2008
Problem - 2805
Find the smallest positive integer $n$ so that $107n$ has the same last two digits as $n$.
This is equivalent to solving the following equation $$n \equiv 107n \pmod{100}$$
or $$n \equiv 7n \pmod{100} \implies 6n \equiv 0 \pmod{100}$$ $$\therefore\quad 6n = 100k \implies n = 50 \cdot \frac{k}{3}$$
where $k$ is an integer. Clearly, the smallest such positive integer is $\boxed{50}$.