Question

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

Answer 1

Let $n$ be 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

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

If $n=6q+1,$ then

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

which is divisible by $6$.

If $n=6q+2,$ then

$n(n+1)(n+2)=(6q+2)(6q+3)(6q+4)=12(3q+1)(2q+1)(2q+3),$

which is divisible by $6$.

Similarly , $n(n+1)(n+2)$ is divisible by $6$ if $n=6q+3$ or, $6q+4$ or, $6q+5.$