Intermediate competition-style problems (ages 13 to 16)

Answer: 40

1. The last two digits of n² depend only on the last two digits of n, so work with n = 10a + b, where b is the units digit and a the tens digit.
2. n² ends in 1, so b² ends in 1: b = 1 or b = 9.
3. b = 1: n² = 100a² + 20a + 1. Its tens digit is the units digit of 2a, which must be 2, so a ends in 1 or 6: n ends in 11 or 61.
4. b = 9: n² = 100a² + 180a + 81. Its tens digit is the units digit of 18a + 8, i.e. of 8a + 8, which must be 2, so 8a ends in 4: a ends in 3 or 8: n ends in 39 or 89.
5. Check: 11² = 121, 39² = 1521, 61² = 3721, 89² = 7921. ✓
6. So 4 numbers in every block of 100, and 1 to 1000 is 10 blocks: 4 × 10 = 40.

Why this works: Working modulo 100 means you only ever look at the last two digits. Expanding (10a + b)² shows exactly which digit of n controls which digit of n².

Answer: D, 37

1. N has twelve 1s. Split it as 111111 followed by 111111: N = 111111 × 1000000 + 111111 = 111111 × 1000001.
2. Split 111111 the same way: 111111 = 111 × 1000 + 111 = 111 × 1001.
3. 111 = 3 × 37 and 1001 = 7 × 11 × 13.
4. 1000001 = 101 × 9901, and neither 101 nor 9901 has a prime factor below 100 (101 is prime; 9901 is prime).
5. So the prime factors of N below 100 are 3, 7, 11, 13 and 37. In particular 31 and 41 do not divide N.
6. The largest is 37 (D).

Why this works: Numbers made of repeated digits split along their pattern: a block of 1s of length ab is (block of length a) × (1 000…01 …). 1001 = 7 × 11 × 13 is a factorisation worth knowing.

Answer: 8

1. Rearrange ab = 2(a + b) + 20 as ab − 2a − 2b = 20.
2. Add 4 to both sides so the left factorises: ab − 2a − 2b + 4 = 24, i.e. (a − 2)(b − 2) = 24.
3. a and b are positive, so a − 2 ≥ −1 and b − 2 ≥ −1. Both brackets negative would need (−1)(−24), impossible, and one negative makes the product negative. So both brackets are positive.
4. Positive factor pairs of 24: 1 × 24, 2 × 12, 3 × 8, 4 × 6, and each in either order.
5. That gives (a, b) = (3, 26), (26, 3), (4, 14), (14, 4), (5, 10), (10, 5), (6, 8), (8, 6).
6. Check one: 6 × 8 = 48 and 2(6 + 8) + 20 = 48. ✓ There are 8 ordered pairs.

Why this works: Adding the right constant makes xy + px + qy factorise as (x + q)(y + p) − pq. Then a Diophantine equation becomes ‘list the factor pairs’.

Answer: E, 6

1. Work modulo 7 (remainders on dividing by 7).
2. Powers of 2: 2, 4, 8 ≡ 1. So 2³ ≡ 1 and powers of 2 cycle with length 3.
3. 100 = 3 × 33 + 1, so 2¹⁰⁰ = (2³)³³ × 2 ≡ 1 × 2 = 2.
4. Powers of 3: 3, 2, 6, 4, 5, 1, so 3⁶ ≡ 1 and the cycle has length 6.
5. 100 = 6 × 16 + 4, so 3¹⁰⁰ ≡ 3⁴ = 81 = 77 + 4 ≡ 4.
6. Sum: 2 + 4 = 6, so the remainder is 6 (E).

Why this works: Once some power of a number leaves remainder 1, higher powers repeat in a cycle, so a huge exponent only matters through its remainder on dividing by the cycle length.

