Let ‘n’ be any positive integer.

Then from Euclidean’s lemma n = bq + r (now consider b = 3)

⇒ n = 3q + r (here 0 ≤ r < 3) which means the possible values of ‘r’ = 0 or 1 or 2

Now consider r = 0 then ‘n’ = 3q (divisible by 3) and n + 2 = 3q + 2 (not divisible by 3) n + 4 = 3q + 4 (not divisible by 3)

Case – II: For r = 1

n = 3q + 1 (not divisible by 3) n + 2 = 3q + 1 + 2 = 3q + 3 = 3(q + l) divisible by 3

n + 4 = 3q + 1 + 4 = 3q + 5 not divisible by 3

Case – III: For r = 2,

n = 3q + 2 not divisible by 3

n + 2 = 3q + 2 + 2 = 3q + 4, not divisible by 3

n + 4 = 3q + 2 + 4 = 3q + 6 = 3(q + 2) divisible by 3

So in all above three cases we observe, only one of either (n) or (n + 1) or (n + 4) is divisible by 3.

**Hence proved.**