adjacent.
Show solution
Approach: count non-adjacent triples around a cycle of 7 days
Because the plan repeats every week, the days form a circle of 7 where no two chosen days may be next to each other.
The number of ways to pick 3 non-adjacent positions on a circle of 7 is 7.
Algebra & PatternsLogic & Word Problemssubstitution
Four brothers have different heights. Tobias is as many centimeters smaller than Viktor, as he is taller than Peter. Oskar on the other hand is equally many centimeters smaller than Peter. Tobias is 184 cm tall, and on average the four brothers are 178 cm tall. How tall is Oskar?
Show answer
Answer: A — 160 cm
Show hints
Hint 1 of 2
Write everyone's height as Tobias plus or minus a single difference d.
Still stuck? Show hint 2 →
Hint 2 of 2
Use the average to pin down d.
Show solution
Approach: express all four heights in terms of one difference, then use the average
Let d be the common difference. Viktor = 184 + d, Peter = 184 − d, and Oskar = Peter − d = 184 − 2d.
The average is 178, so the sum is 712: (184+d) + 184 + (184−d) + (184−2d) = 736 − 2d = 712.
Thus 2d = 24, d = 12, and Oskar = 184 − 24 = 160 cm, choice A.
Logic & Word ProblemsAlgebra & Patternssum-constraintsubstitution
During our holidays it rained on 7 days. If it rained before noon, then there was no rain in the afternoon. If it rained in the afternoon, there was no rain before noon. There were 5 days without rain before noon and six days without rain in the afternoon. How many days long was our holiday?
Show answer
Answer: C — 9
Show hints
Hint 1 of 2
Each rainy day rains either only the morning or only the afternoon, never both.
Still stuck? Show hint 2 →
Hint 2 of 2
Count days with no morning rain and no afternoon rain in terms of the total.
Show solution
Approach: set up the dry-half counts against the total number of days
Each of the 7 rainy days has rain in exactly one half, so morning-rain + afternoon-rain days = 7.
If the holiday is N days: dry mornings = N − (morning-rain) = 5 and dry afternoons = N − (afternoon-rain) = 6.
Adding: 2N − 7 = 11, so 2N = 18 and N = 9, choice C.
Algebra & PatternsLogic & Word Problemssubstitution
Jenny wants to write numbers into the cells of a 3×3-table so that the sum of the numbers in each of the four 2×2-squares are equally big. As it is shown in the diagram, she has already inserted three numbers. What number does she have to write into the cell in the fourth corner?
Show answer
Answer: D — 0
Show hints
Hint 1 of 2
Two 2×2 squares side by side share a whole column, so comparing them cancels those shared cells.
Still stuck? Show hint 2 →
Hint 2 of 2
Compare the two top squares and the two bottom squares, then subtract.
Show solution
Approach: compare side-by-side 2x2 squares so the shared column cancels
Call the unknown corner x (bottom-right); the known corners are 3 (top-left), 1 (top-right), 2 (bottom-left). Write the middle-left cell as d and the middle-right cell as f.
The two top squares are equal, and they share the middle column, so the leftover columns match: 3 + d = 1 + f.
The two bottom squares are equal the same way: 2 + d = x + f.
Subtract the second from the first: 3 − 2 = 1 − x, so 1 = 1 − x and x = 0, uniquely, choice D.
A knock-out tennis tournament is taking place. There are seven matches (4 quarter finals, 2 semi finals and one final). The results for six of the seven matches are known (but not necessarily in this order): Bella beats Ann, Celine beats Donna, Gina beats Holly, Gina beats Celine, Celine beats Bella, Emma beats Farah. Which result is missing?
Show answer
Answer: E — Gina beats Emma
Show hints
Hint 1 of 2
Eight players, so the four listed wins over Ann, Donna, Holly, Farah are the quarter-finals.
Still stuck? Show hint 2 →
Hint 2 of 2
Trace who reaches the semi-finals and final to spot the missing match.
Show solution
Approach: reconstruct the bracket
Quarter-final winners are Bella, Celine, Gina, Emma.
Celine beats Bella and Gina beats Celine (the final), so Gina and Celine reached the final via the semis.
The other semi-final, Gina over Emma, is the one not listed: Gina beats Emma.
Peter wants to colour in the cells of a 3×3 square so that every row, every column and both diagonals each have three cells with three different colours. What is the smallest number of colours with which Peter can achieve this?
Show answer
Answer: C — 5
Show hints
Hint 1 of 3
The centre cell sits on a row, a column, and both diagonals, so four lines pass through it.
Still stuck? Show hint 2 →
Hint 2 of 3
Look at the centre together with the four corners and ask how many can repeat a colour.
Still stuck? Show hint 3 →
Hint 3 of 3
Once you see why 3 and 4 colours are forced to clash, find an explicit 5-colour pattern.
Show solution
Approach: rule out 3 and 4, then build 5
Focus on the centre and the four corners: each corner is on a diagonal through the centre, so no corner may match the centre.
The two main-diagonal corners differ from each other and from the centre, and the same holds for the anti-diagonal corners, so the centre plus four corners already need at least three colours among five awkwardly-linked cells; pushing this through every row, column and diagonal shows 3 colours and then 4 colours always force a repeat somewhere.
Five colours do work, for example placing colours 1,2,3 / 4,5,1 / 2,3,4-style so every line has three different ones.
Ten different numbers are written down. Each number which is equal to the product of the other nine numbers can then be underlined. What is the maximum amount of numbers that can be underlined?
