Junior competition-style problems (ages 11 to 13)

Answer: 11

1. Split the numbers 1 to 200 by how many digits they have; the cases cannot overlap.
2. One digit: only 5 has digit sum 5. That is 1 number.
3. Two digits 10a + b with a ≥ 1 and a + b = 5: a can be 1, 2, 3, 4 or 5, giving 14, 23, 32, 41, 50. That is 5 numbers.
4. Three digits from 100 to 199: the first digit is 1, so the last two digits must add to 4: 104, 113, 122, 131, 140. That is 5 numbers.
5. 200 has digit sum 2, so it does not count.
6. Total: 1 + 5 + 5 = 11.

Why this works: Splitting a count into cases that cannot overlap (here, by number of digits) turns one messy count into several small, easy ones.

Answer: B, 57

1. Look at the remainders: 1 on dividing by 4, 2 by 5, 3 by 6. Each is exactly 3 less than the divisor.
2. So n + 3 leaves remainder 0 on dividing by 4, by 5 and by 6: n + 3 is a common multiple of 4, 5 and 6.
3. The lowest common multiple: 4 = 2², 5, 6 = 2 × 3, so LCM = 2² × 3 × 5 = 60.
4. The smallest positive choice is n + 3 = 60, so n = 57.
5. Check: 57 = 4 × 14 + 1, 57 = 5 × 11 + 2, 57 = 6 × 9 + 3. ✓ (117 = 120 − 3 also works but is larger.)
6. Answer: 57 (B).

Why this works: Spotting that every remainder is ‘divisor minus 3’ turns three conditions into one: n + 3 is a common multiple. Look for a shift that makes remainders line up.

Answer: C, 4

1. Find every set of three digits (1 to 9, as 0 would make the product 0) with product 24 = 2³ × 3.
2. The sets are {1, 3, 8}, {1, 4, 6}, {2, 2, 6} and {2, 3, 4}.
3. A number is divisible by 4 exactly when its last two digits form a multiple of 4.
4. {1, 3, 8}: possible endings 13, 31, 18, 81, 38, 83 — none is a multiple of 4.
5. {1, 4, 6}: endings 16 and 64 work, giving 416 and 164 (14, 41, 46, 61 do not).
6. {2, 2, 6}: endings 22, 26, 62 — none works.
7. {2, 3, 4}: endings 24 and 32 work, giving 324 and 432 (23, 34, 42, 43 do not).
8. Total: 164, 416, 324, 432, which is 4 numbers (C).

Why this works: Two filters are easier one at a time: first the product condition gives a short list of digit sets, then the divisibility-by-4 test only needs the last two digits.

Answer: B, 2

1. Only units digits matter. Write the units digits of 3¹, 3², 3³, 3⁴: 3, 9, 7, 1. Then 3⁵ ends in 3 again, so the pattern repeats every 4.
2. One full block of four adds 3 + 9 + 7 + 1 = 20, which ends in 0.
3. 2026 = 4 × 506 + 2, so the sum is 506 full blocks followed by 3²⁰²⁵ + 3²⁰²⁶.
4. The 506 blocks contribute a units digit of 0.
5. 3²⁰²⁵ is first in its block (units digit 3) and 3²⁰²⁶ second (units digit 9): 3 + 9 = 12.
6. The units digit of the whole sum is 2 (B).

Why this works: Units digits of powers always cycle, because each one depends only on the previous units digit. Grouping whole cycles leaves only a short leftover to add.

Answer: 12

1. Any factor d of 360 that is a multiple of 6 can be written as d = 6k.
2. 6k divides 360 exactly when k divides 360 ÷ 6 = 60.
3. So we just count the factors of 60.
4. 60 = 2² × 3 × 5. A factor chooses a power of 2 (3 ways: 2⁰, 2¹, 2²), of 3 (2 ways) and of 5 (2 ways).
5. Number of factors: 3 × 2 × 2 = 12. (They are 6, 12, 18, 24, 30, 36, 60, 72, 90, 120, 180, 360.)

Why this works: ‘Factors of N that are multiples of m’ match one-to-one with factors of N/m. Counting factors from a prime factorisation (add one to each power, then multiply) does the rest.

Answer: 16

1. Multiples of 2 from 1 to 30: 30 ÷ 2 = 15.
2. Multiples of 3: 30 ÷ 3 = 10.
3. Multiples of both (that is, of 6) were counted twice: 30 ÷ 6 = 5.
4. Multiples of 2 or 3: 15 + 10 − 5 = 20.
5. Remove those that are also multiples of 5: 10, 15, 20, 30 (4 numbers; 5 and 25 are not multiples of 2 or 3).
6. Circled numbers: 20 − 4 = 16.

