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TMUA 2017 · Paper 2 · Question 5 of 20

TMUA 2017 Paper 2 Question 5

Proof and counterexample — What a single counterexample can refute. Try it first; the answer and a full worked solution are below.

TMUA 2017 · Paper 2Proof and counterexampleWhat a single counterexample can refute8 options
Consider the following three statements:

1   10p2 + 1 and 10p2 − 1 are both prime when p is an odd prime.
2   Every prime greater than 5 is of the form 6n+1 for some integer n.
3   No multiple of 7 greater than 7 is prime.

The result 91 = 7 × 13 can be used to provide a counterexample to which of the above statements?

  1. Anone of them
  2. B1 only
  3. C2 only
  4. D3 only
  5. E1 and 2 only
  6. F1 and 3 only
  7. G2 and 3 only
  8. H1, 2 and 3
Show the answer and worked solution
answer · B
  1. Anone of them
  2. B1 only
  3. C2 only
  4. D3 only
  5. E1 and 2 only
  6. F1 and 3 only
  7. G2 and 3 only
  8. H1, 2 and 3
Look for a place where 91 has to appear. Taking p = 3 in statement 1 gives 10p2 − 1 = 89 and 10p2 + 1 = 91 = 7×13, which is not prime, so 1 falls. Statement 2 needs a prime greater than 5 that is not of the form 6n+1 — 91 is not even prime, so it cannot serve (and in any case 91 = 6×15+1). Statement 3 is true, and 91 is an example of it rather than against it, since 91 is a multiple of 7 and is not prime. Only 1.