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

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

Solution:

To Prove: the product of three consecutive positive integers is divisible by 6.

Proof: 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

$\Rightarrow n(n+1)(n+2)=6 q(6 q+1)(6 q+2)$, which is divisible by 6

If n = 6q + 1

$\Rightarrow n(n+1)(n+2)=(6 q+1)(6 q+2)(6 q+3)$

$\Rightarrow n(n+1)(n+2)=6(6 q+1)(3 q+1)(2 q+1)$

Which is divisible by 6

If n = 6q + 2

$\Rightarrow n(n+1)(n+2)=(6 q+2)(6 q+3)(6 q+4)$

$\Rightarrow n(n+1)(n+2)=12(3 q+1)(2 q+1)(2 q+3)$

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.