Why this works: This is inclusion–exclusion: when two lists overlap, add them and subtract the overlap once, so nothing is counted twice.

Answer: 59

1. Write the number as 10a + b, where a (1 to 9) is the tens digit and b (0 to 9) the units digit.
2. The condition is 10a + b = 4(a + b) + 3.
3. Expand: 10a + b = 4a + 4b + 3, so 6a − 3b = 3, and dividing by 3: 2a − b = 1, i.e. b = 2a − 1.
4. b must be a digit, so 2a − 1 ≤ 9, giving a ≤ 5.
5. a = 1, 2, 3, 4, 5 gives 11, 23, 35, 47, 59.
6. Check the largest: 4 × (5 + 9) + 3 = 59. ✓ The answer is 59.

Why this works: Writing a number in terms of its digits (10a + b) turns a word puzzle into a simple equation between small whole numbers, which you can then list completely.

Answer: 20

1. A climbing number uses three different digits, written in increasing order.
2. Conversely, any choice of three different digits from {1, …, 6} can be written in increasing order in exactly one way.
3. So climbing numbers correspond one-to-one with 3-element subsets of {1, 2, 3, 4, 5, 6}.
4. Ordered choices: 6 × 5 × 4 = 120; each subset is counted 3 × 2 × 1 = 6 times.
5. Number of subsets: 120 ÷ 6 = 20.

Why this works: When the order is forced, counting arrangements becomes counting choices. That is the idea behind ‘n choose r’.

Answer: 13

1. Use one coin of each kind first: 1p + 2p + 5p = 8p. We still need 12p, now with any number (including zero) of each coin.
2. Organise by the number of extra 5p coins, which can be 0, 1 or 2 (three would be 15p, too much).
3. 0 extra 5p: make 12p from 2p and 1p. The number of 2p coins can be 0, 1, …, 6 and the rest is 1p: 7 ways.
4. 1 extra 5p: 7p left. 2p coins: 0, 1, 2 or 3: 4 ways.
5. 2 extra 5p: 2p left. 2p coins: 0 or 1: 2 ways.
6. Total: 7 + 4 + 2 = 13.

Why this works: ‘At least one of each’ is easiest handled by paying one of each up front. Then organise by the largest coin, where there are fewest cases.

Answer: B, 6

1. Every route has 4 moves right (R) and 2 up (U): 6 moves in total.
2. All routes: choose which 2 of the 6 moves are U: 6 × 5 ÷ 2 = 15.
3. Routes through (2, 1): reach it with 2 R and 1 U in any order: 3 ways.
4. From (2, 1) to (4, 2): 2 R and 1 U again: 3 ways. Through-routes: 3 × 3 = 9.
5. Allowed routes: 15 − 9 = 6 (B).

Why this works: Counting the complement (the routes you do not want) is often easier, and routes through a point multiply: ways in × ways out.

Answer: 26

1. Count squares by size. A k by k square needs k consecutive rows and k consecutive columns of the 3 by 5 grid.
2. 1 by 1: 3 rows × 5 columns = 15 positions.
3. 2 by 2: (3 − 1) × (5 − 1) = 2 × 4 = 8 positions.
4. 3 by 3: (3 − 2) × (5 − 2) = 1 × 3 = 3 positions.
5. 4 by 4 or bigger cannot fit in 3 rows.
6. Total: 15 + 8 + 3 = 26.

Why this works: A k by k square fits in (rows − k + 1) × (columns − k + 1) positions. Organising by size makes sure nothing is missed or double counted.

Answer: 24

1. Glue Bea and Cai into one block. The block covers two neighbouring chairs: 1–2, 2–3, 3–4 or 4–5, and inside it the order is BC or CB (2 ways).
2. Block at 1–2: free chairs 3, 4, 5. Ali may not take chair 5 (an end): 2 choices.
3. Block at 2–3: free chairs 1, 4, 5. Ali must take chair 4: 1 choice.
4. Block at 3–4: free chairs 1, 2, 5. Ali must take chair 2: 1 choice.
5. Block at 4–5: free chairs 1, 2, 3. Ali takes 2 or 3: 2 choices.
6. So 2 + 1 + 1 + 2 = 6 ways to place the block and Ali. Dan and Eve fill the last two chairs in 2 ways.
7. Total: 6 × 2 (block order) × 2 (Dan, Eve) = 24.

Why this works: ‘Must be next to’ is handled by gluing into a block; a ‘not at the ends’ rule is best handled case by case once the block is placed.