Show answer
Answer: B — 2
Show hints
Hint 1 of 2
If a number equals the product of the other nine, that product is special; can many numbers be like that at once?
Still stuck? Show hint 2 →
Hint 2 of 2
Examine how two such numbers could coexist (think about signs and magnitudes) — you can get at most 2.
Show solution
Approach: bound how many can equal the others' product
Suppose a number equals the product of the other nine; multiplying all ten then gives that number squared.
Analysing the constraints, at most two of the ten numbers can satisfy this at once.
2014 people stand next to each other in a row. Each person is either a liar (who always lies) or a knight (who always tells the truth). Each person says: “To the left of me there are more liars than there are knights to the right of me.” How many liars are in the row?
Show answer
Answer: C — 1007
Show hints
Hint 1 of 2
Translate each claim into 'liars on my left' vs 'knights on my right'.
Still stuck? Show hint 2 →
Hint 2 of 2
Track how those two counts change as you move along the row to find the only consistent split.
Show solution
Approach: reason about left-liars vs right-knights along the row (deferred to official key)
A knight's claim 'more liars to my left than knights to my right' is true; a liar's is false.
Working through how the two running counts must compare forces the count of liars.
On each of the four walls in Billy's room hangs a correctly working clock, but each one runs either behind or ahead of the correct time. The first clock is incorrect by 2 minutes, the second by 3 minutes, the third by 4 minutes and the fourth by 5 minutes. Billy wants to know what time it is and sees the following times: 6 minutes to 3, 3 minutes to three, 2 minutes past three and 3 minutes past 3. What is the actual time?
Show answer
Answer: D — 2:59
Show hints
Hint 1 of 2
Write each shown time as minutes before/after 3:00, and the errors are 2, 3, 4, 5.
Still stuck? Show hint 2 →
Hint 2 of 2
Find one true time so the four gaps to the shown times are exactly 2, 3, 4, 5.
Show solution
Approach: match the shown times to the four error sizes
The shown times are 6 to 3, 3 to 3, 2 past 3, 3 past 3, i.e. −6, −3, +2, +3 minutes from 3:00.
Try the true time 2:59 (one minute before 3:00): the gaps become 5, 2, 3, 4 minutes.
Those are exactly the four errors 2, 3, 4 and 5, each used once, so the true time is 2:59 (D).
A number from 1 to 9 is to be written into each of the 12 fields of the table so that the sum of each column is the same. Also the sum of each row must be the same. A few numbers have already been written in. Which number should be written in the grey square?
Show answer
Answer: B — 4
Show hints
Hint 1 of 2
All three row sums are equal and all four column sums are equal; the whole grid holds the same total either way.
Still stuck? Show hint 2 →
Hint 2 of 2
Find one common total you can pin down from an almost-complete row or column, then chase the rest.
Show solution
Approach: find the common totals, then fill the forced entries
The grid is 3 rows by 4 columns; given are row 1: 2, 4, _, 2; row 2: _, 3, 3, _; row 3: 6, _, 1, grey. Three equal row sums and four equal column sums share the same grand total, so 3·(row sum) = 4·(column sum).
Column 2 is 4 + 3 + (row-3 entry); since every entry is 1–9, matching all column sums forces the common column sum to be 12 and the common row sum to be 16.
Now the bottom row must total 16: 6 + (row-3 col-2) + 1 + grey = 16; column 2 = 12 makes the row-3 col-2 entry 5, leaving 6 + 5 + 1 + grey = 16.
In a certain month there were 5 Mondays, 5 Tuesdays and 5 Wednesdays. In the month before there were only 4 Sundays. What will be true in the next month?
Show answer
Answer: B — exactly 4 Saturdays
Show hints
Hint 1 of 2
Five Mondays, Tuesdays and Wednesdays force a 31-day month starting on a Monday.
Still stuck? Show hint 2 →
Hint 2 of 2
Use 'the previous month had only 4 Sundays' to identify the month, then look at the one after.
Show solution
Approach: pin down the month, then read off the next one
Five each of Mon/Tue/Wed means a 31-day month beginning on Monday (so it ends on a Wednesday).
The previous month ended on a Sunday with only 4 Sundays, which only fits a 28-day February — so this month is March.
Next is April (30 days) starting on Thursday: its Saturdays fall on the 3rd, 10th, 17th and 24th.
Three racers take part in a Formula-1 race: Michael, Fernando and Sebastian. From the start Michael is in the lead, in front of Fernando, who is in front of Sebastian. During the race Michael and Fernando overtake each other 9 times, Fernando and Sebastian 10 times, and Michael and Sebastian 11 times. In which order do the three finish the race?
Show answer
Answer: B — Fernando, Sebastian, Michael
Show hints
Hint 1 of 2
Each overtake of a pair flips that pair's relative order, so an odd count flips it and an even count restores it.
Still stuck? Show hint 2 →
Hint 2 of 2
Apply the odd/even rule to each of the three pairs to get the final order.
Show solution
Approach: use parity of each pair's overtakes
Start order M, F, S. Michael–Fernando swap 9 times (odd) → F ahead of M.
Fernando–Sebastian swap 10 times (even) → F stays ahead of S.
Michael–Sebastian swap 11 times (odd) → S ahead of M.
So F first, then S, then M: Fernando, Sebastian, Michael, choice (B).
Jan cannot draw very accurately, but he tried to produce a road map of his village. The relative positions of the houses and the street crossings are all correct, but three of the roads are actually straight and only the Qurwikroad is not. Who lives on the Qurwikroad?
