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Problem-solving strategy

Working backwards

If you know where something ends, undo the steps one at a time. A process that branches going forwards is often forced going backwards: the number before 50 in a ‘double or add 3’ chain can only be 25 or 47.

In games, working backwards from the end labels every position as winning or losing. In algebra, it means starting from what you want to show and asking what would make it true.

When to try it

  • You are told the final state, total or answer and asked about the start.
  • A game ends in a simple position (no counters left).
  • Each step can be undone uniquely, or in only a couple of ways.

Watch out: When working backwards in algebra, each step must be reversible (squaring both sides is not), so check your answer in the original problem.

Two worked examples

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

Problem J02

Junior · Number theoryMultiple choice

What is the smallest positive whole number that leaves remainder 1 when divided by 4, remainder 2 when divided by 5 and remainder 3 when divided by 6?

Hint

Each remainder is 3 less than the divisor. What happens if you add 3 to the number?

Second hint

So n + 3 is a multiple of 4, 5 and 6. What is their lowest common multiple?

Full worked solution

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.

Where it leads: When the remainders do not line up so neatly, the Chinese remainder theorem still guarantees a solution as long as the divisors share no common factor.

Strategy: Working backwards, Parity and remainders

Problem I03

Intermediate · Number theoryShort answer

How many ordered pairs of positive whole numbers (a, b) satisfy ab = 2(a + b) + 20?

Hint

Move everything to one side and add 4 to both sides so the left side factorises.

Second hint

ab − 2a − 2b + 4 = 24, so (a − 2)(b − 2) = 24.

Full worked solution

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’.

Where it leads: Adding a constant to complete a product (Simon’s favourite factoring trick) turns many equations into divisor counts.

Strategy: Working backwards

Practise: 114 problems that use working backwards

Other strategies

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

All strategy guides · Extension & competition maths