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Problem-solving strategy

Parity and remainders

Odd and even (parity) is the simplest invariant, and remainders on division by small numbers are the next simplest. A sum of two numbers is even when they are both odd or both even; a square leaves remainder 0 or 1 on division by 3 or 4; the units digit of a power repeats in a short cycle.

Working with remainders rules out whole families of answers at once and makes huge numbers small.

When to try it

  • Huge numbers or powers appear (units digits, last two digits, remainders).
  • You suspect something is impossible: try remainders mod 2, 3, 4, 5, 8 or 9.
  • The problem involves alternating colours, turns or sides.

Watch out: A remainder test can rule things out, but passing it proves nothing: a number with the right remainder still need not be a square.

Two worked examples

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

Problem J37

Junior · LogicMultiple choice

In a year that is not a leap year, 1 March is a Tuesday. On what day of the week is 25 December?

Hint

Count the days from 1 March to 25 December and find the remainder when you divide by 7.

Second hint

From 1 March to 25 December is 299 days; 299 = 7 × 42 + 5.

Full worked solution

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.

Where it leads: Calendar questions are arithmetic modulo 7; Zeller’s congruence turns any date into a weekday by formula.

Strategy: Parity and remainders

Problem I30

Intermediate · ProbabilityMultiple choice

Two different cards are drawn from nine cards numbered 1 to 9. What is the probability that their sum is odd?

Hint

An odd sum needs one odd card and one even card.

Second hint

One odd (5 choices) and one even (4 choices) out of C(9, 2) = 36 pairs.

Full worked solution

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).

Where it leads: Parity arguments make many probability questions short: only odd/even matters here, not the actual numbers.

Strategy: Parity and remainders

Practise: 45 problems that use parity and remainders

Other strategies

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

All strategy guides · Extension & competition maths