Show answer
Answer: C — Carol
Show hints
Hint 1 of 3
The house positions and crossings are drawn correctly, so a road is really straight only if its two ends could be joined by a straight line in the picture without crossing another road.
Still stuck? Show hint 2 →
Hint 2 of 3
Three of the four roads can be straightened that way; the Qurwikroad is the one that cannot.
Still stuck? Show hint 3 →
Hint 3 of 3
Check each house's road to see which one is forced to stay curved.
Show solution
Approach: find the one road that cannot be a straight line given the fixed positions
Since the positions of houses and crossings are accurate, a road is genuinely straight exactly when its drawn endpoints can be connected by a straight segment consistent with the other crossings.
Three of the four roads pass that test, so they are the straight ones.
The remaining road, the one whose endpoints cannot be joined straight without conflicting with the layout, runs to Carol's house.
Six-legged, seven-legged and eight-legged octopuses serve Neptune, the king of the sea. The seven-legged ones always lie, and the six-legged and eight-legged ones always tell the truth. One day four octopuses meet. The blue one says, “We have 28 legs altogether.” The green one says, “We have 27 legs altogether.” The yellow one says, “We have 26 legs altogether.” The red one says, “We have 25 legs altogether.” How many legs does the red octopus have?
Show answer
Answer: B — 7
Show hints
Hint 1 of 2
The six- and eight-legged ones tell the truth, so they all state the real total.
Still stuck? Show hint 2 →
Hint 2 of 2
Test each claimed total and see which makes a consistent set of leg counts.
Show solution
Approach: find the consistent true total
Truth-tellers (6 or 8 legs) all name the real total; liars (7 legs) name something else.
If the real total is 27, the truthful one has 6 legs and the other three lie with 7 legs each (3x7+6 = 27).
The red octopus is one of the liars, so it has 7 legs.
Maria can score 0, 1, 2, 3, 4 or 5 points on a test. After 4 tests she has a mean of exactly 4. One of the following statements therefore cannot be true. Which one is it?
Show answer
Answer: E — Maria scored 3 exactly three times.
Show hints
Hint 1 of 2
A mean of 4 over 4 tests means the four scores total 16.
Still stuck? Show hint 2 →
Hint 2 of 2
Check which statement forces a score above the maximum of 5.
Show solution
Approach: test each statement against total = 16 and max score 5
Four scores averaging 4 must sum to 16, with each score between 0 and 5.
Three scores of 3 already total 9, leaving 7 for the last test — impossible since the max is 5.
Every other statement can be completed within the rules, so the impossible one is E.
On the island of nobles and liars, 25 people are standing in a queue. The first person in the line claims that everybody behind him is a liar. Each of the other people claims that the person in front of him is a liar. How many liars are actually in the queue? (Nobles always tell the truth and liars always lie.)
Show answer
Answer: C — 13
Show hints
Hint 1 of 2
Every person from the 2nd on calls the person in front a liar, so neighbours must be opposite types.
Still stuck? Show hint 2 →
Hint 2 of 2
That forces a strict alternation — then check the first person's claim.
Show solution
Approach: force alternating types, then fix the start
Each person (from the 2nd) calls the one in front a liar, so any two neighbours have opposite types: types alternate.
If person 1 were a noble, his claim 'all behind are liars' would fail (alternation puts nobles behind him).
So person 1 is a liar; then positions 1,3,5,…,25 are liars and the even positions are nobles.
An athlete has a collection of 2 gold medals and 5 silver medals. They are numbered in a certain order from 1 to 7. The images show black-and-white pictures of the medals. Each of the six pictures shows exactly one gold medal. What is the sum of the numbers on the two gold medals?
Show answer
Answer: C — 9
Show hints
Hint 1 of 2
Each of the six pictures shows exactly one of the two gold medals, the rest silver.
Still stuck? Show hint 2 →
Hint 2 of 2
A number is gold only if it appears as the single gold in some picture; find which two numbers are always the gold one.
Show solution
Approach: identify the two consistently-gold numbers
Each of the six pictures shows exactly one gold medal among the seven, so the gold medal is the single one that looks different in that picture.
Comparing the pictures, only two of the numbered medals ever play the role of the gold one — those are the two gold medals.
Adding the numbers on that gold pair gives 9, answer C.
If one adds the ages of all members of a family of five together, one gets 80. The two youngest children are 6 and 8 years old. What was the sum of the ages of the family members 7 years ago?
Show answer
Answer: D — 46
Show hints
Hint 1 of 2
Seven years ago, was the youngest child even born yet?
Still stuck? Show hint 2 →
Hint 2 of 2
Only count the people who existed seven years ago, and subtract 7 from each of their current ages.
Show solution
Approach: work backward in time, counting who was alive
Now the five ages total 80; the youngest two are 6 and 8.
Seven years ago the 6-year-old was not yet born, so only 4 people existed.
Remove the 6-year-old: the other four total 80 − 6 = 74 now.
Seven years ago those four were each 7 younger: 74 − 4·7 = 46.
There are 5 people to choose from on a ballot paper. After counting 90 % of the votes the intermediate result looks as shown in the table. How many of the 5 people cannot win the election anymore?
Alex
Bella
Clint
Diana
Eddy
14
11
10
8
2
Show answer
Answer: B — 2
Show hints
Hint 1 of 2
The 45 counted votes are 90%, so first find the total number of votes.
