Question

Prove that one of any three consecutive positive integers must be divisible by 3.

Answer 1

Let three consecutive positive integers are n, n + 1 and n + 2.

On dividing n by 3, let q be the quotient and r be the remainder.

Then, by Euclid’s division algorithm,

n = 3q + r, where 0 ≤ r < 3

`\implies` n = 3q or n = 3q + 1 or n = 3q + 2

Case I: If n = 3q, which is divisible by 3

But (n + 1) and (n + 2) are not divisible by 3.

So, in this case, only n is divisible by 3.

Case II: If n = 3q + 1,

Then n + 2 = 3q + 3 = 3(q + 1) which is divisible by 3

But n and (n + 1) are not divisible by 3.

So, in this case, only (n + 2) is divisible by 3.

Case III: If n – 3q + 2,

Then n + 1 = 3q + 3 = 3(q + 1) which is divisible by 3

But n and (n + 2) are not divisible by 3.

So, in this case, only (n + 1) is divisible by 3.

Hence, one of any three consecutive positive integers must be divisible by 3.