Senior logic problems (ages 16 to 18)

Answer: A, A, B, F

1. Suppose E is a truth-teller. Then at most one person is truthful, so E is the only one, and A lies. A’s false claim means B and E are the same type, so B is truthful — two truth-tellers. Contradiction: E is a liar.
2. E lies, so at least two people tell the truth.
3. A says B ≠ E. Since E is a liar, this says ‘B is truthful’, so A and B are the same type.
4. C says A ≠ B, which is false: C is a liar.
5. F says B ≠ C, i.e. ‘B is truthful’ (C lies), so F is the same type as B. D says A ≠ F, which is false (A = B = F): D is a liar.
6. If A, B, F also lied, nobody would be truthful, contradicting E’s lie. So A, B, F tell the truth, and B’s ‘exactly three liars’ (C, D, E) checks out.
7. The truth-tellers are A, B and F (A).

Why this works: Once one liar is fixed, statements like ‘X and Y are different’ become ‘X is truthful’ or ‘X lies’, and the types propagate. Count-type claims then decide the last case.

Where it leads: Statements of the form ‘X and Y are different types’ are XOR equations; systems of them are solved by linear algebra mod 2.

Answer: 16

1. Pair up 1 to 30 into the 15 pairs that add to 31: {1, 30}, {2, 29}, …, {15, 16}.
2. Choosing one number from each pair, e.g. 1 to 15, gives 15 numbers with no two adding to 31 (the largest sum is 29). So 15 is not enough.
3. With 16 numbers and only 15 pairs, two chosen numbers lie in the same pair (pigeonhole principle), and those two add to 31.
4. So 16 always works and 15 does not.
5. Answer: 16.

Why this works: The pigeonhole principle needs well-chosen ‘holes’: here the 15 pairs. An example with one fewer shows the bound is exactly right.

Where it leads: 15 numbers (one from each pair, e.g. 1 to 15) are not enough, so 16 is the minimum: bound plus example.

Answer: 16

1. Call a pile size losing if the player about to move loses with best play. 0 is losing (no counter to take: the previous player took the last).
2. Moves remove 1, 2 or 4, which leave remainders 1, 2 or 1 on dividing by 3, never 0. So from a multiple of 3 every move leads to a non-multiple of 3.
3. From a non-multiple of 3, removing 1 or 2 always reaches a multiple of 3.
4. So by induction the losing positions are exactly the multiples of 3, and from any other pile the first player wins by moving to one.
5. The second player wins for 3, 6, …, 48: 16 starting sizes.

Why this works: Classify positions as winning or losing from the end backwards; a pattern (here, multiples of 3, because 1, 2, 4 leave remainders 1, 2, 1) usually appears quickly and can then be proved.

Where it leads: 4 ≡ 1 (mod 3), so the move 4 acts like the move 1 modulo 3, and the game behaves like ‘remove 1 or 2’.

Answer: A, Box 1

1. Label 4 says label 3 is false, so labels 3 and 4 contradict each other: exactly one of them is true.
2. Only one label is true in total, so labels 1, 2 and 5 are all false.
3. Label 2 (‘not in box 1’) is false, so the prize is in box 1.
4. Check the others: label 1 (even box) false ✓; label 5 (box 4) false ✓; label 3 (box 2 or 5) false, so label 4 is true — exactly one true label. ✓
5. The prize is in box 1 (A).

Why this works: A pair of statements that contradict each other always contains exactly one truth. Using that to fill the ‘one true label’ quota makes all the others false at once.

Where it leads: Puzzles where exactly one statement is true are solved by trying each candidate for the true statement in turn.

Answer: 29

1. The date is DDMMYYYY with the year 20ab, i.e. digits D D M M 2 0 a b.
2. Reading backwards must give the same string, so the day is ‘ba’ and the month is ‘02’ (February).
3. So every palindromic date is in February, and its day is the year’s last two digits reversed. Each valid February day gives exactly one year: day ba → year 20ab (e.g. 21 February → 2012).
4. Days 01 to 28 always exist. Day 29 needs year 2092, which is a leap year, so 29 February 2092 exists.
5. Days 30 and 31 do not exist in February.
6. So there are 29 such dates (from 10 February 2001 to 29 February 2092).

Why this works: A palindrome condition fixes half of the string from the other half, so only the free half needs counting, subject to the calendar’s rules.

Where it leads: Palindromic dates are rare and depend on the date format: in MMDDYYYY the count is different.

Answer: 63

1. Statement (1) says N is a multiple of 6, which is even; statement (5) says N is odd. They cannot both be true.
2. Exactly four are true, so exactly one is false, and it must be (1) or (5). Hence (2), (3) and (4) are true: N has 6 divisors, N > 50 and its digits add to 9.
3. If (1) is true, N is a multiple of 6 above 50 with digit sum 9: 54, 72 or 90. Their divisor counts are 8, 12 and 12, not 6. So (1) is false.
4. So (5) is true: N is odd, above 50, with digit sum 9: 63 or 81.
5. 63 = 3² × 7 has (2 + 1)(1 + 1) = 6 divisors; 81 = 3⁴ has 5.
6. N = 63.