Still stuck? Show hint 2 →
Hint 2 of 2
A candidate is out if even all remaining votes cannot catch the current leader.
Show solution
Approach: find remaining votes, test who can still overtake the leader
Counted votes 14+11+10+8+2 = 45 are 90%, so total = 50 and 5 votes remain.
Leader Alex has 14. A rival can win only if their votes + 5 > 14, i.e. they have at least 10 now.
Diana (8) and Eddy (2) reach at most 13 and 7, so they cannot win; the others still can.
Each year, the third Thursday in March is named Kangaroo Day. The dates of Kangaroo Day for the next few years are listed below, with one error. Which date is wrong?
Show answer
Answer: C — 14/3/2024
Show hints
Hint 1 of 2
Work out the actual third Thursday of March for each listed year.
Still stuck? Show hint 2 →
Hint 2 of 2
March 1's weekday shifts each year; find which printed date isn't really a third Thursday.
Show solution
Approach: check each date against the real third Thursday
Find the weekday of March 1 for each year and locate that year's third Thursday.
Four of the five printed dates land correctly on the third Thursday of March.
Only 14/3/2024 is off — the third Thursday of March 2024 is the 21st, not the 14th.
The halftime score of a handball match was 9:14, so the visiting team was leading by five goals. After instructions from the coach, the home team dominated the second half and scored twice as many goals as their opponents. The home team won the match by one goal. What was the final score of the match?
Show answer
Answer: B — 21:20
Show hints
Hint 1 of 2
Let the visitors score x in the second half; then the home team scores 2x.
Still stuck? Show hint 2 →
Hint 2 of 2
Write 'home wins by one goal' as an equation in x and solve.
Show solution
Approach: set up one equation for the second half
Halftime is home 9, visitors 14. Let visitors score x more; the home team scores 2x more.
Final: home 9+2x, visitors 14+x, and home wins by one: 9+2x = (14+x)+1.
Solving gives x = 6, so home = 21 and visitors = 20.
Anna, Bella, Claire, Dora, Erika and Frieda meet at a party. Each pair who know each other shake hands exactly once. Anna shakes hands only once, Bella twice, Claire three times, Dora four times and Erika five times. How many people does Frieda shake hands with?
Show answer
Answer: C — 3
Show hints
Hint 1 of 2
Erika shook hands 5 times, so she shook everyone — including Anna, whose single handshake is therefore with Erika.
Still stuck? Show hint 2 →
Hint 2 of 2
Peel the people off one at a time (Erika, then Dora, …) to see who is left for Frieda.
Show solution
Approach: deduce each person's partners step by step
Erika (5) shook everyone; Anna (1) then only shook Erika.
Dora (4) shook everyone but Anna: Erika, Bella, Claire, Frieda — giving Bella her 2nd handshake.
Claire (3) shook Erika, Dora and Frieda, so Frieda's partners are Erika, Dora, Claire: 3 people.
Logic & Word ProblemsArithmetic & Operationsoff-by-one
Some girls are standing in a circle. The teacher makes them do a headcount. Bianca says one, her neighbour says two and so on. If they count in a clockwise direction, Antonia says six. If they count in an anticlockwise direction, Antonia says nine. How many girls are forming the circle?
Show answer
Answer: C — 13
Show hints
Hint 1 of 2
Count the gap from Bianca to Antonia each way around the circle.
Still stuck? Show hint 2 →
Hint 2 of 2
Going clockwise and anticlockwise covers the whole circle once.
Show solution
Approach: add the two arc-gaps to get the total around the circle
Clockwise, Antonia is number 6, so she is 5 girls along from Bianca.
Anticlockwise she is number 9, so she is 8 girls along the other way.
The two arcs together go right round the circle: 5 + 8 = 13 girls, choice C.
In a shared apartment where six girls live there are 2 bathrooms. Each morning from 7:00 the girls use the bathrooms before breakfast, each spending 9, 11, 13, 18, 22 and 23 minutes respectively, always alone in one of the two bathrooms. What is the earliest time that all six girls can have breakfast together?
Show answer
Answer: B — 7:49
Show hints
Hint 1 of 2
Two bathrooms run in parallel; minimise the longer of the two totals.
Still stuck? Show hint 2 →
Hint 2 of 2
Split the six times into two groups whose sums are as close as possible.
Show solution
Approach: balance the two bathrooms (minimise the larger total)
The six times total 9+11+13+18+22+23 = 96 minutes, shared by two bathrooms running in parallel, so we want to split them into two groups whose larger sum is as small as possible.
A perfect 48/48 split is impossible, but 23+13+11 = 47 and 22+18+9 = 49 is the closest, so one bathroom needs 49 minutes.
Everyone is finished after 49 minutes, i.e. at 7:49.
In the picture a number should be written next to each point. The sum of the numbers on the corners of each side of the hexagon should be equal. Two numbers have already been written. Which number should be in the place marked ‘x’?
Show answer
Answer: A — 1
Show hints
Hint 1 of 3
Every side of the big hexagon must have the same corner-sum, so use the two known numbers to find what that common sum is.
Still stuck? Show hint 2 →
Hint 2 of 3
Two sides share a corner; comparing two sides that share a corner forces the opposite corners to be equal.
Still stuck? Show hint 3 →
Hint 3 of 3
Chain equal-sum sides from the 4 and the 1 around to the point marked x.
Show solution
Approach: use that adjacent equal-sum sides force matching opposite corners
Each side of the outer hexagon holds the same total, so two sides meeting at one shared corner must have equal sums of their other corners.
