Prove that the product of three consecutive positive integers is divisible by 6.
Let n is any positive integer.
Since any positive integer is of the form 6p or, 6p + 1 or, 6p + 2 or, 6p + 3 or, 6p + 4 or, 6p + 5.
Case 1:
If n = 6p then
n(n + 1)(n + 2) = 6p (6p + 1)(6p + 2), which is divisible by 6.
Case 2:
If n = 6p + 1, then
n(n + 1)(n + 2) = (6p + 1)(6p + 2) (6p + 3) ⇒ 6 (6p + 1)(3p + 1)(2p + 1), which is divisible by 6.
Case 3:
If n = 6p + 2, then
n(n + 1)(n + 2) = (6p + 2) (6p + 3)(6p + 4) ⇒ 12 (3p + 1)(2p + 1)(2p + 3), which is divisible by 6.
Similarly, n(n + 1)(n + 2) is divisible by 6 if n = 6p + 3 or, 6p + 4 or, 6p + 5.