Intermediate logic problems (ages 13 to 16)

Answer: B, C and D

1. “X is a truth-teller” means the speaker and X are the same type; “X is a liar” means they are opposite types.
2. B says A is truthful, so A and B are the same type.
3. D says B lies: D is the opposite of B. E says D lies: E is the opposite of D, so the same as B. C says E lies: C is the opposite of E, so the same as D.
4. So A, B, E are one type and C, D the other.
5. If A, B, E were truthful, only C and D would lie — 2 liars, but A says 4. Contradiction.
6. So A, B, E lie (3 liars, and A’s ‘four’ is false, as it must be), and the truth-tellers are C and D (B).

Why this works: Statements about other people’s types give ‘same’ or ‘opposite’ links. Following the chain splits everyone into two camps, and one counting statement decides which camp tells the truth.

Where it leads: Encoding ‘same type / opposite type’ links turns knight–knave puzzles into colouring a graph with two colours.

Answer: 2

1. 10 games, each giving out 2 points (2 for a win, or 1 + 1 for a draw): 20 points in total.
2. Let the bottom team have p points. The others have different, larger totals, so at least p + 1, p + 2, p + 3, p + 4.
3. Then 5p + 10 ≤ 20, so p ≤ 2.
4. p = 2 is possible with totals 2, 3, 4, 5, 6. Example with teams T1–T5: T1 beats T2; T2 beats T3; T3 beats T1 and T4; T4 beats T1 and T5; T5 beats T1, T2 and T3; T2 and T4 draw.
5. Totals: T1 2, T2 2 + 1 = 3, T3 4, T4 4 + 1 = 5, T5 6 — all different.
6. The bottom team can have at most 2 points.

Why this works: A bound from totals (the points in the system are fixed) plus one example that meets the bound is a complete answer: that is how most ‘largest possible’ problems are closed.

Where it leads: Making the smallest value large means bunching the totals together: 2, 3, 4, 5, 6 is as tight as distinct totals can be.

Answer: B, 3

1. In a non-leap year, month lengths modulo 7 are: Jan 3, Feb 0, Mar 3, Apr 2, May 3, Jun 2, Jul 3, Aug 3, Sep 2, Oct 3, Nov 2.
2. Add them up to get how many weekdays after 1 January each month starts: Jan 0, Feb 3, Mar 3, Apr 6, May 1, Jun 4, Jul 6, Aug 2, Sep 5, Oct 0, Nov 3, Dec 5.
3. Count repeats: 3 occurs three times (Feb, Mar, Nov); 0, 5 and 6 occur twice; the others once.
4. The weekday of 1 January only relabels the days, so every non-leap year has the same pattern.
5. The largest number is 3 (B).

Why this works: Only month lengths modulo 7 matter. Once the pattern of shifts is written down, the day of 1 January just relabels the days, so every non-leap year behaves the same.

Where it leads: In a leap year the shifts change from March onwards; the most common value then appears only twice, and in different months.

Answer: 4

1. Compute terms: 2027 → 4 + 0 + 4 + 49 = 57 → 25 + 49 = 74 → 49 + 16 = 65 → 36 + 25 = 61 → 36 + 1 = 37.
2. Continue: 37 → 58 → 89 → 145 → 42 → 20 → 4 → 16 → 37, so the terms cycle with length 8 from the 6th term (37).
3. The cycle is 37, 58, 89, 145, 42, 20, 4, 16 (positions 1 to 8 of the cycle), starting at term 6.
4. Term 100 is 100 − 6 = 94 places after term 6. 94 = 8 × 11 + 6, so it is 6 places along the cycle from 37: the 7th entry.
5. The 7th entry is 4, so the 100th term is 4.

Why this works: Digit-square sums quickly drop below 1000 and then must eventually repeat. Once a cycle appears, a far-off term only needs its position modulo the cycle length.

Where it leads: Every starting number ends either at 1 (a ‘happy number’) or in the cycle 4, 16, 37, 58, 89, 145, 42, 20.