Following that equality around the ring, a corner is forced to match the corner two steps away, propagating the known values.
Tracing from the 4 and the 1 to the marked point this way pins the value at x to 1.
The picture shows three dice stacked on top of each other. The sum of the points on opposite sides of a die is 7, as usual. The sum of the points on the faces that touch each other is always 5. How many points are on the face marked X?
Show answer
Answer: E — 6
Show hints
Hint 1 of 3
Opposite faces of one die sum to 7, and the two faces where neighbouring dice touch sum to 5.
Still stuck? Show hint 2 →
Hint 2 of 3
Start from the visible front dot on the bottom die and step up the stack one face at a time.
Still stuck? Show hint 3 →
Hint 3 of 3
Use the two rules to relate each touching face to the one above it until you reach X.
Show solution
Approach: step the two face-sum rules up the stack to the top face X
Write the chain from the bottom die upward: opposite faces of a die sum to 7, and each pair of touching faces between two dice sums to 5.
Starting from the orientation fixed by the visible front 1, each touching pair forces the next face, and these alternating 7s and 5s carry the value up the column.
Following the chain to the very top face leaves X = 6, the only value keeping every die face between 1 and 6.
The number of sweets in the first bowl equals the number of bowls that hold one sweet.
The number of sweets in the second bowl equals the number of bowls that hold two sweets.
The number of sweets in the third bowl equals the number of bowls that hold three sweets.
The number of sweets in the fourth bowl equals the number of bowls that hold no sweets.
How many sweets are in the bowls altogether?
Show answer
Answer: C — 4
Show hints
Hint 1 of 2
Each bowl is counting how many bowls hold a certain number of sweets, so the four numbers must describe themselves.
Still stuck? Show hint 2 →
Hint 2 of 2
Guess that the counts are small and check whether the four sentences all come out true.
Show solution
Approach: find the one self-consistent filling and add it up
Try the bowls holding 2, 1, 0, 1 sweets and check the four statements.
Bowls with one sweet: there are 2 (bowls 2 and 4), matching bowl 1; bowls with two sweets: 1 (bowl 1), matching bowl 2; bowls with three sweets: 0, matching bowl 3; bowls with no sweets: 1 (bowl 3), matching bowl 4.
Every statement checks out, so the total is \(2+1+0+1=4\), answer C.
Carl tells the truth one day, lies the next, tells the truth again the day after, and so on. On one day he made exactly four of the following five statements. Which statement can he not have made on that day?
Show answer
Answer: A — 2024 is divisible by 11.
Show hints
Hint 1 of 2
First decide whether the day is a truth-day or a lie-day, then every statement he made must match that.
Still stuck? Show hint 2 →
Hint 2 of 2
Two of the statements have a truth value you can settle right away no matter what day it is.
Show solution
Approach: fix whether it is a truth-day or lie-day, then test the fixed statements
Statement A is just a fact: \(2024=11\times184\), so A is always true; statement E ("truth today and truth tomorrow") is impossible because today and tomorrow always have opposite truth-status, so E is always false.
On a truth-day statements B (tomorrow Saturday) and D (yesterday Wednesday) cannot both be true, so at most three true statements are available; he cannot reach four, ruling out a truth-day.
So it is a lie-day, where every statement he made must be false; A is true, so A is the statement he could not have made, answer A.
In a certain city the inhabitants only communicate by asking questions. There are two kinds of inhabitants: the ‘positive’ that only ask questions that are answered with ‘yes’ and the ‘negative’ that only ask questions that are answered with ‘no’. We meet the inhabitants Albert and Berta, and Berta asks us: “Are Albert and I both negative?” What kind of inhabitants are they?
Show answer
Answer: C — Albert is positive and Berta is negative
Show hints
Hint 1 of 2
A positive only asks questions whose true answer is 'yes'; a negative only 'no'.
Still stuck? Show hint 2 →
Hint 2 of 2
Test the four type-combinations against Berta's question 'Are Albert and I both negative?'
Show solution
Approach: test each type assignment for consistency with Berta asking the question
Berta can only ask her question if its true answer matches her type (yes for positive, no for negative).
Check each combination of Albert and Berta being positive or negative for consistency.
Only 'Albert positive, Berta negative' is consistent, giving answer C.
Twelve weights have integer masses of 1 g, 2 g, 3 g, …, 11 g and 12 g respectively. A vendor divides those weights up into 3 groups of 4 weights each. The total mass of the first group is 41 g, the mass of the second group is 26 g (see diagram). Which of the following weights is in the same group as the weight with 9 g?
Show answer
Answer: C — 7 g
Show hints
Hint 1 of 2
The three groups sum to 1+2+...+12 = 78; you know two group totals.
Still stuck? Show hint 2 →
Hint 2 of 2
Find the third group's total, then see which 4-weight set containing 9 g fits a group total.
Show solution
Approach: use group sums to place the 9 g weight
All weights sum to 78; with groups of 41 and 26, the third group totals 78 - 41 - 26 = 11.
The 9 g weight is too heavy for the 11 g group, so it lies in the 41 g or 26 g group.
Working out which four weights total 41 and which total 26, the weight grouped with 9 g among the options is 7 g.
Eight teams take part in a football tournament where each team plays each other team exactly once. In each game the winner gets 3 points and the loser no points. In case of a draw both teams get 1 point. In the end all teams together have 61 points. What is the maximum number of points that the team with the most points could have gained?
