Prove that $3^x + 3^{x-2}$ ends with $0$ for any integer $x > 1$ I think that $3^x+3^{x-2}$ ends in a $0$ (i.e. is divisible by $10$) $\forall x \in \Bbb Z, x > 1$.
Examples:
$3^2+3^{2-2}=9+1=10 \\
3^3+3^{3-2}=27+3=30 \\
3^4+3^{4-2}=81+9=90 .$
In fact, I wrote a quick Python program and left it on overnight, it reported every number in the domain working.
I don't know a proof for this, though, and I also don't know if it's already a theorem or something with a fancy name that I just happened to stumble across.
Also, I don't know any really good tags for this. If you know one, please comment or edit the post.

Let $x \geq 2$ be an integer. Note that $$3^x+3^{x-2}=3^{x-2}(3^2+1)=3^{x-2} \cdot 10$$
