Question

Prove that the product of three consecutive positive integer is divisible by 6.

Answer 1

Let, n be any positive integer. Since any positive integer is of the form 6q or 6q + 1 or 6q + 2 or, 6q + 3 or 6q + 4 or 6q + 5.

If n = 6q, then

n(n + 1)(n + 2) = (6q + 1)(6q + 2)(6q + 3)

= 6[(6q + 1)(3q + 1)(2q + 1)]

= 6m, which is divisible by 6?

If n = 6q + 1, then

n(n + 1)(n + 2) = (6q + 2)(6q + 3)(6q + 4)

= 6[(6q + 1)(3q + 1)(2q + 1)]

= 6m, which is divisible by 6

If n = 6q + 2, then

n(n + 1)(n + 2) = (6q + 2)(6q + 3)(6q + 4)

= 6[(3q + 1)(2q + 1)(6q + 4)]

= 6m, which is divisible by 6.

If n = 6q + 3, then

n(n + 1)(n + 2) = (6q + 3)(6q + 4)(6q + 5)

= 6[(6q + 1)(3q + 2)(2q + 5)]

= 6m, which is divisible by 6.

If n = 6q + 4, then

n(n + 1)(n + 2) = (6q + 4)(6q + 5)(6q + 6)

= 6[(6q + 4)(3q + 5)(2q + 1)]

= 6m, which is divisible by 6.

If n = 6q + 5, then

n(n + 1)(n + 2) = (6q + 5)(6q + 6)(6q + 7)

= 6[(6q + 5)(q + 1)(6q + 7)]

= 6m, which is divisible by 6.

Hence, the product of three consecutive positive integer is divisible by 6.