I am looking to show that $n$, $n + 1$, or $n + 2$ are divisible by 3 I am seeking to prove the following:
If $n \in \mathbb{Z}$, then exactly one of the following is true: $\frac{n}{3} \in \mathbb{Z}, \frac{n + 1}{3} \in \mathbb{Z}, \frac{n + 2}{3} \in \mathbb{Z}$.
I believe this is pretty self evident, and hence have no idea where to start. Any tips would be greatly appreciated!

Induction proof:
$0$ is divisible by $3$.
If one of $n,n+1,n+2$ is divisible by $3$ then one of $(n+1)+2,(n+1)+1,(n+1)$ is divisble by $3$.
Prove the negative case from the positive case, or prove it be induction, too.
The more general theorem is that if $a,b$ are integers with $b\neq 0$ then there exists $q,r$ so that:
$$a=bq+r,\,r<|b|$$
This is called the division algorithm.
