Junior logic problems (ages 11 to 13)

Answer: D, Dev

1. Amy: not the dog, not the fish, and (allergic) not the cat, so Amy has the rabbit.
2. Who can have the dog? Not Amy, not Ben (clue), not Dev (clue), so Cara has the dog.
3. Left: the cat and the fish for Ben and Dev.
4. Dev does not have the cat, so Dev has the fish (and Ben the cat, which fits Ben’s clue).
5. Dev owns the fish (D).

Why this works: In a logic grid, look for the person (or pet) with only one option left, fix it, and let that knock out options elsewhere. Each clue is used when it bites.

Where it leads: When two people share the same restrictions, they must share the same small set of options: a pigeonhole idea inside a logic grid.

Answer: B, B and E

1. “X is a truth-teller” is true exactly when speaker and X are the same type. So E and B are the same type, and C and D are the same type.
2. Suppose C and D are truth-tellers. D says C and E are the same type, so E is truthful, hence B too. Then at most A lies, but B (truthful) says exactly three lie. Contradiction.
3. So C and D are liars.
4. D’s statement is false: C and E are different types, so E is a truth-teller, and therefore B is too.
5. B is truthful, so exactly three lie: with C, D lying, A must be the third liar.
6. Check A: A says B and C are the same type; B truthful, C liar, so it is false, as a liar’s must be. ✓
7. The truth-tellers are B and E (B).

Why this works: Pick the statement that links two people and test both cases. A truth-teller’s statement must be true and a liar’s false; a consistent assignment is one where every statement checks out.

Where it leads: Statements about how many people are liars are self-referential; checking each possible count is often quicker than checking each person.

Answer: C, Sunday

1. Count the days from 1 March to 25 December.
2. Full months from 1 March to 1 December: 31 + 30 + 31 + 30 + 31 + 31 + 30 + 31 + 30 = 275 days.
3. From 1 December to 25 December: 24 more days. Total 299 days.
4. Days of the week repeat every 7: 299 = 7 × 42 + 5, so the weekday moves on 5 places.
5. Tuesday + 5: Wednesday, Thursday, Friday, Saturday, Sunday.
6. 25 December is a Sunday (C).

Why this works: Days of the week repeat every 7, so only the remainder on division by 7 matters. Starting from 1 March avoids the leap-day question altogether.

Where it leads: Calendar questions are arithmetic modulo 7; Zeller’s congruence turns any date into a weekday by formula.

Answer: 51

1. Count zeros place by place.
2. Units place: 10, 20, …, 300 end in 0: 30 zeros.
3. Tens place (only three-digit numbers have one that can be 0): 100–109 (10 numbers), 200–209 (10), and 300 (1): 21 zeros.
4. Hundreds place is never 0.
5. Total: 30 + 21 = 51.

Why this works: Counting place by place is cleaner than counting number by number, because each place follows a simple repeating pattern.

Where it leads: 0 behaves differently from other digits because we never write leading zeros.

Answer: 5

1. Each pair of teams plays once: 6 × 5 ÷ 2 = 15 games.
2. A game with a winner hands out 3 points in total; a draw hands out 1 + 1 = 2.
3. If all 15 games had winners the total would be 45.
4. Each draw lowers the total by 1. The total was 40, so 45 − 40 = 5 games were draws.
5. Check: 10 wins × 3 + 5 draws × 2 = 30 + 10 = 40. ✓ Answer 5.

Why this works: Instead of tracking every team, look at what each game adds to the total. The total then tells you how many games were of each kind.

Where it leads: If all games had winners the total would be 45; each draw lowers it by 1.

Answer: D, 5th

1. Ola is 1st and Ned 6th, so places 2 to 5 are for Kai, Mo, Lu and Pip.
2. Mo is directly behind Kai and Lu two places behind Kai, so Kai, Mo, Lu are three consecutive places in that order.
3. Inside places 2 to 5 this block is either 2, 3, 4 (Pip 5th) or 3, 4, 5 (Pip 2nd).
4. Pip did not come 2nd, so the block is Kai 2nd, Mo 3rd, Lu 4th.
5. Pip finished 5th (D).

