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, 6q + 5.
If n = 6q,
n (n + 1) (n + 2) = 6q (6q + 1) (6q + 2), which is divisible by 6
If n = 6q + 1
n (n + 1) (n + 2) = (6q + 1) (6q + 2) (6q + 3)
n (n + 1) (n + 2) = 6 (6q + 1) (3q + 1) (2q + 1) Which is divisible by 6
If n = 6q + 2
n (n + 1) (n + 2) = (6q + 2) (6q + 3) (6q + 4)
n (n + 1) (n + 2) = 12 (3q + 1) (2q + 1) (2q + 3),
Which is divisible by 6.
Similarly we can prove others.
Hence it is proved that the product of three consecutive positive integers is divisible by 6.