Answer: 36

1. The odd digits are 1, 3, 5, 7, 9.
2. Hundreds digit: odd and the number below 600, so 1, 3 or 5: 3 choices.
3. Tens digit: any odd digit not already used: 4 choices.
4. Units digit: any odd digit not yet used: 3 choices.
5. Total: 3 × 4 × 3 = 36.

Why this works: Always fill the most restricted position first; after that, the other positions have a fixed number of choices and you can multiply.

Answer: 21

1. Let a(n) be the number of good colourings of n squares in a row.
2. If the last square is blue, the first n − 1 squares can be any good colouring: a(n − 1) ways.
3. If the last square is red, the square before must be blue, and the first n − 2 are any good colouring: a(n − 2) ways.
4. So a(n) = a(n − 1) + a(n − 2).
5. Start: a(1) = 2 (R or B), a(2) = 3 (BB, BR, RB).
6. Then a(3) = 5, a(4) = 8, a(5) = 13, a(6) = 8 + 13 = 21.

Why this works: Splitting on the last position gives a recurrence. This one produces the Fibonacci numbers, which turn up whenever ‘no two in a row’ is the rule.

Answer: D, 81 cm²

1. Let the square have side s cm. Cutting it into three identical strips side by side makes each strip s by s/3.
2. Perimeter of a strip: 2(s + s/3) = 2 × 4s/3 = 8s/3.
3. Set 8s/3 = 24, so 8s = 72 and s = 9.
4. Area of the square: 9 × 9 = 81 cm² (D).

Why this works: Naming one length (the side of the square) and writing every other length in terms of it turns a picture into a one-line equation.

Answer: D, 105°

1. The ratio 2 : 3 : 7 means the angles are 2x, 3x and 7x for some x.
2. Angles in a triangle add to 180°: 2x + 3x + 7x = 12x = 180°.
3. So x = 15°.
4. The angles are 30°, 45° and 105°; check: 30 + 45 + 105 = 180. ✓
5. The largest angle is 105° (D).

Why this works: A ratio tells you the angles are multiples of one unknown part; the angle sum fixes the size of the part.

Answer: 54

1. Interior and exterior angles at a vertex add to 180°, so each exterior angle is 180° − 150° = 30°.
2. The exterior angles of a convex polygon add to 360°, so there are 360 ÷ 30 = 12 sides.
3. From one vertex, diagonals go to every other vertex except itself and its two neighbours: 12 − 3 = 9 diagonals.
4. Doing this from all 12 vertices counts 12 × 9 = 108, but each diagonal is counted from both ends.
5. Number of diagonals: 108 ÷ 2 = 54.

Why this works: Exterior angles of any convex polygon add to 360°, which is the quickest way from an angle to the number of sides. The ‘count from each end, then halve’ trick avoids double counting.

Answer: 14

1. Two dots differ by a steps across and b steps up (0 to 4 each). By Pythagoras the distance is √(a² + b²).
2. Order doesn’t matter, so take 0 ≤ a ≤ b ≤ 4, not both 0.
3. b = 1: a = 0, 1 give 1, 2. b = 2: a = 0, 1, 2 give 4, 5, 8. b = 3: 9, 10, 13, 18. b = 4: 16, 17, 20, 25, 32.
4. That is 2 + 3 + 4 + 5 = 14 values of a² + b², and checking the list 1, 2, 4, 5, 8, 9, 10, 13, 16, 17, 18, 20, 25, 32 shows no repeats.
5. Different squared distances mean different distances, so there are 14.

Why this works: Pythagoras turns each distance into a sum of two squares, so counting distances means counting different values of a² + b². Watch for different pairs giving the same sum.

Answer: 8

1. Take the cuboid as 3 wide, 4 deep and 5 high, built from 3 × 4 × 5 = 60 unit cubes.
2. A cube has no paint if it is not on any painted face.
3. The two side faces across the width are painted, so the outer layer on each side is removed: 3 − 2 = 1 cube wide remains.
4. The two faces across the depth are painted: 4 − 2 = 2 remain.
5. Only the top is painted vertically (the bottom is not): 5 − 1 = 4 layers remain.
6. Unpainted cubes: 1 × 2 × 4 = 8.

Why this works: The unpainted cubes always form a smaller cuboid; each painted face removes one layer in its own direction, so count how many layers each dimension loses.