Why this works: Fix the most constrained pieces first (the three runners who must be consecutive). Then only a couple of cases remain, and the last clue picks one.

Where it leads: Placing the largest rigid block first is the quickest way through ordering puzzles.

Answer: A, Ada only

1. Bea says Ada is a liar, so Bea and Ada are different types (if Bea is truthful Ada lies; if Bea lies Ada is truthful).
2. So Cy’s statement ‘Ada and Bea are the same type’ is false: Cy is a liar.
3. If Ada were a liar, Bea would be the only truth-teller, making Ada’s statement true: a contradiction.
4. So Ada tells the truth, Bea and Cy lie, and indeed exactly one is truthful. Answer: Ada only (A).

Why this works: Pinning down the relationship between two people first (they must differ) settles Cy at once and leaves just one case to test.

Where it leads: Truth-teller puzzles are propositional logic: each person gives an equation ‘T_X ⇔ statement’. Computers solve huge versions with SAT solvers.

Answer: 15

1. 100 days = 14 complete weeks + 2 extra days.
2. The 14 complete weeks contain exactly 14 Sundays.
3. The 2 extra days can include one more Sunday (if the year starts on a Saturday or a Sunday).
4. The greatest number is 15.

Why this works: Splitting into complete weeks plus a remainder shows which days can get one extra appearance.

Where it leads: How many Friday the 13ths can a year have? Using the same remainder idea, every year has at least one and at most three.

Answer: 12 hours

1. The clock gains 5 minutes per real hour.
2. To be an hour (60 minutes) ahead it must gain 60 minutes: 60 ÷ 5 = 12 hours.
3. At midnight the correct time is 12:00 and the clock shows 1:00.

Why this works: A steady gain builds up in proportion to time, so ‘how long until it gains X’ is a single division.

Where it leads: A clock that gains 12 hours shows the right time again (on a 12-hour face): here after 144 hours, i.e. 6 days.

Answer: 3

1. If every month had at most 2 birthdays, there would be at most 24 students. There are 30, so some month has at least 3.
2. But 4 is not certain: the birthdays could be 3 in each of six months and 2 in each of the other six (18 + 12 = 30).
3. So the largest certain number is 3.

Why this works: The pigeonhole principle gives the guarantee, and an even spread shows nothing larger is guaranteed: both halves are needed.

Where it leads: In general, n objects in k boxes force some box to hold at least ⌈n/k⌉. A related surprise: with only 23 people, two probably share a birthday.

Answer: 99

1. Pages 1 to 9 use 9 digits, leaving 189 − 9 = 180 digits.
2. Two-digit pages use 2 digits each: 180 ÷ 2 = 90 pages, namely 10 to 99.
3. So the book has 99 pages.

Why this works: Working backwards from the total, one group of page numbers at a time, finds the last page.

Where it leads: Some totals are impossible: no book needs exactly 190 digits. Can you see why? (After page 99, every page adds 3.)

Answer: 18

1. 7th from the front means 6 people are in front of Pat.
2. 12th from the back means 11 people are behind Pat.
3. Total: 6 + 1 + 11 = 18. (Not 7 + 12 = 19, which counts Pat twice.)

Why this works: Adding the two positions counts Pat twice; counting the people on each side avoids the trap.

Where it leads: This ‘off by one’ error, also called the fence-post error, is one of the most common bugs in computer programs.

Answer: C, No, it is impossible

1. If each of 5 people shakes 3 hands, adding up everyone’s count gives 5 × 3 = 15.
2. Each handshake involves two people, so it is counted twice: the sum must be even.
3. 15 is odd, so this is impossible (C).

Why this works: Double counting shows the total of everyone’s handshakes is always even. A parity check rules the arrangement out without trying any.

Where it leads: This is the handshake lemma of graph theory: the number of people who shake an odd number of hands is always even.

Answer: 24

1. Exactly one statement is false. Test each possibility.
2. (4) false: a multiple of 4 and of 6 (so of 12), greater than 20, at most 30: 24, which is indeed not a square. ✓
3. (1) false: a square multiple of 6 above 20: the first is 36, too big. (2) false: a square multiple of 4 above 20, not a multiple of 6: none up to 30 (only 16 and 4 are square multiples of 4, both too small). (3) false: a square multiple of 12 up to 20: none.
4. So the number is 24.

