Senior competition-style problems (ages 16 to 18)

Answer: D, 4 × 1013² = 4 104 676

1. Factorise: 2026 = 2 × 1013, with 1013 prime.
2. A whole number is a perfect cube exactly when every prime in its factorisation has an exponent that is a multiple of 3.
3. 2026k must therefore contain 2³ and 1013³ at least. 2026 already has 2¹ and 1013¹, so k must supply 2² and 1013².
4. Nothing else is needed, and any other prime in k would have to appear cubed, making k bigger. So k = 2² × 1013².
5. Then 2026k = 2³ × 1013³ = 2026³. ✓ k = 4 × 1013² = 4 104 676 (D).

Why this works: Perfect powers are recognised from prime factorisations: n is a perfect m-th power exactly when every exponent is a multiple of m.

Answer: 29

1. We need 3²⁰²⁶ mod 100. Find a power of 3 that is ≡ 1 (mod 100).
2. 3⁵ = 243 ≡ 43. Square: 3¹⁰ ≡ 43² = 1849 ≡ 49.
3. Square again: 3²⁰ ≡ 49² = 2401 ≡ 1 (mod 100). So the last two digits repeat every 20 powers.
4. 2026 = 20 × 101 + 6, so 3²⁰²⁶ ≡ 3⁶ (mod 100).
5. 3⁶ = 729, so the last two digits are 29.

Why this works: Once a power is ≡ 1 mod 100 the last two digits cycle. Repeated squaring (3⁵, 3¹⁰, 3²⁰) finds the cycle quickly.

Answer: 579

1. Whether 7 divides n² + n + 1 depends only on n mod 7.
2. Test n = 0, 1, 2, 3, 4, 5, 6: n² + n + 1 = 1, 3, 7, 13, 21, 31, 43. Only n ≡ 2 (7) and n ≡ 4 (21) are multiples of 7.
3. Numbers from 1 to 2026 of the form 7k + 2: need 7k + 2 ≤ 2026, so k = 0, 1, …, 289: 290 numbers.
4. Of the form 7k + 4: 7k + 4 ≤ 2026 gives k ≤ 288.9, so k = 0 to 288: 289 numbers.
5. Total: 290 + 289 = 579.

Why this works: Divisibility by 7 of a polynomial in n depends only on n mod 7, so a check of 7 cases settles every n; then it is a counting exercise.

Answer: C, 130

1. 1000 = 2³ × 5³, so its divisors are 2^a5^b with a, b from 0 to 3.
2. A divisor is a perfect square when both exponents are even: a, b ∈ {0, 2}.
3. The square divisors are 1, 4, 25, 100.
4. Their sum is 1 + 4 + 25 + 100 = 130, which also equals (1 + 2²)(1 + 5²) = 5 × 26.
5. Answer: 130 (C).

Why this works: Sums over divisors factorise: the sum of 2^a5^b over the allowed a and b is (sum of allowed 2-powers) × (sum of allowed 5-powers).

Answer: 30

1. Factorise: x² − y² = (x − y)(x + y) = 2025. Put u = x − y and v = x + y, so uv = 2025.
2. Conversely x = (u + v)/2 and y = (v − u)/2, which are integers exactly when u and v have the same parity.
3. 2025 is odd, so any factor pair (u, v) has both factors odd: every factor pair gives a solution, and different pairs give different (x, y).
4. 2025 = 3⁴ × 5² has (4 + 1)(2 + 1) = 15 positive divisors, so 15 ordered pairs (u, v) with u, v > 0.
5. Negative pairs (−u, −v) also multiply to 2025: another 15.
6. Total: 30 integer solutions (for example u = 1, v = 2025 gives x = 1013, y = 1012).

Why this works: A difference of squares factorises, and the change of variables (u, v) ↔ (x, y) is one-to-one as long as u and v have the same parity.

