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Problem-solving strategy

Spot the pattern and generalise

When a problem is too big, shrink it. Try 1, 2, 3, 4 instead of 2026, write the answers down, and look for a pattern. Then explain why the pattern must continue: that explanation is usually the real solution, often a recurrence or an identity.

Generalising also works the other way: a specific problem can be easier as a special case of something general (replace 2026 by n and look for structure).

When to try it

  • A large number such as 2026 or 100 appears where a small one would do.
  • The process repeats (a sequence, a recurrence, a cycle of last digits).
  • Small cases are quick to compute by hand.

Watch out: Patterns can mislead: points on a circle give 1, 2, 4, 8, 16 regions, then 31. Always find a reason, not just a pattern.

Two worked examples

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

Problem J04

Junior · Number theoryMultiple choice

What is the units digit of 31 + 32 + 33 + … + 32026?

Hint

Write down the units digits of the first few powers of 3. They repeat.

Second hint

The units digits of 3, 32, 33, 34 are 3, 9, 7, 1, and each block of four adds to 20. How many complete blocks are in 2026 terms?

Full worked solution

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.

Where it leads: Cycles of last digits are modular arithmetic; Euler’s theorem says the cycle length always divides 4 for powers of numbers ending in 1, 3, 7 or 9.

Strategy: Spot the pattern and generalise

Problem I02

Intermediate · Number theoryMultiple choice

The number N is written with twelve 1s: N = 111 111 111 111. What is the largest prime factor of N that is less than 100?

Hint

111 111 111 111 = 111 111 × 1 000 001. And 111 111 = 111 × 1001.

Second hint

Factorise further: 111 = 3 × 37, 1001 = 7 × 11 × 13, and 1 000 001 = 101 × 9901.

Full worked solution

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.

Where it leads: Repunits (numbers made of 1s) factor according to the divisors of their length; Rn can only be prime when n is prime.

Strategy: Spot the pattern and generalise

Practise: 118 problems that use spot the pattern and generalise

Other strategies

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

All strategy guides · Extension & competition maths