Why this works: ‘Exactly one is false’ gives four cases; checking each and finding only one survivor proves the answer is unique.

Where it leads: Puzzles where statements refer to how many statements are true can be self-referential and even paradoxical; logicians use them to study truth itself.

Answer: 1

1. A move turns two coins: two heads become tails (−2), two tails become heads (+2), or one of each swaps (0).
2. So the number of heads always changes by an even amount: it stays odd (it starts at 5).
3. So 0 heads is impossible. 1 head is reachable: turn coins 1 and 2, then coins 3 and 4.
4. The smallest possible number is 1.

Why this works: An invariant (here, the parity of the number of heads) proves something can never happen, without trying every sequence of moves.

Where it leads: Invariants prove that the ‘15 puzzle’ with two tiles swapped cannot be solved, and that some chessboards with squares removed cannot be tiled by dominoes.

Answer: 36

1. At the end, one champion remains, so 36 players must be knocked out.
2. Each match knocks out exactly one player, and nobody is knocked out any other way.
3. So exactly 36 matches are played, however the byes are arranged.

Why this works: Counting losers instead of rounds turns a messy bracket into a one-line argument.

Where it leads: Looking for a quantity that changes by exactly one with each step is a powerful counting idea: it also shows a bar of chocolate with n squares needs n − 1 snaps to break into squares.

Answer: 8

1. From the centre square, every knight move goes two squares in some direction, off the board, so the centre is never reached (and never left).
2. From a corner the knight can reach an edge square, then another corner, and so on around the outside: corner → edge → corner → … visiting all 8 outer squares.
3. So it can visit 8 squares.

Why this works: Checking the moves from the centre shows it is cut off; the other squares form a single loop of knight moves.

Where it leads: The knight’s graph on a 3 × 3 board is an 8-cycle plus an isolated point. On larger boards, a knight’s tour visiting every square once exists for 5 × 5 and up.

Answer: 3

1. All four digits equal d means the time dd:dd.
2. Hours dd ≤ 23 allows d = 0, 1, 2. Minutes dd ≤ 59 is then fine.
3. So 00:00, 11:11 and 22:22: 3 times.

Why this works: The tightest restriction (hours at most 23) decides everything, so check it first.

Where it leads: How many palindromic times (like 12:21) are there in a day? The minutes are fixed by the hours, so count the hours whose reverse is a valid minute.

Answer: 12

1. Adding the three sides counts each corner twice and each midpoint once: 3S = (1 + 2 + … + 6) + (corners) = 21 + corners.
2. The corners add up to at most 4 + 5 + 6 = 15, so 3S ≤ 36 and S ≤ 12.
3. S = 12 works: corners 4, 5, 6 with 3 between 4 and 5, 1 between 5 and 6, 2 between 6 and 4.
4. The largest S is 12.

Why this works: Adding all the lines at once and seeing which numbers are double counted gives a bound; an example then shows the bound is achieved.

Where it leads: The same argument gives the smallest S (corners 1, 2, 3: S = 9). Bound plus construction is the standard shape of an extremal proof.

Answer: B, Friday

1. 30 = 4 × 7 + 2, so every weekday appears 4 times, except the weekdays of the 1st and 2nd (and the 29th and 30th), which appear 5 times.
2. Friday and Saturday appear five times, so the 1st and 2nd are a Friday and a Saturday.
3. The 2nd is the day after the 1st, so the 1st is a Friday (B).

Why this works: Splitting the month into full weeks plus a remainder shows exactly which weekdays get the extra appearances.

Where it leads: In a 31-day month, three weekdays appear five times. Can a month ever have five Mondays and five Fridays?

Answer: 1

1. With 1 or 2 counters, the player to move takes them all and wins. With 3, whatever they take (1 or 2), the opponent takes the rest: 3 is a losing position.
2. Likewise 6 and 9 are losing: whatever the mover takes, the opponent takes enough to make the total taken 3, reaching 3 less.
3. From 10 the first player takes 1, leaving 9, a losing position for her opponent. She then always makes each round total 3.