Answer: 15

1. Each weight is either in the pan or not: 2⁴ = 16 selections, 15 of them non-empty.
2. Could two selections give the same mass? Each weight is heavier than all the smaller ones together: 2 > 1, 5 > 1 + 2, 10 > 1 + 2 + 5.
3. So the heaviest weight used decides which ‘range’ the total is in, and no two selections collide.
4. The masses are 1, 2, 3, 5, 6, 7, 8, 10, 11, 12, 13, 15, 16, 17, 18 g.
5. That is 15 different masses.

Why this works: If each weight is bigger than the sum of all smaller ones, every selection gives a different total. Here 5 > 3 and 10 > 8, so there are no collisions.

Where it leads: With weights 1, 2, 4, 8 (powers of 2) every total from 1 to 15 is possible exactly once: binary notation.

Answer: 473

1. Write the code as abc. A three-digit number is a multiple of 11 when a − b + c is a multiple of 11 (0, 11, …).
2. a + b + c = 14 and a − b + c differ by 2b, so they have the same parity: a − b + c is even, and between −9 and 18, so it is 0.
3. a − b + c = 0 and a + b + c = 14 give 2b = 14: b = 7 and a + c = 7.
4. With a > c and all digits different: 7 0 (770 repeats 7), 6 1, 5 2, 4 3, giving 671, 572, 473.
5. Below 500: only 473. Check: 473 = 11 × 43 and 4 + 7 + 3 = 14. ✓ The code is 473.

Why this works: The alternating-sum test for 11 plus the digit sum gives two equations; noticing that they must have the same parity pins down the middle digit.

Where it leads: Divisibility by 11 uses the alternating digit sum because 10 ≡ −1 (mod 11).

Answer: 3

1. C says A is a knight, so C and A are the same type (a knight says it truly; a knave says it falsely about a knave).
2. Suppose D is a knave. Then B’s statement is true, so B is a knight. If A is a knight, B and C differ, so C is a knave, but C matches A: contradiction. If A is a knave, B and C are the same, so C is a knight, but C matches A: contradiction. So D is a knight.
3. D is a knight, so at least three are knights, and B (who says D is a knave) is a knave. So A, C and D are all knights.
4. Check: A says B and C differ (knave, knight) ✓; C says A is a knight ✓. There are 3 knights.

Why this works: Pairing up people whose types must match or differ cuts the 16 possibilities down quickly; the remaining case split on D finishes it.

Where it leads: Each statement is a logical equation in true/false variables. With n people there are 2ⁿ cases, so organised deduction beats checking them all.

Answer: 12

1. The first row is any order of 1, 2, 3: 6 ways.
2. The second row must differ from the first in every column. Of the 6 orders, only the two cyclic shifts of the first row do this: 2 ways.
3. The third row is then forced (each column needs its missing number), and it is always a valid order.
4. Total: 6 × 2 = 12.

Why this works: Filling row by row and noticing that later rows are tightly constrained gives a short product.

Where it leads: These are Latin squares. There are 576 of size 4 and 161,280 of size 5; sudoku grids are special 9 × 9 Latin squares (about 6.67 × 10²¹ of them).

Answer: D, The second player

1. With 1 counter left, the player to move must take it and loses.
2. From 2, 3 or 4 counters you can leave exactly 1, so those are wins for the mover. From 5, every move leaves 2, 3 or 4: 5 is losing.
3. In the same way the losing positions are 1, 5, 9, 13, 17, 21: one more than a multiple of 4.
4. 21 is a losing position for the first player: whatever she takes, the second player takes enough to make 4 in that round. The second player wins (D).

Why this works: Working backwards from the end labels every pile size as winning or losing; the pattern repeats every 4 because each round can be forced to total 4.

Where it leads: ‘Last counter loses’ is called misère play. For single-pile games it just shifts the pattern by one; for Nim with several piles it changes the strategy only at the very end.

Answer: 22

1. The minute hand turns 6° per minute and the hour hand 0.5° per minute, so the gap grows by 5.5° per minute.
2. In 12 hours (720 minutes) the gap grows by 5.5 × 720 = 3960° = 11 full turns.
3. In each full turn of the gap it passes 90° once and 270° once: 2 right angles per turn.
4. Total: 11 × 2 = 22 (not 24, because the hands overlap only 11 times in 12 hours).

Why this works: Thinking about the gap between the hands (relative motion) instead of each hand separately turns the clock into one steadily turning angle.

