MODBasic Harvard-MIT Intermediate
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 $\boxed{50}$.

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