Answer: 5

1. Split by the number of digits of 2026 in base b.
2. Two digits happen for 46 ≤ b ≤ 2026 (b² > 2026). A two-digit palindrome is ‘aa’ = a(b + 1) with 1 ≤ a < b.
3. So b + 1 divides 2026 = 2 × 1013: b + 1 = 1013 (a = 2, b = 1012) or b + 1 = 2026 (a = 1, b = 2025). (b + 1 = 2 is too small.)
4. Three digits happen when b² ≤ 2026 < b³, i.e. 13 ≤ b ≤ 45. A palindrome ‘a c a’ means 2026 = a(b² + 1) + cb.
5. Checking these bases: b = 13 gives digits 11, 12, 11 (11 × 170 + 12 × 13 = 2026); b = 14 gives 10, 4, 10 (10 × 197 + 56); b = 45 gives 1, 0, 1 (2025 + 1). No other base in 13–45 works.
6. For b ≤ 12 the number has four or more digits, and none of those representations is a palindrome.
7. Total: bases 13, 14, 45, 1012, 2025: 5.

Why this works: Base-b digits come from repeated division, and fixing the number of digits turns ‘palindrome’ into a small equation in b. Two-digit palindromes are always multiples of b + 1.

Answer: D, 2026

1. 2026² = 2² × 1013², so n! needs at least two factors of the prime 1013 (the 2s are easy).
2. The prime 1013 appears in n! once for each multiple of 1013 that is at most n (1013² is far bigger than any n here).
3. Two factors need two multiples of 1013: 1013 and 2026. So n ≥ 2026.
4. 2026! contains 1013 and 2026 = 2 × 1013, and plenty of 2s, so it is divisible by 2026².
5. The smallest n is 2026 (D). (1013! up to 2025! contain 1013 only once.)

Why this works: Legendre’s idea: the power of a prime p in n! counts multiples of p, p², … up to n. For a large prime only the multiples of p itself matter.

Answer: 171

1. Let t(n) be the number of tilings of a 2 by n board.
2. Look at the left-hand column. Either a vertical domino covers it (leaving 2 by (n − 1)), or two horizontal dominoes cover the first two columns, or a 2 by 2 square does (each leaving 2 by (n − 2)).
3. So t(n) = t(n − 1) + 2 t(n − 2).
4. Start: t(0) = 1 (the empty board) and t(1) = 1 (one vertical domino).
5. Then t(2) = 3, t(3) = 5, t(4) = 11, t(5) = 21, t(6) = 43, t(7) = 85.
6. t(8) = 85 + 2 × 43 = 171.

Why this works: Tilings of strips satisfy linear recurrences found by asking how the first column is covered. Here t(n) = (2ⁿ⁺¹ + (−1)ⁿ)/3, the Jacobsthal numbers.

Answer: 68

1. A good string splits into runs of equal letters, each of length 1 or 2, alternating A and B.
2. Once the first letter is chosen (2 ways), the letters are forced by the run lengths, so count sequences of 1s and 2s adding to 8.
3. Let c(n) be the number of sequences of 1s and 2s adding to n: the last part is 1 or 2, so c(n) = c(n − 1) + c(n − 2), with c(1) = 1, c(2) = 2.
4. c(3) = 3, c(4) = 5, c(5) = 8, c(6) = 13, c(7) = 21, c(8) = 34.
5. Total: 2 × 34 = 68.

Why this works: Describing strings by run lengths separates ‘which letter’ (fixed once the first is chosen) from ‘how long each run is’, a composition count.

Answer: 120

1. An angle inscribed in a circle is obtuse exactly when it stands on an arc greater than a semicircle; so a triangle is obtuse when its three vertices lie inside an arc smaller than a semicircle.
2. On a regular 12-gon, neighbouring vertices are 30° apart, so ‘less than a semicircle’ means the three vertices lie among 6 consecutive vertices (spanning at most 150°).
3. Count each obtuse triangle from its first vertex going clockwise: choose that vertex (12 ways), then the other two from the next 5 vertices: C(5, 2) = 10.
4. Each obtuse triangle has exactly one such first vertex, so there are 12 × 10 = 120.
5. Check: of C(12, 3) = 220 triangles, 60 are right-angled (a diameter, 6 ways, and a third vertex, 10 ways), and 220 − 120 − 60 = 40 acute.
6. Answer: 120.

Why this works: An inscribed angle is obtuse when it stands on an arc greater than a semicircle. Counting from a canonical starting vertex avoids counting the same triangle several times.