Answer: D, 2025

1. If n = p^a q^b … (prime factorisation), n has (a + 1)(b + 1)… divisors.
2. 15 = 15 or 15 = 3 × 5, so n = p¹⁴ or n = p⁴ q² for different primes p, q.
3. n is a multiple of 15, so both 3 and 5 divide n. That rules out p¹⁴ (one prime only) and forces {p, q} = {3, 5}.
4. The two options are 3⁴ × 5² = 81 × 25 = 2025 and 3² × 5⁴ = 9 × 625 = 5625.
5. The smaller is 2025 (D). (225 = 3²5² is a trap: it has only 3 × 3 = 9 divisors.)

Why this works: The divisor-count formula turns ‘exactly k divisors’ into ‘write k as a product’. To make n small, give the largest powers to the smallest primes.

Answer: 10

1. Each zero at the end needs one factor 10 = 2 × 5, so count the 2s and the 5s.
2. 1 × 3 × 5 × … × 99 is a product of odd numbers, so it has no factor 2. The only 2s come from 2¹⁰: ten of them.
3. Factors of 5 in the odd product: the odd multiples of 5 up to 99 are 5, 15, 25, …, 95, which is 10 numbers.
4. 25 and 75 each contain 5 twice, adding 2 more: twelve 5s in total.
5. Pairs 2 × 5: min(10, 12) = 10, so there are 10 zeros.

Why this works: Trailing zeros count pairs 2×5. Usually 2s are plentiful and 5s are scarce; here the odd product has no 2s, so the 2s run out first.

Answer: 11

1. Write the number as 100a + 10b + c with a from 1 to 9 and b, c from 0 to 9.
2. The condition is 100a + 10b + c = 19(a + b + c) = 19a + 19b + 19c.
3. Rearrange: 81a = 9b + 18c, and divide by 9: 9a = b + 2c.
4. b + 2c is at most 9 + 18 = 27, so a is 1, 2 or 3.
5. a = 1: b + 2c = 9, with c = 0 to 4 (b = 9, 7, 5, 3, 1): 5 numbers (e.g. 190 = 19 × 10).
6. a = 2: b + 2c = 18, with c = 5 to 9 (b = 8, 6, 4, 2, 0): 5 numbers.
7. a = 3: b + 2c = 27 needs b = c = 9: 1 number, 399 = 19 × 21.
8. Total: 5 + 5 + 1 = 11.

Why this works: Digit problems collapse to small linear equations once the number is written in place-value form; bounding the digits keeps the case list short.

Answer: C, 120

1. LEVELS has six letters: L twice, E twice, V once, S once.
2. All arrangements: 6! ÷ (2! × 2!) = 720 ÷ 4 = 180 (divide out swaps of identical letters).
3. Arrangements with the two Es together: glue them into one block EE, leaving five items L, L, V, S, EE.
4. Those arrange in 5! ÷ 2! = 60 ways.
5. Not together: 180 − 60 = 120 (C).

Why this works: ‘Not together’ = all − together, and ‘together’ is counted by gluing. Dividing by factorials of repeated letters removes arrangements that look identical.

Answer: 40

1. 5 is in the subset, so 4 and 6 cannot be (no two consecutive numbers).
2. The rest splits into two separate pieces, {1, 2, 3} and {7, 8, 9, 10}, which cannot interact because 4 and 6 are out.
3. From {1, 2, 3} with no two consecutive: ∅, {1}, {2}, {3}, {1, 3}: 5 choices.
4. From {7, 8, 9, 10}: ∅, four single numbers, {7, 9}, {7, 10}, {8, 10}: 8 choices.
5. The choices are independent, so multiply: 5 × 8 = 40.

Why this works: Fixing one element splits the problem into independent pieces, and the counts multiply. The counts 5 and 8 are Fibonacci numbers again.

