How to show $n(n+1)(2n+1) \equiv 0 \pmod 6$? I've been asked to show that:
$n(n+1)(2n+1) \equiv 0 \pmod 6$
I found in a previous question that:
$n(n+1)$ was divisible by $2$ and resulted in an even number e.g
$n(n+1) \equiv 0 \pmod 2$
so I figured I needed to find:
$(2n+1) \equiv 0 \pmod 3$ in order to complete
$n(n+1)(2n+1) \equiv 0 \pmod 6$
but I am unsure on how to find $(2n+1) \equiv 0 \pmod 3$
Is this the right way to find the mod 6 and if so could you tell me how I could find
$(2n+1) \equiv 0 \pmod 3?$

The numbers $2n$, $2n+1$, and $2n+2$ are three consecutive numbers, so one of them is divisible by $3$. If $2n+1$ is not, then one of $2n$ or $2n+2$ is. But if $2n$ is divisible by $3$, so is $n$. And if $2n+2$ is divisible by $3$, so is $n+1$. Thus one of $n$, $2n+1$, and $n+1$ is divisible by $3$.