Where it leads: The hands point in exactly opposite directions 11 times in 12 hours too. Do the hour, minute and second hands ever all coincide except at 12 o’clock? (No.)

Answer: 48

1. ABC − CBA = 99(A − C) = 297, so A − C = 3. The middle digit B cancels.
2. C ≥ 1 and A ≤ 9, so (A, C) = (4, 1), (5, 2), …, (9, 6): 6 pairs.
3. B can be any digit except A and C: 8 choices.
4. Total: 6 × 8 = 48.

Why this works: Algebra on the digits shows the difference depends only on A − C, so the condition splits into two independent choices.

Where it leads: Reverse-and-subtract with three digits always gives a multiple of 99. Continue by reversing and adding the result and you often reach 1089: try to prove when.

Answer: C, It is always odd

1. Replacing a and b by |a − b| changes the total by (a + b) − |a − b| = 2 × min(a, b), an even number.
2. So the parity of the total never changes. At the start the total is 55, which is odd.
3. So the last number is always odd: (C). It is not always 1: for example you can end with 3 or 5.
4. So the true statement is it is always odd.

Why this works: An invariant (the parity of the sum) survives every move, whatever choices are made, so it pins down a property of the final number.

Where it leads: Finding invariants (or monovariants, which only change one way) is the standard way to answer ‘can this process ever reach …?’ questions.

Answer: D, 21! − 1

1. a + b + ab + 1 = (a + 1)(b + 1). So if you add 1 to every number on the board, a move multiplies two of these ‘plus one’ values together.
2. So the product of (x + 1) over the board never changes. At the start it is 2 × 3 × … × 21 = 21!.
3. At the end there is one number N with N + 1 = 21!, so N = 21! − 1 (D), whatever the order.

Why this works: Spotting the factorisation reveals an invariant product, which fixes the final answer regardless of the choices made.

Where it leads: The operation a * b = a + b + ab is just multiplication in disguise (shift by 1). Recognising a known operation under a change of variable is a powerful trick.

Answer: 16

1. Two numbers have a sum divisible by 7 when their remainders are (0, 0), (1, 6), (2, 5) or (3, 4).
2. From 1 to 30: remainder 0: 4 numbers; 1: 5; 2: 5; 3: 4; 4: 4; 5: 4; 6: 4.
3. Largest set with no such pair: all of remainder 1 (5), all of remainder 2 (5), all of remainder 3 or 4 (4), and at most one multiple of 7 (1): 15 numbers.
4. So 15 is not enough, but any 16 numbers must contain a good pair: the answer is 16.

Why this works: The pigeonhole principle with carefully chosen ‘holes’ (pairs of remainders) gives the guarantee; an explicit 15-number set shows nothing smaller works.

Where it leads: Erdős–Ginzburg–Ziv: among any 2n − 1 whole numbers, some n of them have a sum divisible by n.

Answer: 3

1. Red and blue form a block ‘red, blue’ in houses (1, 2), (2, 3) or (3, 4). Green is in house 1 or 4.
2. Green in house 1: the block is in (2, 3) or (3, 4). Block (3, 4) puts yellow in house 2, next to green: not allowed. Block (2, 3) leaves yellow in 4: G R B Y. ✓
3. Green in house 4: the block is in (1, 2) or (2, 3). Block (1, 2) puts yellow in 3, next to green: not allowed. Block (2, 3) leaves yellow in 1: Y R B G. ✓
4. Two arrangements fit, but in both the blue house is number 3.

Why this works: Even when a puzzle has more than one solution, the question may have a unique answer: check every arrangement that fits.

Where it leads: Deciding what is certain versus merely possible is the core of constraint reasoning, from timetables to diagnosing faults.

Answer: 361

1. Guesses 123 and 456 show the code has two digits from {1, 2, 3} and one from {4, 5, 6}.
2. If 2 were in the code: guess 264 says it is the only one of 2, 6, 4 and sits first; guess 512 says 5 and 1 are out; so the code uses 2, 3 and one of 4, 5, 6, but 4, 5, 6 are all ruled out. Contradiction. So 2 is out, and the code contains 1 and 3.
3. Guess 512: 1 is in the code, not in place 2; guess 123: 1 is not in place 1. So 1 is in place 3. Then 5 is out, so the third digit is 4 or 6.
4. Guess 264: exactly one of 6 (in place 2) or 4 (in place 3) is right; place 3 is taken by 1, so 6 is in place 2. Then 3 is in place 1 (and indeed not in place 3).
5. The code is 361.