Answer: 18

1. A path to (4, 4) has 4 R steps and 4 U steps. Changing direction 3 times means it is made of 4 straight runs.
2. Runs alternate, so the pattern is R, U, R, U or U, R, U, R.
3. For R U R U: the 4 R steps split into 2 non-empty runs: 1 + 3, 2 + 2 or 3 + 1 (3 ways). The 4 U steps likewise: 3 ways.
4. So 3 × 3 = 9 paths start with R.
5. By symmetry 9 paths start with U.
6. Total: 18.

Why this works: Describing a path by its runs rather than its steps turns ‘count turns’ into ‘split a number into positive parts’, which is a stars-and-bars count.

Answer: D, 495

1. The first digit is at least 1 and digits never decrease, so every digit is at least 1: 0 never appears.
2. The number is determined by how many 1s, 2s, …, 9s it uses (4 digits in total), because the order is forced.
3. So count the ways to choose 4 digits from 1–9 with repetition allowed: x₁ + x₂ + … + x₉ = 4 with each x ≥ 0.
4. Stars and bars: 4 stars and 8 bars, C(12, 4) = 12 × 11 × 10 × 9 ÷ 24 = 495.
5. Answer: 495 (D).

Why this works: When order is forced, you only choose a multiset. Choosing k items from n types with repetition gives C(n + k − 1, k).

Answer: 44

1. Give each child 1 sweet first. Now share the other 6 so that each child gets at most 3 more.
2. Without the upper limit: 6 identical sweets to 4 children is stars and bars, C(6 + 3, 3) = C(9, 3) = 84.
3. Remove the shares where some child gets 4 or more extra. Pick that child (4 ways), give them 4, and share the remaining 2 freely: C(2 + 3, 3) = C(5, 3) = 10. That is 4 × 10 = 40.
4. Two children cannot both get 4 extra (that needs 8 > 6), so nothing was removed twice.
5. Answer: 84 − 40 = 44.

Why this works: Lower limits are handled by handing them out first; upper limits by inclusion–exclusion on the cases that break them.

Answer: 200

1. Any three dots make a triangle unless they lie on one straight line.
2. Choices of three dots from 12: C(12, 3) = 220.
3. Collinear triples in rows: 3 rows of 4 dots, C(4, 3) = 4 each: 12.
4. In columns: 4 columns of 3 dots, 1 each: 4.
5. On diagonals of slope 1 or −1: lines with 3 dots start in the first or second column of the bottom (or top) row: 2 each way, 4 in total. No other slope passes through 3 dots in a 3 by 4 array.
6. Collinear triples: 12 + 4 + 4 = 20, so triangles: 220 − 20 = 200.

Why this works: Triangles = triples − collinear triples. The work is in finding every line with three or more dots, including the slanted ones.

Answer: 102

1. Committees of 4 from 9 people: C(9, 4) = 126.
2. Remove the all-boy ones, C(5, 4) = 5, and the all-girl one, C(4, 4) = 1: 120 mixed committees.
3. Now remove mixed committees containing both Asa and Bo. With both in, choose 2 more from the other 7: C(7, 2) = 21.
4. Of those 21, the ones with no girl use 2 of the 3 other boys: C(3, 2) = 3. They are all-boy, already removed. So 21 − 3 = 18 mixed ones remain to remove.
5. Answer: 120 − 18 = 102.

Why this works: Apply one restriction at a time and be careful not to subtract the same committee twice — the all-boy committees with Asa and Bo were already gone.

Answer: C, 63

1. Put A = (0, 0), B = (12, 0), C = (12, 12), D = (0, 12). Then E = (12, 6) is the midpoint of BC and F = (3, 12) has DF = 3.
2. The square has area 12 × 12 = 144.
3. Corner triangle ABE: legs AB = 12, BE = 6, area ½ × 12 × 6 = 36.
4. Corner triangle ECF: legs EC = 6, CF = 12 − 3 = 9, area ½ × 6 × 9 = 27.
5. Corner triangle FDA: legs FD = 3, DA = 12, area ½ × 3 × 12 = 18.
6. Triangle AEF = 144 − 36 − 27 − 18 = 63 (C).

