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!
By the Division Algorithm $\, n+2 = 3q+r,\,$ for some $\, r\in\{0,1,2\}$ 

Therefore we conclude that  $\ n+\! \underbrace{r'}_{\large 2\,-\,r}\!\! = 3q,\, $ for some $\,r'\in \{0,1,2\}$
If $n \in \mathbb{Z} \to$ $n = 3k$ or $3k+1$ or $3k+2$, hence one of the statements must be true.