Why this works: Each clue is a constraint; combining clues about which digits are present before worrying about positions keeps the reasoning manageable.

Where it leads: Knuth showed every code in the classic Mastermind game (4 positions, 6 colours, repeats allowed) can be found in at most 5 guesses.

Answer: 57

1. One-digit hours h = 1 to 9: the time h:ab reads h a b, a palindrome when b = h, with a = 0 to 5: 6 times per hour, 54 in total.
2. Two-digit hours 10, 11, 12: hh:ab is a palindrome when a = second hour digit and b = first: 10:01, 11:11, 12:21. 3 times.
3. Total: 54 + 3 = 57.

Why this works: Splitting by the number of digits makes the palindrome condition a simple rule in each case.

Where it leads: On a 24-hour clock the count is different: hours 13 to 15 give 13:31, 14:41, 15:51, while 16:61 is not a time. Try counting it.

Answer: 1

1. Bouton’s rule: the player to move loses exactly when the ‘Nim-sum’ (binary addition without carrying) of the heaps is 0.
2. 3 = 011, 4 = 100, 5 = 101. Nim-sum: 011 ⊕ 100 ⊕ 101 = 010 = 2, not 0, so the first player can win.
3. A winning move makes the Nim-sum 0: change a heap h to h ⊕ 2 if that is smaller. 3 → 1 works (1 < 3); 4 → 6 and 5 → 7 are increases, not allowed.
4. So the only winning move is to take 2 from the heap of 3: 1 move.

Why this works: The binary columns stay balanced after a move to Nim-sum 0, and any move from such a position unbalances them, so the winner can always restore the balance.

Where it leads: The Sprague–Grundy theorem says every impartial game is equivalent to a Nim heap, so this binary trick solves a huge family of games.

Answer: 2037

1. Each normal year shifts the weekday of 1 January by 1, each leap year by 2. The target year must start on a Thursday and must not be a leap year.
2. Shifts after 2026, 2027, 2028 (leap), …: running totals 1, 2, 4, 5, 6, 7 (reaching 2032), but 2032 is a leap year, so its calendar differs from February.
3. Continuing: 2033: 9, 2034: 10, 2035: 11, 2036: 12, 2037: 14 ≡ 0 (mod 7), and 2037 is not a leap year.
4. The answer is 2037.

Why this works: Tracking the weekday shift modulo 7 year by year, while respecting leap years, finds the first exact match.

Where it leads: Calendars repeat every 6, 11 or 28 years depending on where the year sits in the leap cycle; the full Gregorian calendar repeats every 400 years (146097 days, exactly 20871 weeks).

Answer: 15

1. People liking tea or coffee = 25 + 30 − (both) ≤ 40.
2. So both ≥ 55 − 40 = 15.
3. 15 is possible: 10 like only tea, 15 like both, 15 like only coffee. The smallest number is 15.

Why this works: Inclusion–exclusion turns the overlap into a single inequality; an example shows the bound is reached.

Where it leads: The largest possible overlap is 25 (everyone who likes tea also likes coffee). Bounds like these are called Bonferroni inequalities in probability.

Answer: 7

1. There are C(7, 2) = 21 calls. In a round at most 3 calls can happen, since 7 people make at most 3 pairs.
2. So at least 21/3 = 7 rounds are needed.
3. 7 rounds are enough: seat the friends at the corners of a regular heptagon; in round k, the person at corner k sits out and the others pair up along chords parallel to the ‘opposite’ side. Every pair meets exactly once.
4. The answer is 7.

Why this works: A counting bound shows at least 7 rounds; a symmetric construction shows 7 suffice.

Where it leads: This is a round-robin schedule (a 1-factorisation of the complete graph). Sports leagues use the same ‘circle method’ to build fixture lists.

Answer: 26

1. With 1 disc: A → B → C: 2 moves.
2. With n discs: move the top n − 1 discs A → C, move the big disc A → B, move n − 1 discs C → A, big disc B → C, then n − 1 discs A → C. So m(n) = 3m(n − 1) + 2.
3. m(1) = 2, m(2) = 8, m(3) = 26.
4. The answer is 26 (= 3³ − 1).

