Mod operator and divisibility by $3$? I wanted to prove that $n(n+1)(2n+1)$ is always divisible by three, for this I used the following approach:
$n = 0, 1\pmod2$
for $n = 0$;
$$n(n+1)(2n+1) \equiv 0\pmod3$$
for $n = 1$;
$$n(n+1)(2n+1) \equiv 0\pmod3$$
But now I am unsure if this the correct way of doing because of the following eg:
Let $k = \text{an odd number}\equiv 1\mod3$
for $k = 1$;
$2k + 1 \equiv 0\pmod3$, therefore for every odd number $2k + 1$ is divisible by $3$, but this is not true.
So my question is the approach I am using to prove $n(n+1)(2n+1)$ divisible by $3$ is wrong or only the second example is wrong?, and why?

Every number $n$ is congruent to $0$, $1$, or $2$ mod $3$. 
If $n \equiv 0 \pmod 3$, then $3 \mid n$. 
If $n \equiv 1 \pmod 3$, then $2n \equiv 2 \pmod  3$ which implies $2n + 1 \equiv 3 \pmod 3$ which implies $2n + 1 \equiv 0 \pmod  3$. 
If $n \equiv 2 \pmod 3$, then $n + 1 \equiv 3 \pmod 3$ which implies $n + 1 \equiv 0 \pmod 3$. 
Hence for all for all $n$, $n(n + 1)(2n + 1)$ is divisible by $3$.
