Prove that the Product of Three Consecutive Positive Integers is Divisible by 6

Answer: Let the 3 consecutive positive integers be n, n+1, n+2.

Any number divided by 3 leaves the remainder 0,1 or 2.

So, n= 3p, 3p+1 or 3p+2 where p is any integer.

If n=3p n is divisible by 3

If n = 3p + 1, ⇒ n + 2 = 3p + 1 + 2 = 3p + 3 = 3(p + 1) is divisible by 3

If n = 3p + 2, ⇒ n + 1 = 3p + 2 + 1 = 3p + 3 = 3(p + 1) is divisible by 3

Therefore , n(n+1)(n+2) is divisible by 3

When a number is divisible by 2, the remainder obtained is 0 or 1

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

If n = 2q ⇒ n and n + 2 = 2q + 2 = 2(q + 1) are divisible by 2.

If n = 2q + 1 ⇒ n + 1 = 2q + 1 + 1 = 2q + 2 = 2 (q + 1) is divisible by 2.

Therefore , n(n+1)(n+2) is divisible by 2.

Since n(n+1)(n+2) is divisible by both 2 and 3. It is divisible by 6.

The product of three consecutive positive integers is divisible by 6, proved.