Why this works: A slanted triangle inside a rectangle is best found by subtraction: the pieces left over are right-angled and easy.

Answer: B, 6

1. Hypotenuse: √(20² + 21²) = √(400 + 441) = √841 = 29.
2. Area: ½ × 20 × 21 = 210. Half-perimeter: s = (20 + 21 + 29)/2 = 35.
3. Joining the incentre to the vertices splits the triangle into three triangles of height r, so area = r × s.
4. r = 210 ÷ 35 = 6.
5. Check with the right-angle shortcut r = (a + b − c)/2 = (20 + 21 − 29)/2 = 6. ✓ Answer 6 (B).

Why this works: Splitting the triangle into three triangles from the incentre gives Area = r × s, where s is the half-perimeter. It works for every triangle.

Answer: 8

1. The perpendicular from the centre to a chord bisects it, making a right-angled triangle: (half-chord)² + d² = r².
2. Chord 6: half-chord 3, so d = √(25 − 9) = 4.
3. Chord 8: half-chord 4, so d = √(25 − 16) = 3.
4. Chords on opposite sides of the centre: 4 + 3 = 7 apart.
5. Chords on the same side: 4 − 3 = 1 apart.
6. Sum of the two possible distances: 7 + 1 = 8.

Why this works: The perpendicular from the centre bisects a chord, giving a right-angled triangle with the radius as hypotenuse. ‘Two possible answers’ is a cue to draw both configurations.

Answer: B, 8 + 8√2

1. Extend four sides of the octagon to form a square. The four cut-off corners are right-angled isosceles triangles, each with hypotenuse 2 (a side of the octagon).
2. A right-angled isosceles triangle with hypotenuse 2 has legs 2/√2 = √2.
3. The square’s side is √2 + 2 + √2 = 2 + 2√2, so its area is (2 + 2√2)² = 4 + 8√2 + 8 = 12 + 8√2.
4. The four corners have area 4 × ½ × √2 × √2 = 4.
5. Octagon: 12 + 8√2 − 4 = 8 + 8√2 (B), about 19.3.

Why this works: A regular octagon is a square with its corners snipped off. Building up to a simpler shape and subtracting is often easier than splitting into pieces.

Answer: 8

1. Heron’s formula: half-perimeter s = (10 + 17 + 21)/2 = 24.
2. Area = √(s(s − a)(s − b)(s − c)) = √(24 × 14 × 7 × 3) = √7056 = 84.
3. For each side, altitude = 2 × area ÷ side, so the longest side has the shortest altitude.
4. Altitude to 21: 168 ÷ 21 = 8. (To 17: about 9.9; to 10: 16.8.)
5. Shortest altitude: 8.

Why this works: Every altitude times its base gives twice the same area, so altitudes are inversely proportional to the sides they meet.

Answer: D, 1/8

1. The water fills a cone at the bottom (the point), with the same shape as the whole cone.
2. Its height is half the full height, so every length (height and radius) is half: the scale factor is k = 1/2.
3. Volumes of similar solids scale by k³: (1/2)³ = 1/8.
4. Check with the formula: ⅓π(r/2)²(h/2) = (1/8) × ⅓πr²h. ✓
5. The water is 1/8 of the volume (D).

Why this works: For similar solids, lengths scale by k, areas by k² and volumes by k³. Half the height holds only an eighth of the volume.

Answer: 16

1. Lattice points on the circle are integer pairs (x, y) with x² + y² = 65.
2. Squares up to 65: 0, 1, 4, 9, 16, 25, 36, 49, 64. Pairs adding to 65: 1 + 64 and 16 + 49 (65 − 0, 65 − 4, 65 − 9, 65 − 25, 65 − 36 are not squares).
3. So {|x|, |y|} = {1, 8} or {4, 7}.
4. Each pair gives 2 orders and 4 sign patterns: 8 points, e.g. (1, 8), (8, 1), (−1, 8), ….
5. Total: 8 + 8 = 16.