Why this works: The largest disc forces a rigid plan; the smaller discs repeat the same problem three times, giving a recurrence.

Where it leads: m(n) = 3ⁿ − 1: this version visits every one of the 3ⁿ legal arrangements of the discs exactly once, tracing a path through all of them.

Answer: 4

1. If there were 10 or more blue balls, you could pick 10 blue ones with no red: so there are at most 9 blue, i.e. at least 11 red.
2. Similarly at most 14 red balls.
3. So the number of red balls is 11, 12, 13 or 14, and each of these satisfies both conditions.
4. That is 4 possibilities.

Why this works: ‘Any k contain a red’ is the same as ‘fewer than k are blue’: turning a statement about all selections into a count is the key step.

Where it leads: Translating ‘every subset of size k has property P’ into a statement about the whole set is the heart of extremal combinatorics (Turán, Ramsey).

Answer: 3

1. Each person who is not the tallest must be shown to be shorter than someone, so must lose at least one comparison.
2. Each comparison has exactly one loser, so at least 3 comparisons are needed.
3. 3 suffice: compare A with B, the winner with C, and that winner with D.
4. The answer is 3.

Why this works: Counting ‘losers’ (like counting knocked-out players in a tournament) gives a lower bound that a simple method meets.

Where it leads: Finding both the tallest and the shortest of n needs about 3n/2 comparisons, not 2n; finding the second tallest needs n + ⌈log₂ n⌉ − 2.

Answer: 1

1. A knight cannot say ‘I am a knave and …’ (the first part would be false). So A is a knave.
2. A’s statement is then false. Its first part (‘I am a knave’) is true, so the second part must be false: B is a knave.
3. B says A is a knight, which is false: consistent with B being a knave. C says B is a knave, which is true, so C is a knight.
4. Exactly 1 knight (C).

Why this works: A statement containing ‘I am a knave’ immediately fixes the speaker as a knave; then the rest of the statement must be what makes it false.

Where it leads: A compound statement ‘P and Q’ is false when at least one part is false. Careful handling of ‘and’, ‘or’ and ‘not’ is the start of formal logic.

Answer: D, Press the centre and the four edge-middle lamps

1. Order does not matter and double presses cancel, so a plan is a set of lamps; a lamp ends on when it is switched an odd number of times.
2. Check D (centre and four edge-middles, a plus shape). Centre: switched by all 5 presses: odd, on. Each edge-middle: by itself and the centre: 2, off. Each corner: by its two edge-middle neighbours: 2, off. ✓
3. The others fail: A also lights the four edge-middles; B lights the corners; C lights the corners and leaves the edge-middles switched 3 times (on).
4. So the answer is press the centre and the four edge-middle lamps (D). (Checking all 512 sets shows it is the only one.)

Why this works: Working with parities (odd or even numbers of switches) turns the puzzle into simple counting, independent of the order of presses.

Where it leads: This is the game ‘Lights Out’. It is linear algebra over the field with two elements: on the 3 × 3 board every pattern is reachable in exactly one way, but on 4 × 4 or 5 × 5 boards some patterns are impossible.

Answer: E, It cannot be done

1. The moves turn cards 1-2-3, 2-3-4 or 3-4-5. Order does not matter, and using a move twice cancels, so each move is effectively used once or not at all. Every card must be turned an odd number of times.
2. Card 1 is turned only by 1-2-3, so that move must be used. Card 5 is turned only by 3-4-5, so that move must be used.
3. Card 2 is turned by 1-2-3 and 2-3-4. To be turned an odd number of times, 2-3-4 must not be used. But then card 3 is turned twice (by 1-2-3 and 3-4-5): even. Contradiction.
4. So it cannot be done (E), however many moves you make.

Why this works: Reducing the problem to ‘which moves, each used once or not’ leaves only 8 cases, and parity forces a contradiction without trying them all.

Where it leads: With 6 cards it can be done (turn 1-2-3 and 4-5-6). Which numbers of cards work? This is linear algebra over the numbers mod 2, the same maths as error-correcting codes.