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Divisibility · 3 marks

If n is divisible by 60, then it is divisible by both 5 and 12.

Answer: Converse: If n is divisible by both 5 and 12, then it is divisible by 60. Both are true (5 and 12 have no common factor other than 1).

Step-by-step solution

Given: n is a positive integer
To find: Frame the converse of this proposition. Then, determine if each of the two statements is true or not. Justify the true statements and give a counterexample for each false statement.

Idea: Unlike 4 and 6, the numbers 5 and 12 share no prime factor. So if 5 divides 12k, the 5 must come from k, which makes n a multiple of 60.

  1. Converse: If n is divisible by both 5 and 12, then it is divisible by 60.½ mark
  2. Proposition — true. If n = 60k, then n = 5 × 12k and n = 12 × 5k, so n is divisible by 5 and by 12.1 mark
  3. Converse — true. Since 12 divides n, write n = 12k. Since 5 divides n, the prime 5 appears in the prime factorisation of 12k. 12 = 2 × 2 × 3 has no factor 5, so 5 must divide k.1 mark
  4. So k = 5m and n = 12 × 5m = 60m: n is divisible by 60. Both statements are true.½ mark
Converse: If n is divisible by both 5 and 12, then it is divisible by 60. Both the proposition and its converse are true.

Check: Multiples of 12: 12, 24, 36, 48, 60, 72, … — the first one ending in 0 or 5 (divisible by 5) is 60, and the next is 120 = 2 × 60.

Answer to write in the exam

Converse: If n is divisible by both 5 and 12, then it is divisible by 60.

Proposition: n = 60k = 5(12k) = 12(5k). True.

Converse: n = 12k; 5 | 12k and 5 ∤ 12 (12 = 2 × 2 × 3), 5 prime ⇒ 5 | k

k = 5m ⇒ n = 60m. True.

∴ Both the proposition and its converse are true.

Common mistakes that cost marks

  • Using the idea from 24 = 4 × 6 here and calling the converse false. 5 and 12 have no common factor, so the converse holds.
  • ‘Proving’ the converse by checking 60, 120 and 180 only. A general argument is needed.

How this can come in the exam

MCQ (1 mark)

For which pair (p, q) is ‘if n is divisible by both p and q, then n is divisible by pq‘ true for every positive integer n?

  1. (4, 10)
  2. (6, 9)
  3. (7, 9)
  4. (8, 12)
Show answer

(C) (7, 9)
7 and 9 have no common factor other than 1. The others fail: 20 (for 4, 10), 18 (for 6, 9), 24 (for 8, 12).

Assertion–Reason (1 mark)

Assertion (A): Every number divisible by both 3 and 8 is divisible by 24.
Reason (R): 3 and 8 have no common factor other than 1.

  1. Both Assertion (A) and Reason (R) are true and R is the correct explanation of A.
  2. Both Assertion (A) and Reason (R) are true but R is not the correct explanation of A.
  3. Assertion (A) is true but Reason (R) is false.
  4. Assertion (A) is false but Reason (R) is true.
Show answer

(A) Both Assertion (A) and Reason (R) are true and R is the correct explanation of A.
Because 3 and 8 share no factor, n = 8k with 3 | 8k forces 3 | k, so 24 | n. R explains A.

Try one yourself

Is ‘If n is divisible by both 4 and 9, then n is divisible by 36′ true? Give a reason.

Show answer

True. n = 9k; 4 divides 9k and 9 = 3 × 3 has no factor 2, so 4 divides k. Then n = 36m.

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