Let q be quotient and r be the remainder.

On applying Euclid’s algorithm, i.e. dividing n by 3,

**we have **

n = 3q + r 0 ≤ r ˂ 3

⇒ n = 3q + r r = 0, 1 or 2

⇒ n = 3q or n = (3q + 1) or n = (3q + 2)

**Case 1: **If n = 3q, then n is divisible by 3.

**Case 2:** If n = (3q+1), then (n+2) = 3q + 3 = 3(q + 1), which is clearly divisible by 3.

In this case, (n+2) is divisible by 3.

**Case 3:** If n = (3q+2), then (n+4) = 3q + 6 = 3(q + 2), which is clearly divisible by 3.

In this case, (n+4) is divisible by 3.

**Hence,** one and only one out of n, (n+2) and (n+4) is divisible by 3.