Question

Prove that one and only one out of n, n + 2 and n + 4 is divisible by 3, where n is any positive integer.

Answer 1

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 + r, where r = 0, 1, 2

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

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

But n + 2 and n + 4 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 + 4 are not divisible by 3.

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

Case III: If n = 3q + 2,

Then (n + 4) = 3q + 6 = 3(q + 2),

Which is divisible by 3 but n and (n + 2) are not divisible by 3.

So, in this case, only (n + 4) is divisible by 3.

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