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TMUA 2021 · Paper 2 · Question 10 of 20

TMUA 2021 Paper 2 Question 10

Proof and counterexample — Counterexamples · smallest failing case. Try it first; the answer and a full worked solution are below.

TMUA 2021 · Paper 2Proof and counterexampleCounterexamples · smallest failing case7 options
The first seven terms of a sequence of positive integers are: u1 = 15,   u2 = 21,   u3 = 30,   u4 = 37,  u5 = 44,   u6 = 51,   u7 = 59 Consider the following statement about this sequence:

(×)   If n is a prime number, then un is a multiple of 3 or un is a multiple of 5.

What is the smallest value of n that provides a counterexample to (×)?

  1. A1
  2. B2
  3. C3
  4. D4
  5. E5
  6. F6
  7. G7
Show the answer and worked solution
answer · E
  1. A1
  2. B2
  3. C3
  4. D4
  5. E5
  6. F6
  7. G7
Only prime values of n can break the statement, since for any other n the "if" part is false and nothing is claimed. That rules out n = 1 and n = 4 straight away. Checking the primes in order: u2 = 21 = 3× 7 is a multiple of 3, and u3 = 30 is a multiple of both. But u5 = 44 has digit sum 8 and does not end in 0 or 5, so it is a multiple of neither. The smallest counterexample is n = 5.