Problem - 4230
Let $m$ and $n$ be two distinct positive integers. Find the minimal value of $(m+n)$ such that the last three digits of $2017^m$ and $2017^n$ are equal.
This is equivalent to solve $2017^m \equiv 2017^n \pmod{1000}$. Assuming $n > m$, this equation can be re-written as $$2017^m(2017^{n-m}-1)\equiv 0 \pmod{1000}$$ or, $$17^{n-m}\equiv 1\pmod {1000}$$
This means $(n-m)$ is the order of $17$ modulo $1000$ which, by # 4230, is $100$. Accordingly, the answer to the original problem is $\boxed{102}$.