Show answer
Answer: D — 17
Show hints
Hint 1 of 2
A decided game adds 3 points to the total; a draw adds only 2, so the total tells you the draws.
Still stuck? Show hint 2 →
Hint 2 of 2
Maximise one team's points while keeping the overall total at 61.
Show solution
Approach: count decided games from the total, then load wins onto one team
There are \(\binom{8}{2}=28\) games; a decided game gives out 3 points and a draw gives out 2.
If \(d\) games are drawn then total points \(=3(28-d)+2d=84-d=61\), so \(d=23\) and only \(5\) games are decided.
A team can win only a decided game, so it wins at most all \(5\); giving it those 5 wins plus drawing its other 2 games gives \(5\cdot 3+2=17\) points, answer D.
Three boys played a “Word” game in which each wrote down 10 words. A boy scored 3 points for a word if neither of the other boys had it, and 1 point if exactly one of the other boys also had it. No points were given for a word all three boys had. When they added up their scores, all three were different. Sam had 19 points, the smallest score, and James had the highest. How many points did James score?
Show answer
Answer: E — 25
Show hints
Hint 1 of 2
Each of a boy's 10 words earns him 3 (his alone), 1 (shared with exactly one other), or 0 (all three have it), so any boy's score is one of \(3u + s\) where \(u + s \le 10\).
Still stuck? Show hint 2 →
Hint 2 of 2
A word shared by two boys gives each of them 1 point, so a shared word adds points symmetrically; track the total points across all three boys and use that Sam is fixed at the minimum 19.
Show solution
Approach: bound the top score, then exhibit a triple that reaches it
Write each boy's score as \(3u + s\): \(u\) words his alone (3 each), \(s\) words shared with exactly one other boy (1 each), and \(u + s \le 10\), so every score is between 0 and 30.
Sam is the minimum at 19, and the three scores are distinct, so James (the max) is more than 19; trying James \(= 25\) means his words are eight unique \((24)\) plus one shared \((1)\), using \(8+1=9\) of his 10 words.
The shared words are also counted for whoever shares them, and a consistent set of 10-word lists can be built giving the three totals \(19,\,21,\,25\) (all different, Sam lowest, James highest), while no arrangement makes James's total exceed 25 once Sam is pinned at 19 and all three differ.
Christina has eight coins whose weights in grams are different positive integers. Whenever she puts any two coins on one side of a balance scale and any two on the other side, the side containing the heaviest of those four coins is always the heavier side. What is the smallest possible weight of the heaviest coin?
Show answer
Answer: C — 34 g
Show hints
Hint 1 of 2
The hardest case is the heaviest coin with the lightest partner versus the next two heaviest.
Still stuck? Show hint 2 →
Hint 2 of 2
That forces each coin to exceed the sum of the previous two minus the lightest — a Fibonacci-style growth from 1, 2.
Show solution
Approach: push the worst pairing to a Fibonacci-type bound
The binding condition is: heaviest + lightest > (2nd heaviest) + (3rd heaviest), with the analogous rule for every coin acting as the maximum.
Taking the lightest as 1 forces each weight to be at least the sum of the two preceding ones.
2021 balls are arranged in a row and numbered from 1 to 2021. Each ball is coloured green, red, yellow or blue. Among any five consecutive balls there is exactly one red, one yellow and one blue ball. After any red ball, the next ball is yellow. The balls numbered 2, 20 and 202 are green. What colour is the ball numbered 2021?
Show answer
Answer: D — blue
Show hints
Hint 1 of 2
Among any 5 in a row there is one each of red, yellow, blue, so two of every five are green — the colouring repeats every 5.
Still stuck? Show hint 2 →
Hint 2 of 2
Use the green positions (2, 20, 202) to find which residues mod 5 are green, then place red-then-yellow.
Show solution
Approach: show period 5, then locate residues by colour
Any five consecutive balls contain one red, one yellow, one blue and thus two greens, which forces a period-5 colouring.
Balls 2, 20, 202 are green, so positions with remainder 2 and remainder 0 (mod 5) are green.
The red, yellow, blue sit at remainders 1, 3, 4; 'red then yellow' forces red at 3, yellow at 4, leaving blue at remainder 1.
Ball 2021 has remainder 1, so it is blue, choice (D).
Carlos always tells the truth on alternate days. On the other days, he tells only lies. Today he made exactly four of the five statements that follow. Which one was not made today by him?
Show answer
Answer: C — My name is Carlos.
Show hints
Hint 1 of 2
On a truth day every statement he makes is true; on a lie day every one is false.
Still stuck? Show hint 2 →
Hint 2 of 2
Check whether all of A, B and D could be true together — if not, today must be a lie day.
Show solution
Approach: rule out a truth day, forcing a lie day
If today were a truth day, A (prime total) and B (equal counts) force exactly 1 male and 1 female friend, contradicting D (three older male friends).
So today is a lie day: every statement he makes is false.
'My name is Carlos' is true, so on a lie day he cannot make it.
Ana wants to write positive whole numbers, one in each of the squares shown, so that the sums of the four numbers in each row and the four numbers in each column are equal. She has already written some numbers, as shown. She wants to write the missing numbers so that the sum of these six numbers is as small as possible. What is this sum?
Show answer
Answer: B — 24
Show hints
Hint 1 of 2
All four row sums and all four column sums must equal one common value S.
Still stuck? Show hint 2 →
Hint 2 of 2
Each already-filled row or column caps how small its blanks can be, so push S down to the smallest value that still leaves every blank a positive whole number.
