Question

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

Answer 1

Let three consecutive positive integers be n,
n + 1 and n + 2

Whenever a number is divided by 3, the remainder obtained is either 0 or 1 or 2
∴ n = 3p or 3p + 1 or
3p + 2, where p is some integer.
If n = 3p,
then n is divisible by 3.
If n = 3p + 1,
then,
n + 2 = 3p + 1 + 2 = 3p + 3 = 3(p + 1) is divisible by 3.
If n = 3p + 2,
then
n + 1 = 3p + 2 + 1 = 3p + 3 = 3(p + 1) is divisible by 3

So, we can say that one of the numbers among n,
n + 1 and n + 2 is always divisible by 3.