Prove that if $n$ is a product of two consecutive integers, its units digits must be $0,2, $or $6$ 
Prove that if $n$ is a product of two consecutive integers, its units digits must be $0,2,$ or $6$.

I'm having a hard time with the $0,2,6$ and part but here is what I have so far.
Since $n$ is the product of two consecutive integers then one must be odd and the other must be even.
Let $m=2k+1$ represent the odd integer where $k \in \mathbb{Z}$, and let $q=2k$ represent the even integer where $k \in \mathbb{Z}$. Then
$$n=m \times q=(2k+1)(2k)=(2k)^2+2k$$
This tells me that $n$ is even. So how can I show that units digit must be $0,2,$ or $6$?
 A: When you multiply two numbers, the units place of the product is determined only by the units' places of the two numbers. So it is sufficient to check the cases one by one.
$$
0 \times 1 \text{ gives } 0 \\
1 \times 2 \text{ gives } 2\\
2 \times 3 \text{ gives } 6\\
3 \times 4 \text{ gives } 2\\
4 \times 5 \text{ gives } 0\\
5 \times 6 \text{ gives } 0\\
6 \times 7 \text{ gives } 2\\
7 \times 8 \text{ gives } 6\\
8 \times 9 \text{ gives } 2\\
9 \times 0 \text{ gives } 0\\
$$
A: You can also work over all the cases of $k$ modulo $5$.


*

*If $k=5t$, then 
$$(2(5t))^2+2(5t)=100t^2+10t=10(10t^2+t)$$
so our number ends with a $0$.

*If $k=5t+1$, then
$$(2(5t+1))^2+2(5t+1)=100t^2+40t+4+10t+2=10(10t^2+5t)+6$$
so our number ends with a $6$.

*If $k=5t+2$, then
$$(2(5t+2))^2+2(5t+2)=100t^2+80t+16+10t+4=10(10t^2+9t+20)$$
so our number ends with a $0$.

*If $k=5t+3$, then
$$(2(5t+3))^2+2(5t+3)=100t^2+120t+36+10t+6=10(10t^2+13t+40)+2$$
so our number ends with a $2$.

*If $k=5t+4$, then
$$(2(5t+4))^2+2(5t+4)=100t^2+160t+64+10t+8=10(10t^2+17t+70)+2$$
so our number ends with a $2$.
A: As has already been pointed out, we can find all the information we need by looking at $\mathbb{Z}/5\mathbb{Z}$. A simple check yields that $$\overline{0}\cdot\overline{1}=\overline{0}$$ $$\overline{1}\cdot\overline{2}=\overline{2}$$
$$\overline{2}\cdot\overline{3}=\overline{1}$$
$$\overline{3}\cdot\overline{4}=\overline{2}$$
$$\overline{4}\cdot\overline{0}=\overline{0}$$
Since the only other remainder in possible for $\overline{2}$ is odd, we needn't check it, but a quick check of $3\times 2$ yields that we actually mean $6$. The group is cyclic and we're therefore done.
