Prove that exactly one out of every three consecutive integers is divisible by 3.

Given: - Let n, n + 1, n + 2 are three consecutive positive integers

To prove: - Exactly one out of every 3 consecutive positive integers is divisible by 3.

Proof: - We now that n is of the form

3q, 3q + 1, 3q + 2

1^st case: - When n = 3q

Here n is divisible by 3, but n + 1, n + 2 are not divisible by 3

2^nd case : - When n = 3q + 1

Here (n + 2) = 3q + 1 + 2

= 3q + 3 = 3 (q + 1)

3(q + 1) is divisible by 3, but n, n + 1 are not divisible by 3

3^rd case: - When n = 3q + 2

Here n + 1 = 3q + 2 + 1 = 3q + 3

= 3(q + 1) is divisible by 3, but n, n + 1 are not divisible by 3

Exactly one out of every 3 consecutive positive integers is divisible by 3.

Hence proved.