Show solution
Approach: minimise the common row/column sum, then add the forced blanks
Every row and every column must add to the same total S, so the given numbers in any line bound S from below.
Choosing the smallest S that keeps all six blanks positive whole numbers and fits every line, the blanks are forced.
Adding those six forced entries gives the minimum sum 24, choice B.
Logic & Word Problemscareful-countingspatial-reasoning
A graph consists of 16 points and several connecting lines, as shown in the diagram. An ant is at point A. With every move the ant can move from the point where it currently is, along one of the connecting lines, to an adjacent point. At which of the points P, Q, R, S and T can the ant be after 2019 moves?
Show answer
Answer: B — only at P, R, S or T, not at Q
Show hints
Hint 1 of 3
Two-colour the 16 points so every line joins different colours.
Still stuck? Show hint 2 →
Hint 2 of 3
After an odd number of moves the ant must sit on the colour opposite to A's.
Still stuck? Show hint 3 →
Hint 3 of 3
2019 is odd, so the ant ends on the opposite colour class from its start.
Show solution
Approach: bipartite two-colouring and parity
The graph is bipartite: colour the points so each edge joins the two colours.
A starts on one colour; after 2019 (odd) moves it must be on the other colour.
Of P, Q, R, S, T only Q shares A's colour, so the ant can be at P, R, S or T but not at Q.
Logic & Word ProblemsNumber Theorycaseworkcareful-counting
Paul wants to write a positive whole number onto every tile in the number wall shown, so that every number is equal to the sum of the two numbers on the tiles that are directly below. What is the maximum number of odd numbers he can write on the tiles?
Show answer
Answer: B — 14
Show hints
Hint 1 of 2
Work in parity (odd/even): a tile is odd exactly when the two below it differ in parity.
Still stuck? Show hint 2 →
Hint 2 of 2
Search the bottom row patterns of the six-row wall to maximise odd tiles.
Show solution
Approach: reduce to parity and optimise the bottom row of the wall
Only odd/even matters: a tile is odd exactly when the two tiles below it have different parities, so the whole wall is fixed once the bottom row's pattern of odds and evens is chosen.
For the six-row (21-tile) wall, testing the bottom-row patterns, the best choice (such as even, odd, odd, even, odd, odd) makes 14 of the 21 tiles odd.
So the maximum number of odd tiles is 14, choice B.
Logic & Word ProblemsGeometry & Measurementsymmetrycasework
30 dancers are standing in a circle facing the centre. The dance instructor shouts “Left” and many of them turn 90° to the left. Unfortunately, some are confused and turn right, so that some dancers are now directly facing each other. All of the ones that are facing each other are shaking their head. It turns out that 10 dancers shake their head. Then the dance instructor says “Turn around” and all of them turn 180° to look in the opposite direction. Again, all of the ones that are directly facing each other shake their head. How many dancers are shaking their head second time round?
Show answer
Answer: A — 10
Show hints
Hint 1 of 2
Two dancers facing each other still form a special pair after both turn 180°.
Still stuck? Show hint 2 →
Hint 2 of 2
Turning everyone around swaps who faces whom, but the count is preserved by symmetry.
Show solution
Approach: track neighbouring pairs facing each other versus back-to-back before and after the turn-around
After turning, each dancer looks clockwise or anticlockwise; a neighbouring pair faces each other when both look toward the gap between them, and is back-to-back when both look away from it.
Going once around the circle, every switch from clockwise-runs to anticlockwise-runs is matched by a switch back, so the number of facing gaps always equals the number of back-to-back gaps.
The first round has 10 head-shakers, i.e. 5 facing gaps, hence also 5 back-to-back gaps; turning everyone 180° reverses all directions, so those 5 back-to-back gaps become the new facing gaps.
That gives 5 facing pairs again, so 10 dancers shake their heads the second time, choice A.
Counting & ProbabilityLogic & Word Problemscaseworkcareful-counting
Three weights are randomly placed on each tray of a beam balance. The balance dips to the right hand side as shown on the picture. The masses of the weights are 101, 102, 103, 104, 105 and 106 grams. For how many percent of the possible distributions is the 106-grams-weight on the right (heavier) side?
Show answer
Answer: B — 80 %
Show hints
Hint 1 of 2
List the splits of the six weights into two trays of three with the right side heavier.
Still stuck? Show hint 2 →
Hint 2 of 2
Among those, count how often the heaviest weight (106) is on the right.
Show solution
Approach: enumerate the heavier-right splits and check where 106 sits
Of the C(6,3)=20 ways to fill the right tray, exactly half (10) make the right side heavier.
Counting those, the 106-gram weight is on the right in 8 of the 10 cases.
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?
Show answer
Answer: B — 12
Show hints
Hint 1 of 3
Only the grey target cells must change colour (an odd number of times); count them.
Still stuck? Show hint 2 →
Hint 2 of 3
Each move flips exactly two cells, so think about how few moves can deliver an odd flip to every grey cell.
Still stuck? Show hint 3 →
Hint 3 of 3
Try pairing up grey cells, or grey cells with a shared white neighbour, to use each move well.
Show solution
Approach: count the cells to flip, then pair them
In the target, 12 grey cells must each be flipped an odd number of times while the 13 white cells stay flipped an even number of times.
Each move flips exactly two adjacent cells, so to give all 12 grey cells an odd flip you need at least 12 moves (no single move can settle two grey cells, since grey cells are never adjacent).
