Question

Show that one and only one of n, n + 1 and n + 2 is divisible by 3, where n is a positive integer.

Answer 1

Given: Let n be a positive integer.

By Euclid’s division lemma divide n by 3: n = 3q + r with q an integer and remainder r ∈ {0, 1, 2}.

Case r = 0: n = 3q, so n is divisible by 3; n + 1 = 3q + 1 and n + 2 = 3q + 2 are not divisible by 3.

Case r = 1: n = 3q + 1, so n is not divisible by 3; n + 1 = 3q + 2 is not divisible by 3; n + 2 = 3q + 3 = 3(q + 1) is divisible by 3.

Case r = 2: n = 3q + 2, so n and n + 2 are not divisible by 3; n + 1 = 3q + 3 = 3(q + 1) is divisible by 3.

In each possible case exactly one of n, n + 1 and n + 2 is divisible by 3; hence one and only one of the three consecutive integers is divisible by 3.