Why this works: Find a pair of statements that clash; that tells you where the false one must be, and the rest become hard facts to filter with.

Where it leads: Self-referential statement puzzles reduce to checking which statement is false; contradictions prune the cases quickly.

Answer: 8

1. 0 counters: the player to move has lost, so 0 is L. A pile is W if some square move reaches an L pile, otherwise L.
2. 1 → 0: W. 2 → 1 only: L. 3 → 2: W. 4 → 0: W. 5 → 4 or 1 (both W): L. 6 → 5: W. 7 → 6 or 3: L. 8 → 7: W. 9 → 0: W. 10 → 9, 6, 1: L.
3. Continuing: 11 W, 12 L, 13 W, 14 W, 15 L, 16 W, 17 L, 18 W, 19 W, 20 L.
4. L piles (second player wins): 2, 5, 7, 10, 12, 15, 17, 20: 8.

Why this works: Working backwards from the end of the game labels every position; a position is losing exactly when all moves lead to winning positions.

Where it leads: For this ‘subtract a square’ game there is no simple pattern, and it is known that the losing positions have density zero (Sarközy’s theorem in disguise).

Answer: 25

1. Chessboard colouring with dark corners: 25 dark squares, 24 light. A domino always covers one dark and one light square.
2. So the removed square must be dark, otherwise 25 dark and 23 light remain. That rules out the 24 light squares.
3. Every dark square works: the board minus one dark square can always be tiled (for example, cut the board into a spiral ‘path’ through all 49 squares; removing a dark square splits it into two paths of even length, each tiled by dominoes).
4. Answer: 25.

Why this works: A colouring invariant rules out half the choices at once; a construction shows the rest all work.

Where it leads: Gomory’s theorem: removing one dark and one light square from an 8 × 8 board always leaves a tileable board, by the same Hamiltonian-cycle argument.

Answer: B, A and C

1. If B is a knight, A and C are knaves. Then exactly one of A, B (namely B) is a knight, so C’s statement is true: impossible for a knave. So B is a knave.
2. B is a knave, so ‘B is a knave or …’ is true: A is a knight.
3. Then exactly one of A, B is a knight, so C’s statement is true and C is a knight. Check B’s statement ‘A and C are knaves’: false, as a knave’s should be. ✓
4. Knights: A and C (B).

Why this works: Testing the most restrictive statement (B’s ‘and’) first quickly leads to a contradiction, and everything else follows.

Where it leads: An ‘or’ statement is true as soon as one part is; an ‘and’ statement is false as soon as one part is. Exploiting this asymmetry is the key to most knight–knave puzzles.

Answer: 5

1. Three numbers have a sum divisible by 3 when their remainders are all equal or all different.
2. k = 4 is not enough: 0, 0, 1, 1 (remainders 0, 0, 1, 1) contain no such triple.
3. k = 5: if some remainder appears 3 or more times we are done. Otherwise each of the three remainders appears at most twice, and with 5 numbers all three remainders must appear: take one of each.
4. So k = 5.

Why this works: The pigeonhole principle, with the two ways a triple can work, shows 5 numbers always succeed; an example shows 4 can fail.

Where it leads: Erdős–Ginzburg–Ziv: among any 2n − 1 whole numbers there are n whose sum is divisible by n. Here n = 3 gives 5.

Answer: C, 5

1. There are 4! = 24 possible orders. Each comparison has 2 outcomes, so 4 comparisons can distinguish at most 16 cases: not enough.
2. 5 comparisons suffice: compare A–B and C–D, then the two winners (now you know the heaviest and a chain of three). Insert the remaining object into the chain of three using 2 comparisons (compare with the middle first).
3. So the answer is 5 (C).

Why this works: An information-counting lower bound (log₂ 24 > 4) plus a clever method meeting it settles the minimum.

Where it leads: Sorting n items needs at least log₂ n! ≈ n log₂ n comparisons; merge sort achieves this order. The exact minimum is known only for small n.

Answer: 6

1. With 6 people, take P. Of the other 5, at least 3 are friends of P or at least 3 are strangers to P (pigeonhole). Say 3 friends.
2. If any two of those 3 are friends, they and P form three mutual friends. If not, the 3 are mutual strangers. Either way we are done (the strangers case is the same with roles swapped).
3. With 5 people it can fail: sit them in a circle, neighbours friends, non-neighbours strangers. No three are mutual friends (no triangle of neighbours) and none are mutual strangers.
4. The smallest size is 6.

Why this works: Pigeonhole on one person’s relationships forces a structure; a symmetric pentagon shows 5 is too few.

Where it leads: This is the Ramsey number R(3, 3) = 6. R(4, 4) = 18, but R(5, 5) is still unknown (between 43 and 46).