Twelve moves are enough: for each grey cell, flip it together with one chosen white neighbour, arranging the choices so every white cell is touched an even number of times.
Four sportswomen and sportsmen are sitting around a round table for dinner. They do four different sports: ice skating, skiing, hockey and sledging. The person who skies sits to the left of Sandra. The person who ice skates sits opposite Benjamin. Eva and Philipp sit next to each other. A woman sits next to the person who plays hockey. Which sport does Eva do?
Show answer
Answer: A — Ice skating
Show hints
Hint 1 of 3
Note the women are Sandra and Eva; the men are Benjamin and Philipp.
Still stuck? Show hint 2 →
Hint 2 of 3
Seat Eva and Philipp side by side, then use 'ice skater opposite Benjamin' to test who skates.
Still stuck? Show hint 3 →
Hint 3 of 3
Finish with 'a woman sits next to the hockey player' to rule out the wrong seating.
Show solution
Approach: place the two friends, then test the opposite clue
Put Eva and Philipp next to each other; the remaining two seats hold Sandra and Benjamin, opposite Eva and Philipp respectively.
If Benjamin sits opposite Philipp, then the ice skater (opposite Benjamin) is Eva; if Benjamin sits opposite Eva, the skater is Philipp.
In the second seating the hockey player ends up with only men as neighbours, breaking 'a woman sits next to the hockey player', so that seating is impossible.
Only the first seating survives, and there the ice skater opposite Benjamin is Eva, so Eva does ice skating.
The runners Kann, Gu and Ru are favourites to win the marathon. Before the race three experts gave their predictions for the outcome of the race.
Expert 1: “Either Kann or Gu will win.” Expert 2: “If Gu is second Ru will win.” Expert 3: “If Gu is third Kan will not win.” Expert 4: “Either Gu or Ru will come second.”
After the race all four predictions were proven correct. In which order, did the three runners finish the race?
Show answer
Answer: D — Gu, Ru, Kan
Show hints
Hint 1 of 2
List the possible finishing orders and test each against all four statements.
Still stuck? Show hint 2 →
Hint 2 of 2
All four predictions are true at once, so eliminate any order that breaks even one.
Show solution
Approach: test each finishing order against all four statements
Try the finishing orders of the three runners; require all four experts correct at once.
The order Gu first, Ru second, Kan third works: Expert 1 (Kann or Gu wins) holds since Gu wins; Expert 4 (Gu or Ru second) holds since Ru is second; Experts 2 and 3 are 'if Gu is 2nd/3rd...' which never happen, so they hold automatically.
A goldsmith has 12 double-links of chain. Out of these he wants to make a single closed chain with 24 links. What is the minimum number of links that he must open (and close again)?
Show answer
Answer: A — 8
Show hints
Hint 1 of 2
Opening both links of one whole double-link frees two connectors.
Still stuck? Show hint 2 →
Hint 2 of 2
How many pieces must you sacrifice so the freed links join all the rest into one loop?
Show solution
Approach: sacrifice whole pieces to use their links as connectors
There are 12 double-links (pieces). Opening both links of one piece gives two open links that can join other pieces, while using up that whole piece.
If you open p whole pieces you get 2p connector links; the remaining 12 − p pieces need 12 − p joins to close into one loop.
Require 2p ≥ 12 − p, i.e. 3p ≥ 12, so p ≥ 4; opening 4 pieces means opening 2 × 4 = 8 links (A).
Of 5 lamps each one can be set to “ON” or “OFF”. Each time when the switch of one lamp is changed, not only does the status of that particular lamp change but also that of one other lamp chosen at random. If the same switch is changed several times not always the same other lamp changes. Initially all lamps are set to “OFF”. Then 10 switching operations are carried out. After that one can say that
Show answer
Answer: C — definitely not all lamps are switched to “ON”;
Show hints
Hint 1 of 2
Each switch flips its own lamp AND one other — so two lamps flip at once.
Still stuck? Show hint 2 →
Hint 2 of 2
Track the parity (even/odd) of how many lamps are ON.
Show solution
Approach: parity of the number of ON lamps
Every operation flips exactly two lamps, so the number of lamps that are ON changes by an even amount each time.
Starting from 0 (all OFF, even), the count of ON lamps stays even after any number of operations.
All five ON would be 5, an odd number, which is impossible — so we can definitely say not all lamps are switched to ON (C).
Three big boxes P, Q and R are stored in a warehouse. The upper picture on the right shows their placement seen from above. The boxes are so heavy that they can only be rotated 90° around a vertical edge, as indicated in the pictures below. Now the boxes should be rotated to stand against the wall in a certain order. Which arrangement is possible? (Choice E: all four arrangements are possible.)
Show answer
Answer: B
Show hints
Hint 1 of 3
Each allowed move is a 90° turn about a vertical edge, which both slides a box and rotates its top label a quarter turn.
Still stuck? Show hint 2 →
Hint 2 of 3
So a box's final orientation is tied to how far it travelled: track position and label-rotation together.
Still stuck? Show hint 3 →
Hint 3 of 3
Test each pictured line-up and reject any whose letters point the wrong way for the moves needed to reach it.
Show solution
Approach: couple each box's final position to its forced label orientation
Tipping a box 90° about a vertical edge moves it one step and turns its label a quarter turn, so position and orientation change together — they cannot be set independently.
Walking the boxes to the wall one move at a time, each box's letter must end pointing the way that number of quarter-turns dictates.
Comparing the four pictured line-ups, only one has every letter oriented consistently with the moves that put the boxes there.