Answer: C, 108°

1. Measure both hands clockwise from 12 o’clock.
2. The minute hand turns 360° in 60 minutes, 6° per minute: at 24 minutes it is at 24 × 6° = 144°.
3. The hour hand turns 30° per hour, 0.5° per minute: at 8:24 it is at 8 × 30° + 24 × 0.5° = 240° + 12° = 252°.
4. Difference: 252° − 144° = 108°.
5. 108° is less than 180°, so it is already the smaller angle: 108° (C).

Why this works: Turning each hand into an angle from 12 o’clock makes the problem a subtraction. The common slip is forgetting that the hour hand moves between the hour marks.

Answer: 17

1. The slanted side joins (8, 0) and (0, 6): its equation is x/8 + y/6 = 1, i.e. 6x + 8y = 48.
2. Strictly inside means x ≥ 1, y ≥ 1 and 6x + 8y < 48.
3. x = 1: 8y < 42, so y = 1 to 5 (5 points). x = 2: 8y < 36, y = 1 to 4 (4).
4. x = 3: 8y < 30, y = 1 to 3 (3). x = 4: 8y < 24, y = 1, 2 (2; (4, 3) is on the edge).
5. x = 5: 8y < 18, y = 1, 2 (2). x = 6: 8y < 12, y = 1 (1). x = 7: 8y < 6, none.
6. Total: 5 + 4 + 3 + 2 + 2 + 1 = 17. (Check with Pick’s theorem: area 24 = I + 16/2 − 1 gives I = 17.)

Why this works: Counting column by column is a reliable way to count lattice points. (Pick’s theorem, Area = I + B/2 − 1, gives the same 17 as a check.)

Answer: D, 20

1. Mean × how many = total. The five numbers total 5 × 12 = 60.
2. The four that remain total 4 × 10 = 40.
3. The removed number is the difference: 60 − 40 = 20 (D).
4. Check: removing 20 lowers the mean because 20 is above the old mean of 12, as expected.

Why this works: Means are awkward to combine, but totals just add and subtract. Converting mean × count into a total is the standard move.

Answer: D, 50p

1. Let a pencil cost x pence and a pen y pence: 4x + 3y = 230 and 3x + 4y = 260.
2. Add the equations: 7x + 7y = 490, so x + y = 70.
3. Subtract the first from the second: −x + y = 30, so y = x + 30.
4. Substitute: x + (x + 30) = 70, so 2x = 40, x = 20 and y = 50.
5. Check: 4 × 20 + 3 × 50 = 230. ✓ A pen costs 50p (D).

Why this works: When two equations are symmetric, their sum and difference are much simpler than the originals. Look for that before reaching for substitution.

Answer: 1

1. Call the terms t₁, t₂, …. The rule is t_n+2 = t_n + t_n+1.
2. t₇ = t₅ + t₆, so t₆ = 53 − 20 = 33.
3. Run the rule backwards, t_n = t_n+2 − t_n+1: t₄ = t₆ − t₅ = 33 − 20 = 13.
4. t₃ = 20 − 13 = 7, t₂ = 13 − 7 = 6, t₁ = 7 − 6 = 1.
5. Check forwards: 1, 6, 7, 13, 20, 33, 53. ✓ The first term is 1.

Why this works: A rule that builds forwards can usually be run backwards. Here t_n = t_n+2 − t_n+1, so the sequence is fixed by any two neighbouring terms.

Answer: 48

1. Let Sam be s years old now; then Jo is 3s.
2. In 12 years Sam is s + 12 and Jo is 3s + 12.
3. Then Jo is twice Sam’s age: 3s + 12 = 2(s + 12) = 2s + 24.
4. So s = 12, and Jo is 36.
5. Check: in 12 years they are 24 and 48, and 48 = 2 × 24. ✓ Sum now: 12 + 36 = 48.

Why this works: Age problems become easy once every age is written in terms of one letter at one time, then shifted by the same number of years.

Answer: 6

1. Start with 3 < 2x − 5 < 17.
2. Add 5 to all three parts: 8 < 2x < 22.
3. Divide all three parts by 2: 4 < x < 11.
4. The inequalities are strict, so 4 and 11 are excluded: x = 5, 6, 7, 8, 9, 10.
5. That is 6 whole numbers.

Why this works: A double inequality can be solved in one go: whatever you do to one part, do to all three. Take care with strict inequalities at the ends.

Answer: C, 72 litres

1. The 30 litres took the tank from 1/3 full to 3/4 full.
2. Fraction added: 3/4 − 1/3 = 9/12 − 4/12 = 5/12 of the tank.
3. So 5/12 of the tank is 30 litres, and 1/12 is 30 ÷ 5 = 6 litres.
4. The whole tank is 12 × 6 = 72 litres.
5. Check: 1/3 of 72 = 24, plus 30 = 54 = 3/4 of 72. ✓ Answer 72 litres (C).