Answer: 20

1. The path starts at (0, 0), where both coordinates are even.
2. From an even–even point, one step makes exactly one coordinate odd.
3. The next step must not make both odd, so it must change the same coordinate again, back to even: the path moves in double steps RR or UU between even–even points.
4. So the path is a route on the grid of even points, from (0, 0) to (6, 6) in steps of 2: 3 double steps right and 3 double steps up.
5. Number of routes: C(6, 3) = 20.

Why this works: Spotting an invariant (you can only be at an odd coordinate for one step at a time) turns the problem into a smaller, familiar one.

Answer: C, 540

1. Each book goes to one of 3 students: 3⁶ = 729 ways, but some leave a student with nothing.
2. Ways where a particular student gets nothing: 2⁶ = 64. Three students: 3 × 64 = 192.
3. Ways where two particular students get nothing (all books to the third) were subtracted twice: there are 3 of them, so add 3 back.
4. No way leaves all three with nothing.
5. Inclusion–exclusion: 729 − 192 + 3 = 540 (C).

Why this works: ‘Every box non-empty’ is the classic inclusion–exclusion count of onto functions: kⁿ − C(k,1)(k − 1)ⁿ + C(k,2)(k − 2)ⁿ − ….

Answer: 282

1. Write every number from 0 to 9999 as four digits d₁d₂d₃d₄ (with leading zeros). 0 has digit sum 0, so it never counts.
2. Count solutions of d₁ + d₂ + d₃ + d₄ = 10 with d_i ≥ 0: stars and bars, C(13, 3) = 286.
3. Remove the ones where some ‘digit’ is 10 or more: that digit is 10 and the others 0: 4 cases. (Two digits ≥ 10 is impossible.)
4. So 286 − 4 = 282 numbers below 10 000.
5. 10 000 has digit sum 1, so the answer is 282.

Why this works: Leading zeros make every number the same length so stars and bars applies; the digit cap of 9 is handled by subtracting the few overflows.

Answer: C, 13

1. Let a(n) be the number of arrangements of 1 to n with every number at most one place from home.
2. Look at number 1. If it stays in position 1, the other n − 1 numbers form the same problem: a(n − 1) ways.
3. If 1 moves to position 2, position 1 must be filled by 2 (no other number may move that far). So 1 and 2 swap, and the rest is the problem for n − 2: a(n − 2) ways.
4. So a(n) = a(n − 1) + a(n − 2), with a(1) = 1 and a(2) = 2.
5. a(3) = 3, a(4) = 5, a(5) = 8, a(6) = 13 (C).

Why this works: Local rules (each item moves at most one place) force the arrangement into fixed points and adjacent swaps, which gives a Fibonacci recurrence.

Answer: B, 5/4

1. Put B = (−6, 0), C = (6, 0) and A = (0, 8): AB = AC = √(36 + 64) = 10. Both centres lie on the axis x = 0.
2. Area = ½ × 12 × 8 = 48 and half-perimeter s = (10 + 10 + 12)/2 = 16, so the inradius r = 48/16 = 3: the incentre is (0, 3).
3. Circumradius R = abc/(4 × area) = (10 × 10 × 12)/192 = 25/4. The circumcentre is 25/4 below A: (0, 8 − 25/4) = (0, 7/4).
4. Distance: 3 − 7/4 = 5/4.
5. Check with Euler’s formula OI² = R(R − 2r) = (25/4)(1/4) = 25/16, so OI = 5/4. ✓ Answer 5/4 (B).

Why this works: Symmetry puts both centres on one line, so the distance is a subtraction. Euler’s formula OI² = R(R − 2r) gives an independent check.

Answer: B, 3 − 2√2

1. Put the right angle at the origin with the legs along the axes (lengths 3 and 4).
2. Inradius of a right triangle: r = (3 + 4 − 5)/2 = 1, so the incircle has centre (1, 1) and radius 1.
3. A circle touching both axes has centre (t, t) and radius t.
4. It touches the incircle from outside, so the distance between centres is 1 + t: √2 (1 − t) = 1 + t.
5. Solve: t(1 + √2) = √2 − 1, so t = (√2 − 1)/(√2 + 1) = (√2 − 1)² = 3 − 2√2.
6. The radius is 3 − 2√2 ≈ 0.17 (B).

