In this example, n is any positive integer.
Proposition P: If n is a perfect square, then it has an odd number of factors.
Converse Q: If n has an odd number of factors, then it is a perfect square.
You may recall that we came across these statements in the previous grade. Which of them are true?

Step-by-step solution

Idea: Pair each factor d of n with its partner n ÷ d. Pairs of two different numbers give factors in twos (an even count). Only a pair that repeats a number, (f, f), adds a single factor — and that happens exactly when n = f × f.

1. Factor–partner pairs. If d is a factor of n, so is n ÷ d, and their product is n. For 12 the pairs are (1, 12), (2, 6), (3, 4): 6 factors.½ mark
2. Q is true. Suppose n has an odd number of factors. If every pair had two different numbers, the factors would come in twos and their number would be even. So some pair must repeat a number: (f, f).1 mark
3. Then n = f × f = f², a perfect square. So Q is true.½ mark
4. P is true. Let n = f². Then (f, f) is a pair that repeats a number. There is only one such pair: if (g, g) were another, g² = f² with g and f positive would force g = f.1 mark
5. Every other factor has a partner different from itself, so those factors come in twos (an even count). Adding the single factor f makes the total odd. So P is true.½ mark
6. Both statements are true, so each implies the other.½ mark

Check: 36 = 6²: factors 1, 2, 3, 4, 6, 9, 12, 18, 36 — nine (odd); pairs (1, 36), (2, 18), (3, 12), (4, 9), (6, 6). 12 is not a square and has 6 factors (even).

Answer to write in the exam

Common mistakes that cost marks

• Proving one direction and claiming both. The pairing argument starting from ‘odd number of factors’ proves Q; P needs its own argument.
• Counting the repeated factor twice: in 36 the pair (6, 6) gives only one factor, 6.
• Forgetting 1 and the number itself when listing factors.

How this can come in the exam

Try one yourself

How many factors does 100 have? Is that number odd or even, and why?

More questions like this

• Consider the following argument.
  Each factor of a number has a ‘partner’ factor such that their product yields the given number, e.g., 5 is a factor of 35, and 5 × 7 = 35. Here, 7 is the partner factor of 5, and vice-versa. Let us focus on factor-partner pairs of numbers, e.g., (1, 12), (2, 6), (3, 4) are the pairs for the number 12.
  If a number has an odd number of factors, there must be a factor-partner pair in which the same number repeats (e.g., the partner factor of 5 in 25). If not, the given number will have an even number of factors since each factor can be paired with its factor pair.
  Thus, if a number has an odd number of factors, then it is a perfect square.
  What does this argument prove? Statement P or Q?
• This only proves Q. It doesn’t prove that every square number has an odd number of factors (Proposition P).
  Is P true?
• (i) n has an odd number of factors
  which implies
  (ii) the existence of a factor-partner pair in which the same number repeats, say (f, f)
  which implies
  (iii) n is a square number: n = f × f
  We start with (iii). Clearly, (iii) implies (ii). Does (ii) imply (i)?
• For example, if the number of such pairs is two — say (f, f) and (g, g) — what can we say about the number of factors?
• Proposition P: If two triangles have the same area, then they are congruent.
  Converse Q: If two triangles are congruent, then they have the same area.
  Determine if these statements are true or not. Justify the true statements and give a counterexample for each false statement.

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