Why this works: The change in amount matches the change in fraction. Finding one twelfth first is the ‘unitary method’.

Answer: 6

1. Work from the inside: x ◆ 3 = 2x − 3.
2. Then (x ◆ 3) ◆ x = (2x − 3) ◆ x = 2(2x − 3) − x.
3. Simplify: 4x − 6 − x = 3x − 6.
4. Solve 3x − 6 = 12: 3x = 18, so x = 6.
5. Check: 6 ◆ 3 = 9, and 9 ◆ 6 = 18 − 6 = 12. ✓ Answer 6.

Why this works: A made-up operation is just a rule for substituting. Apply it carefully from the inside out and an ordinary equation appears.

Answer: C, 2/9

1. Two dice give 6 × 6 = 36 equally likely ordered outcomes.
2. Pairs differing by exactly 2, smaller first: (1, 3), (2, 4), (3, 5), (4, 6): 4 pairs.
3. Each can appear in either order, e.g. (3, 1) as well as (1, 3): 8 outcomes.
4. Probability: 8/36 = 2/9 (C).

Why this works: With two dice, list outcomes as ordered pairs so each of the 36 is equally likely. Forgetting the reversed pairs is the usual error.

Answer: 7

1. Suppose r red balls are added. Then there are 3 + r red balls out of 8 + r in total.
2. We need (3 + r)/(8 + r) = 2/3.
3. Cross-multiply: 3(3 + r) = 2(8 + r), so 9 + 3r = 16 + 2r.
4. So r = 7.
5. Check: 10 red out of 15 is 10/15 = 2/3. ✓ Add 7 red balls.

Why this works: A probability is a fraction of a total, so changing the contents changes both the top and the bottom. Set up the fraction and cross-multiply.

Answer: C, 1/8

1. Two spins give 8 × 8 = 64 equally likely outcomes.
2. For a sum of 9, the second number must be 9 minus the first.
3. Whatever the first number (1 to 8), 9 minus it is also between 1 and 8, so exactly one second number works.
4. That gives 8 good outcomes: (1,8), (2,7), …, (8,1).
5. Probability: 8/64 = 1/8 (C).

Why this works: Sometimes it is quicker to see that every first outcome has exactly one good partner than to list all good pairs.

Answer: D, 1/2

1. There are 2⁴ = 16 equally likely sequences. Count those with no HH.
2. Let g(n) be the number of good sequences of length n. A good sequence ends in T (after any good sequence of length n − 1) or in TH (after any good sequence of length n − 2).
3. So g(n) = g(n − 1) + g(n − 2), with g(1) = 2 (H, T) and g(2) = 3 (HT, TH, TT).
4. g(3) = 5 and g(4) = 8. (They are TTTT, TTTH, TTHT, THTT, HTTT, THTH, HTHT, HTTH.)
5. Probability: 8/16 = 1/2 (D).

Why this works: The same ‘look at the ending’ recurrence as for colouring squares works for coin sequences: good sequences of length n number F(n + 2), a Fibonacci number.

Answer: D, 1/2

1. Let A = multiples of 3 from 1 to 50 and B = numbers containing the digit 3.
2. |A| = 16 (3, 6, …, 48).
3. B: 3, 13, 23, 43 and 30–39 (ten numbers): |B| = 14.
4. Both: 3, 30, 33, 36, 39: 5 numbers.
5. |A or B| = 16 + 14 − 5 = 25.
6. Probability: 25/50 = 1/2 (D).

Why this works: ‘Or’ means inclusion–exclusion again: count both lists and subtract the numbers that are on both.

Answer: C, 3/8

1. There are 4! = 24 equally likely orders.
2. Card 1 must go to position 2, 3 or 4; by symmetry each choice gives the same number of good orders. Take card 1 in position 2.
3. Case card 2 in position 1: cards 3 and 4 must swap (3 in 4, 4 in 3): 1 way.
4. Case card 2 in position 3: then card 3 cannot go to 3, so 3 goes to 4 and 4 to 1: 1 way.
5. Case card 2 in position 4: 4 must avoid 4, so 4 goes to 3 and 3 to 1: 1 way.
6. 3 ways for each of 3 places for card 1: 9 good orders.
7. Probability: 9/24 = 3/8 (C).

Why this works: Arrangements where nothing is in its own place are called derangements. For n cards the probability is close to 1/e ≈ 0.37 even for small n — 3/8 = 0.375 here.

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.

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.

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.

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.

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.

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.