TITLE: Prove that if $n$ is a product of two consecutive integers, its units digits must be $0,2, $or $6$
QUESTION [8 upvotes]: 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 $6$?

REPLY [2 votes]: 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.