Why this works: Circles tangent to both arms of a right angle have centres on the bisector y = x, so tangency becomes one distance equation. Rationalising the denominator gives the neat form.

Answer: A, 32 − 8π

1. |x| + |y| ≤ 4 is a square turned on its corner, with vertices (±4, 0) and (0, ±4); its diagonals are 8, so its area is 8 × 8 ÷ 2 = 32.
2. x² + y² ≥ 8 is everything outside the circle of radius √8 = 2√2 about the origin.
3. The distance from the origin to the side x + y = 4 is 4/√2 = 2√2, exactly the radius: the circle touches each side and lies inside the square.
4. So the region is the square with the whole disc removed: area 32 − π × 8.
5. Answer: 32 − 8π ≈ 6.9 (A).

Why this works: Recognising |x| + |y| ≤ c as a rotated square and comparing the distance to its sides with the radius tells you whether the shapes overlap before any integration.

Answer: B, 18√2

1. The base is an equilateral triangle of side 6: area (√3/4) × 36 = 9√3.
2. The apex is directly above the centre of the base. The centre is 2/3 of the way along a median; the median has length 3√3, so the centre is 2√3 from each base vertex.
3. Height, by Pythagoras on an edge: h = √(6² − (2√3)²) = √(36 − 12) = √24 = 2√6.
4. Volume = ⅓ × base × height = ⅓ × 9√3 × 2√6 = 6√18 = 18√2.
5. Answer: 18√2 ≈ 25.5 (B).

Why this works: Volume = ⅓ × base × height for any pyramid; the height comes from Pythagoras once you know the apex is above the centroid of the base.

Answer: C, 24√2/7

1. The bisector splits the right angle at C into two 45° angles. Let CD = d.
2. Triangle ACD has sides 6 and d with the 45° angle between them: area ½ × 6 × d × sin 45°.
3. Triangle BCD: area ½ × 8 × d × sin 45°.
4. Together they make triangle ABC, area ½ × 6 × 8 = 24: ½ × 14 × d × (√2/2) = 24, i.e. (7√2/2) d = 24.
5. d = 48/(7√2) = 48√2/14 = 24√2/7.
6. CD = 24√2/7 ≈ 4.85 (C).

Why this works: Adding areas with the ½ab sin C formula is a quick route to any angle-bisector length: here d = 2ab cos(C/2)/(a + b).

Answer: B, √74

1. The shortest route on the surface crosses two faces; unfolding those faces flat turns it into a straight line, the diagonal of a rectangle.
2. Unfolding pairs the edges in three ways: the rectangle is (a + b) by c for any split of the edges 3, 4, 5.
3. (3 + 4) by 5: √(49 + 25) = √74.
4. (3 + 5) by 4: √(64 + 16) = √80. (4 + 5) by 3: √(81 + 9) = √90.
5. The shortest is √74 ≈ 8.6. (√50, the space diagonal, would go through the inside of the box.)
6. Answer: √74 (B).

Why this works: Shortest paths on a surface become straight lines once the surface is flattened. Try every way of flattening and keep the best: here, add the two shortest edges.

Answer: 31

1. Strictly inside means x²/16 + y²/9 < 1, so |x| ≤ 3 and |y| ≤ 2 (x = ±4 or y = ±3 give at least 1).
2. For each x, we need y² < 9(1 − x²/16).
3. x = 0: y² < 9, y = −2 to 2 (5). x = ±1: y² < 8.44 (5 each). x = ±2: y² < 6.75 (5 each).
4. x = ±3: y² < 9 × 7/16 ≈ 3.94, so y = −1, 0, 1 (3 each).
5. Total: 5 + 10 + 10 + 6 = 31.

Why this works: Counting lattice points column by column only needs the height of the curve at each integer x. Be careful at the edges: ‘strictly inside’ excludes (0, ±3) and (±4, 0).

Answer: 14

1. Let the roots be α, β, γ. From x³ − 4x² + x + 6 (Vieta): α + β + γ = 4 and αβ + βγ + γα = 1.
2. Use (α + β + γ)² = α² + β² + γ² + 2(αβ + βγ + γα).
3. So α² + β² + γ² = 4² − 2 × 1 = 14.
4. Check: the cubic factorises as (x + 1)(x − 2)(x − 3), and 1 + 4 + 9 = 14. ✓ Answer 14.

