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Problem-solving strategy

Extremal principle

Look at the biggest, smallest, first or last object in the problem. Extreme objects have extra properties: the largest number in a set cannot be the sum of two larger ones; the person with the most friends has all their friends among a known group; the smallest counterexample cannot have a smaller one.

For ‘largest possible’ or ‘smallest possible’ questions, the extremal answer needs two halves: a bound (it can be no better than this) and an example (and here it is).

When to try it

  • The question asks for a maximum or minimum.
  • You need to prove something exists or that something cannot go on for ever.
  • There is a natural ordering: sizes, positions, distances.

Watch out: Do not stop after finding an example: without the bound, you have only shown the answer is at least that good.

Two worked examples

Try each one first. The hints and the full solution are underneath.

Problem J39

Junior · LogicShort answer

Six teams play in a league. Each pair of teams plays once. A win earns 3 points, a draw 1 point each and a loss 0. The teams scored 40 points in total. How many games were draws?

Hint

How many games are there, and how many points does each game hand out?

Second hint

15 games. A game with a winner gives out 3 points, a draw gives out 2.

Full worked solution

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.

Strategy: Extremal principle, Working backwards

Problem I06

Intermediate · Number theoryShort answer

How many zeros are at the end of the number 1 × 3 × 5 × 7 × … × 99 × 210?

Hint

Each final zero needs one factor 2 and one factor 5. How many of each are there?

Second hint

The odd product has no factor 2, so all the 2s come from 210. Count the 5s in 1 × 3 × … × 99.

Full worked solution

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.

Where it leads: Trailing zeros count the smaller of the powers of 2 and 5; here, unusually, the 2s are scarce.

Strategy: Extremal principle

Practise: 70 problems that use extremal principle

Other strategies

Organised cases · Count the opposite · Working backwards · Invariants · Pigeonhole principle · Parity and remainders · Symmetry · Spot the pattern and generalise · Proof techniques

All strategy guides · Extension & competition maths