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!
 A: 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 $\,r'\in \{0,1,2\}$
A: 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.
A: You can use the Division Algorithm to write
$$
n=3k+r
$$
where $0\leq r<3$.  If $r=0$ then $n$ is divisible by $3$. If $r=1$, then $n+2$ is divisible by $3$.  If $r=2$, then $n+1$ is divisible by $3$.
A: Since $Z$/3 has 3 elements,they are 0,1,2.
That means every integer has one of the form as following: 3k , 3k+1 , 3k+2, k is an integer
so you can divide this problem into three cases.
(1)if n=3k, then n/3=k;
(2)if n=3k+1, then (n+2)/3=k+1;
(3)if n=3k+2, then (n+1)/3=k+1;
Q.E.D
You can also try to prove 3 divides $n(n+1)(n+2)$
That implies 3|n or 3|(n+1) or 3|(n+2)
A: If $n \in \mathbb{Z} \to$ $n = 3k$ or $3k+1$ or $3k+2$, hence one of the statements must be true.