Why this works: Working backwards from the end of the game finds the losing positions (multiples of 3); the winning strategy is to always leave one.

Where it leads: If players may take 1 to k counters, the losing positions are the multiples of k + 1. Games like Nim generalise this with binary arithmetic.

Answer: C, 2

1. If C were a knave, the statement ‘I am a knave or A is a knight’ would be true, which a knave cannot say. So C is a knight.
2. Then C’s statement is true; ‘I am a knave’ is false, so ‘A is a knight’ must be true.
3. A is a knight, so A’s statement is true: B is a knave. Check: B says ‘A and C are both knaves’, which is false. ✓
4. Knights: A and C, so 2 (C).

Why this works: A statement that includes ‘I am a knave’ can never be said by a knave if the whole statement would then be true. That fixes C immediately.

Where it leads: ‘I am a knave’ on its own can be said by nobody: a knight would be lying and a knave telling the truth. This is a version of the liar paradox.

Answer: 5

1. With a jumps right and b left, the frog ends at 5a − 3b, whatever the order.
2. We need 5a − 3b = 1. Try small a: a = 1 gives 3b = 4 (no); a = 2 gives 3b = 9, b = 3.
3. So 2 + 3 = 5 jumps work, for example +5, −3, +5, −3, −3.
4. Any solution has 5a ≡ 1 (mod 3), so a = 2, 5, 8, …, and a = 2 gives the fewest jumps: 5.

Why this works: The order of the jumps does not matter for where the frog ends up, so the problem becomes a whole-number equation.

Where it leads: Because 5 and 3 have no common factor, the frog can reach every whole number. If the jumps were 6 and 4 it could only reach even numbers: this is Bézout’s identity.

Answer: 6

1. Work backwards: undoing ×2 halves an even number; undoing +3 subtracts 3.
2. 50 ← 25 ← 22 ← 11 ← 8 ← 4 ← 1. Forwards: 1 +3 = 4, ×2 = 8, +3 = 11, ×2 = 22, +3 = 25, ×2 = 50: 6 presses.
3. Fewer is impossible: with 5 presses there are only 2⁵ = 32 key sequences, and checking them (or the shortest routes to each number) shows none reaches 50.
4. Answer: 6.

Why this works: Working backwards from the target narrows the choices: 50 has only two possible previous values, and 25 (odd) has only one.

Where it leads: Finding shortest routes through all reachable numbers is breadth-first search, a basic algorithm behind satnavs and puzzle solvers.

Answer: C, Dan

1. The pilot and Ben are both ‘the youngest’, so Ben is the pilot.
2. Cara is older than the teacher, so Cara is not the teacher (no one is older than themselves).
3. The teacher is therefore Dan (C), and Cara is the doctor.

Why this works: Two descriptions of the same person (‘the youngest’) must name the same person; and a comparison with someone rules out being that someone.

Where it leads: Reasoning with clues like these is what a computer does in ‘logic programming’ languages such as Prolog.

Answer: 32

1. The outer sheet carries pages 1 and N (the last page) side by side, then 2 and N − 1 on the back. Each sheet pairs a page p with N + 1 − p.
2. Page 5 is paired with page 28, so 5 + 28 = N + 1.
3. N = 32. (That sheet carries pages 5, 6, 27 and 28.)

Why this works: Every pair of facing pages on a sheet has the same sum, N + 1: an invariant that pins down N from one pair.

Where it leads: Printers call this imposition. The same pairing idea is Gauss’s trick for adding 1 to n.

Answer: 9

1. Let Ella be a years old now.
2. a + 3 = 2(a − 3) = 2a − 6.
3. So a = 9. Check: in three years 12, three years ago 6, and 12 = 2 × 6. ✓

Why this works: Writing every age relative to ‘now’ turns the riddle into a one-line equation.

Where it leads: Try ‘in n years I will be k times as old as I was n years ago’: a = n(k + 1)/(k − 1). When is that a whole number?