Why this works: Symmetric expressions in the roots can be read straight from the coefficients (Vieta’s formulas) without solving the equation.

Answer: C, x = 64

1. Change every logarithm to base 2: log₄ x = (log₂ x)/2 and log₈ x = (log₂ x)/3.
2. Let L = log₂ x. Then L + L/2 + L/3 = 11.
3. L(6 + 3 + 2)/6 = 11L/6 = 11, so L = 6.
4. x = 2⁶ = 64.
5. Check: log₂ 64 + log₄ 64 + log₈ 64 = 6 + 3 + 2 = 11. ✓ x = 64 (C).

Why this works: Change of base, log_b^k x = (log_b x)/k, puts every term in the same unknown.

Answer: 25

1. x² + y² is the squared distance from the origin to the point (x, y), which lies on the line 3x + 4y = 25.
2. The shortest distance from the origin to the line ax + by = c is |c|/√(a² + b²) = 25/√(9 + 16) = 5.
3. It is reached at the foot of the perpendicular, (3, 4): indeed 3 × 3 + 4 × 4 = 25.
4. So the minimum of x² + y² is 5² = 25.
5. (Cauchy–Schwarz gives the same: 25² = (3x + 4y)² ≤ (9 + 16)(x² + y²).) Answer 25.

Why this works: A quadratic expression like x² + y² often has a geometric meaning; minimising distance to a line is the perpendicular.

Answer: A, −2

1. Put x = 2: f(2) + 2f(−1) = 3 × 4 = 12.
2. Put x = −1 (so that 1 − x = 2): f(−1) + 2f(2) = 3.
3. From the second equation, f(−1) = 3 − 2f(2).
4. Substitute into the first: f(2) + 6 − 4f(2) = 12, so −3f(2) = 6.
5. f(2) = −2 (A). (Then f(−1) = 7; check: −2 + 14 = 12 ✓.)

Why this works: Swapping x with 1 − x gives a second equation in the same two unknowns, so a pair of simultaneous equations appears.

Answer: C, 6

1. Let the first term be a and the ratio r, with |r| < 1. Then a/(1 − r) = 12.
2. Squaring each term gives a geometric series with first term a² and ratio r²: a²/(1 − r²) = 48.
3. Divide, using 1 − r² = (1 − r)(1 + r): [a²/((1 − r)(1 + r))] ÷ [a/(1 − r)] = a/(1 + r) = 48/12 = 4.
4. So a = 12(1 − r) and a = 4(1 + r): 12 − 12r = 4 + 4r, giving r = 1/2 and a = 6.
5. Check: 6 + 3 + 1.5 + … = 12 and 36 + 9 + 2.25 + … = 48. ✓ The first term is 6 (C).

Why this works: Factorising 1 − r² = (1 − r)(1 + r) makes the ratio of the two sums simple. Always check |r| < 1 so both series converge.

Answer: C, 3

1. Any solution has |x| = |3 sin x| ≤ 3, so all solutions lie in −3 ≤ x ≤ 3.
2. x = 0 is a solution.
3. For 0 < x ≤ 3 look at g(x) = 3 sin x − x: g(0) = 0 and g′(0) = 3 − 1 = 2 > 0, so g is positive just after 0.
4. g(3) = 3 sin 3 − 3 ≈ 0.42 − 3 < 0, so g crosses zero somewhere in (0, 3). On (0, 3), g″(x) = −3 sin x < 0, so g is concave and can cross zero only once there.
5. g is odd (g(−x) = −g(x)), so there is exactly one negative solution too.
6. Total: 3 solutions (C) (x = 0 and x ≈ ±2.28).

Why this works: Bounding the region (|x| ≤ 3) and using symmetry and concavity turns a transcendental equation into a picture you can trust.

Answer: 111

1. First differences: 7 − 3 = 4 and 13 − 7 = 6. Second difference: 2.
2. For P(x) = ax² + bx + c the second difference is 2a, so a = 1.
3. Then P(1) = 1 + b + c = 3 and P(2) = 4 + 2b + c = 7. Subtracting: 3 + b = 4, so b = 1, and c = 1.
4. P(x) = x² + x + 1. Check P(3) = 13. ✓
5. P(10) = 100 + 10 + 1 = 111.