Why this works: Lattice points on x² + y² = n come from ways to write n as a sum of two squares; symmetry (swaps and signs) multiplies each one by up to 8.

Answer: D, 119

1. Square x − 1/x = 3: x² − 2 + 1/x² = 9.
2. So x² + 1/x² = 11.
3. Square again: x⁴ + 2 + 1/x⁴ = 121.
4. So x⁴ + 1/x⁴ = 119 (D).

Why this works: Squaring x ± 1/x always produces a constant cross term (±2), so you can climb to higher powers without ever finding x.

Answer: 10

1. For x² − 7x + k = 0 the roots add to 7 and multiply to k.
2. The roots add to 7 and differ by 3, so the larger is (7 + 3)/2 = 5 and the smaller is (7 − 3)/2 = 2.
3. k = 5 × 2 = 10.
4. Check: x² − 7x + 10 = (x − 2)(x − 5). ✓ Answer 10.

Why this works: For x² − sx + p = 0 the roots add to s and multiply to p. Using these facts is often faster than the quadratic formula.

Answer: C, 8

1. f(f(x)) = a(ax + b) + b = a²x + ab + b.
2. This equals 9x + 8 for every x, so the coefficients match: a² = 9 and (a + 1)b = 8.
3. a > 0, so a = 3, and then 4b = 8, b = 2.
4. So f(x) = 3x + 2. Check: f(f(x)) = 3(3x + 2) + 2 = 9x + 8. ✓
5. f(2) = 6 + 2 = 8 (C).

Why this works: Two polynomials that agree for every x have the same coefficients, so one identity gives several equations.

Answer: B, x = 9

1. Square both sides: x + 7 = (x − 5)² = x² − 10x + 25.
2. Rearrange: x² − 11x + 18 = 0, i.e. (x − 2)(x − 9) = 0, so x = 2 or x = 9.
3. Squaring can add false solutions, so check both in the original equation.
4. x = 9: √16 = 4 and 9 − 5 = 4. ✓
5. x = 2: √9 = 3 but 2 − 5 = −3. ✗ (A square root is never negative.)
6. Only x = 9 (B).

Why this works: Squaring can create false solutions, because a = b and a = −b square to the same thing. A square root is never negative, so the right side must be ≥ 0.

Answer: 550

1. Let the first term be a and the common difference d. Sum of the first n terms: S_n = (n/2)(2a + (n − 1)d).
2. Terms 11–20 add to S₂₀ − S₁₀ = 500 − 150 = 350.
3. Each term in a block of ten is 10d bigger than the matching term ten places earlier, so each block’s sum is 100d bigger than the one before.
4. 350 − 150 = 100d, so 100d = 200 (d = 2, and from S₁₀ = 150, a = 6).
5. Terms 21–30: 350 + 200 = 550. (Check: S₃₀ = 15(12 + 58) = 1050 = 500 + 550.)

Why this works: Consecutive equal-length blocks of an arithmetic sequence form another arithmetic sequence. Seeing the structure saves solving for the first term at all.

Answer: D, 24

1. Let u = 2x and v = 3y. Then u + v = 24 and xy = uv/6.
2. For a fixed sum, a product is largest when the two parts are equal (AM–GM: uv ≤ ((u + v)/2)² = 144).
3. So uv ≤ 144, with equality when u = v = 12.
4. Then x = 6, y = 4, and xy = 144/6 = 24.
5. The largest value of xy is 24 (D).

Why this works: For a fixed sum, a product is largest when the parts are equal. Choosing parts (2x and 3y) whose sum is fixed makes the idea apply.

