10. Let $m$ and $n$ be positive integers such that $m>n$. If the last three digits of $2012^{m}$ and $2012^{n}$ are identical, find the smallest possible value of $m+n$.
(A) 98
(B) 100
(C) 102
(D) 104
(E) None of the above
10. Answer: (D)
We want to solve $2012^{m} \equiv 2012^{n}(\bmod 1000)$ which is equivalent to
$$
100012^{n}\left(12^{m-n}-1\right)
$$

Since $\left(12^{m-n}-1\right)$ is odd, we must have $812^{n}$ so $n \geq 2$. It remains to check that $12512^{m-n}-1$ i.e. $12^{m-n} \equiv 1(\bmod 125)$. Let $\varphi$ be Euler's phi function. As $\varphi(125)=125-25=100$, by Euler's theorem, we know that the smallest $m-n$ must be a factor of 100 . By checking all possible factors, we can conclude $m-n=100$ and so the smallest possible value for $m+n$ is 104 since $n \geq 2$.