Let a be any positive integer and b = 6.
By division lemma there exists integers q and r such that
a = 6q+r where 0 ≤ r < 6a = 6q , or a = 6q+1 or, a = 6q+2 or, a = 6q+3 or a = 6q+4 or a = 6q+5
Let n is 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.
Case 1:
If n = 6q then
n(n + 1)(n + 2) = 6q (6q + 1)(6q + 2), which is divisible by 6.
Case 2:
If n = 6q + 1, then
n(n + 1)(n + 2) = (6q + 1)(6q + 2) (6q + 3)
= (6q + 1) 2(3q + 1) 3(2q + 1)
= 6(6q + 1)(3q + 1)(2q + 1), which is divisible by 6.
Case 3:
If n = 6q + 2, then
n(n + 1)(n + 2) = (6q + 2)(6q + 3)(6q + 4)
= 2(3q + 1) 3(2q + 1)(6q + 4)
= 6(3q + 1)(2q + 1)(6q + 4), which is divisible by 6.
Case 4:
If n = 6q + 3, then
n(n + 1)(n + 2) = (6q + 3) (6q + 4)(6q + 5)
= 3(2q + 1)2(3q + 2)(6q + 5)
= 6(2q + 1)(3q + 2)(6q + 5), which is divisible by 6
Case 5:
If n = 6q + 4, then
n(n + 1)(n + 2) = (6q + 4) (6q + 5)(6q + 6)
= (6q + 4) (6q + 5) 6(q + 1)
= 6(6q + 4) (6q + 5)(q + 1), which is divisible by 6.
Case 6:
If n = 6q + 5, then
n(n + 1)(n + 2) = (6q + 5) (6q + 6)(6q + 7)
= (6q + 5) 6(q + 1)(6q + 7)
= 6(6q + 5) (q + 1)(6q + 7), which is divisible by 6.
Hence,
n(n + 1)(n + 2) is divisible by 6