Answer: 10

1. |x − 3| + |x + 2| is the distance from x to 3 plus the distance from x to −2 on the number line.
2. For −2 ≤ x ≤ 3 the two distances add to exactly 5, which is less than 11: x = −2, −1, 0, 1, 2, 3 all work (6 numbers).
3. For x > 3: (x − 3) + (x + 2) = 2x − 1 < 11 gives x < 6: x = 4, 5.
4. For x < −2: (3 − x) + (−2 − x) = 1 − 2x < 11 gives x > −5: x = −4, −3.
5. Total: 6 + 2 + 2 = 10 whole numbers (−4 to 5).

Why this works: Reading |x − a| as a distance turns the inequality into a picture: the sum of distances to two points is constant between them and grows by 2 per step outside.

Answer: B, 7/36

1. Two dice: 36 equally likely ordered outcomes.
2. The larger number is at most 4 when both dice are at most 4: 4 × 4 = 16 outcomes.
3. The larger number is at most 3 when both are at most 3: 3 × 3 = 9 outcomes.
4. The larger is exactly 4 in 16 − 9 = 7 outcomes: (4,1), (4,2), (4,3), (4,4), (1,4), (2,4), (3,4).
5. Probability: 7/36 (B).

Why this works: ‘Max ≤ k’ is easy (every die ≤ k), so ‘max = k’ is a difference of two easy counts.

Answer: C, 5/9

1. Cards 1–9: five odd (1, 3, 5, 7, 9) and four even (2, 4, 6, 8).
2. A sum is odd exactly when one card is odd and the other even.
3. Pairs with one of each: 5 × 4 = 20.
4. All pairs of different cards: C(9, 2) = 36.
5. Probability: 20/36 = 5/9 (C).

Why this works: Parity questions reduce to counting odd/even choices. Without replacement, count unordered pairs consistently (or ordered pairs consistently).

Answer: C, π/4

1. A random point in the square is equally likely to be anywhere, so probability = favourable area ÷ 4.
2. Points within 1 of a corner form a quarter disc of radius 1 at that corner, with area π × 1² ÷ 4 = π/4.
3. Neighbouring corners are 2 apart, so quarter discs of radius 1 only touch at the midpoints of the sides: they do not overlap.
4. Favourable area: 4 × π/4 = π.
5. Probability: π/4 ≈ 0.785, answer π/4 (C).

Why this works: For a point chosen uniformly, probability = favourable area ÷ total area. Always check whether the pieces overlap before adding.

Answer: B, 2/5

1. Four coins give 2⁴ = 16 equally likely outcomes.
2. The condition ‘at least one head’ removes only TTTT, leaving 15 equally likely outcomes.
3. Exactly two heads: choose which 2 of the 4 coins, C(4, 2) = 6 outcomes, all of which have at least one head.
4. Probability: 6/15 = 2/5 (B). (Without the condition it would be 6/16 = 3/8, the trap.)

Why this works: Conditioning shrinks the set of possible outcomes. Count favourable outcomes inside the new, smaller set.

Answer: B, 5/54

1. Three rolls give 6³ = 216 equally likely outcomes.
2. A strictly increasing outcome uses three different numbers in increasing order.
3. Every choice of three different numbers from 1–6 gives exactly one increasing order, so count choices: C(6, 3) = 20.
4. Probability: 20/216 = 5/54 (B).

Why this works: Of the 6 orders of three different numbers exactly one is increasing, so you could also say: P(all different) × 1/6 = (120/216) × (1/6).

Answer: C, 1/6

1. The first red ball on the third draw means: blue, then blue, then red.
2. First draw blue: 6 of 10 balls, probability 6/10.
3. Second blue: 5 blue left of 9, probability 5/9.
4. Third red: 4 red of the 8 left, probability 4/8.
5. Multiply: 6/10 × 5/9 × 4/8 = 120/720 = 1/6 (C).

Why this works: Without replacement, multiply conditional probabilities: each fraction reflects what is left in the bag at that moment.

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.

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.

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.

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.

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.

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.