Question

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

Answer 1

HERE YOUR ANSWER...HOPE IT WILL HELP YOU.. Sol; Let us three consecutive  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. 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. So that n, n + 1 and n + 2 is always divisible by 3. ⇒ n (n + 1) (n + 2) is divisible by 3.   Similarly, whenever a number is divided 2 we will get the remainder is 0 or 1. ∴ n = 2q or 2q + 1, where q is some integer. If n = 2q, then n and n + 2 = 2q + 2 = 2(q + 1) are divisible by 2. If n = 2q + 1, then n + 1 = 2q + 1 + 1 = 2q + 2 = 2 (q + 1) is divisible by 2. So that n, n + 1 and n + 2 is always divisible by 2. ⇒ n (n + 1) (n + 2) is divisible by 2.   But n (n + 1) (n + 2) is divisible by 2 and 3.   ∴ n (n + 1) (n + 2) is divisible by 6.

Answer 2

if we will take any 3 consecutive positive integers and multiply them the product is always divisible by 6 .
it can be proved with examples .
1×2×3 = 6 , divisible by 6
5×6×7 = 210, divisible by 6