Question

prove that the product of three consecutive positive integers is divisible by 6

Answer 1

Hello!

Let the 3 consecutive integers be n, n + 1 and n + 2.

Let 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.

⇒ n (n + 1) (n + 2) is divisible by 3.
 
Similarly,

Let n = 2q or 2q + 1, where q is some integer.

• If n = 2q, then n and n + 2 = 2q + 2 = 2(q + 1) is divisible by 2.

• If n = 2q + 1, then n + 1 = 2q + 1 + 1 = 2q + 2 = 2 (q + 1) is divisible by 2.

⇒ n (n + 1) (n + 2) is divisible by 2.
 
But n, n + 1, n + 2 are divisible by 2 and 3.

∴ n (n + 1) (n + 2) is divisible by 6.

Cheers!

Answer 2

hey

Answer is in attachment