Question

Prove that one of three consecutive positive integer is divisible by 3.

Answer 1

Let x be any positive integer. By Euclid’s division lemma x = 3q + r, where 0 < r ≤ 3

[∴ r = 0, 1, 2]

Putting r = 0, we get x = 3q + 0 = 3q

which is divisible by 3.

Putting r = 1, we get, x = 3q + 1

which is not divisible by 3.

Putting r = 2, we get x = 3q + 2,

which is not divisible by 3.

So, one of every three consecutive positive integers is divisible by 3.