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 $6$?

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 $2$.