Why this works: Finite differences identify polynomials: a quadratic has constant second differences equal to twice its leading coefficient.

Answer: B, 15/38

1. Pairs of different numbers from 1 to 20: C(20, 2) = 190.
2. The product fails to be a multiple of 6 when it has no factor 2 or no factor 3. Count those bad pairs.
3. No factor 2: both numbers odd: C(10, 2) = 45.
4. No factor 3: both from the 14 non-multiples of 3: C(14, 2) = 91.
5. Neither factor: both from the 7 numbers coprime to 6 (1, 5, 7, 11, 13, 17, 19): C(7, 2) = 21. Bad pairs: 45 + 91 − 21 = 115.
6. Good pairs: 190 − 115 = 75, probability 75/190 = 15/38 (B).

Why this works: ‘Divisible by 6’ fails when a factor 2 or a factor 3 is missing; counting the failures with inclusion–exclusion is cleaner than counting successes directly.

Answer: B, 91/36

1. Let I_k be 1 if face k appears in the three rolls and 0 if not. The number of different faces is I₁ + … + I₆.
2. Face k is missing from all three rolls with probability (5/6)³ = 125/216, so E[I_k] = 1 − 125/216 = 91/216.
3. Expectation adds, even though the I_k depend on each other: E[total] = 6 × 91/216.
4. 6 × 91/216 = 91/36 ≈ 2.53.
5. Answer: 91/36 (B).

Why this works: Linearity of expectation: the expected value of a sum is the sum of expected values, even when the parts are dependent. Indicator variables make counts easy.

Answer: C, 5/9

1. The pair (x, y) is a random point in the unit square, so probabilities are areas.
2. The region |x − y| ≥ 1/3 is two corner triangles: y ≤ x − 1/3 and y ≥ x + 1/3.
3. Each is a right-angled isosceles triangle with legs 1 − 1/3 = 2/3, area ½ × (2/3)² = 2/9.
4. Together: 4/9.
5. P(|x − y| < 1/3) = 1 − 4/9 = 5/9 (C).

Why this works: Two independent uniform choices are one random point in a square, so probabilities become areas; the complement here is two easy triangles.

Answer: C, 341/1024

1. The number of heads in 10 tosses is a multiple of 3 when it is 0, 3, 6 or 9.
2. Count sequences with k heads: C(10, k). C(10, 0) = 1, C(10, 3) = 120, C(10, 6) = 210, C(10, 9) = 10.
3. Favourable sequences: 1 + 120 + 210 + 10 = 341.
4. All sequences: 2¹⁰ = 1024.
5. Probability: 341/1024 ≈ 0.333 (C).

Why this works: Counts of the form ‘k is a multiple of 3’ split the binomial coefficients into three nearly equal groups (a roots-of-unity filter shows why), which is why the answer is close to 1/3.

Answer: C, 1/5

1. Distinct arrangements of BANANA (3 A, 2 N, 1 B): 6!/(3! 2!) = 60, all equally likely.
2. Arrange the non-A letters B, N, N first: 3!/2! = 3 ways.
3. They leave 4 gaps (before, between, after): _ B _ N _ N _.
4. No two As together means the three As go into three different gaps: C(4, 3) = 4 ways.
5. Good arrangements: 3 × 4 = 12, probability 12/60 = 1/5 (C).

Why this works: ‘No two together’ is the gap method: place the other letters first, then choose different gaps for the ones that must be separated.

Answer: A, 19/217 ≈ 8.8%

1. Think of 10 000 people. 1% have the condition: 100 people; 9900 do not.
2. The test detects 95% of those with it: 95 true positives.
3. It gives a positive for 10% of those without it: 990 false positives.
4. All positives: 95 + 990 = 1085, of whom 95 have the condition.
5. Probability: 95/1085 = 19/217 ≈ 8.8% (A).

Why this works: Bayes’ theorem in natural frequencies: when a condition is rare, false positives from the large healthy group can outnumber true positives.

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.

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.

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.

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.

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.

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.