Question

show that exactly one of m, m + 2 and m + 4 is divisible by 3

Answer 1

Let n = 3k, 3k + 1 or 3k + 2.

(i) When n = 3k:

n is divisible by 3.

n + 2 = 3k + 2 ⇒ n + 2 is not divisible by 3.

n + 4 = 3k + 4 = 3(k + 1) + 1 ⇒ n + 4 is not divisible by 3.

(ii) When n = 3k + 1:

n is not divisible by 3.

n + 2 = (3k + 1) + 2 = 3k + 3 = 3(k + 1) ⇒ n + 2 is divisible by 3.

n + 4 = (3k + 1) + 4 = 3k + 5 = 3(k + 1) + 2 ⇒ n + 4 is not divisible by 3.

(iii) When n = 3k + 2:

n is not divisible by 3.

n + 2 = (3k + 2) + 2 = 3k + 4 = 3(k + 1) + 1 ⇒ n + 2 is not divisible by 3.

n + 4 = (3k + 2) + 4 = 3k + 6 = 3(k + 2) ⇒ n + 4 is divisible by 3.

Hence, exactly one of the numbers n, n + 2 or n + 4 is divisible by 3.