Annual Harvard-MIT Mathematics Tournament

Solution:

Answer: $3$. Note $1 = 1^{2} < 2^{2} < 3^{2} < 10 < 4^{2} < \cdots < 9^{2} < 10^{2} = 100$. Divide $2^{n}$ and $5^{n}$ by $10$ repeatedly until each is reduced to a decimal number less than $10$ but at least $1$; call the resulting numbers $x$ and $y$. Since $(5^{n})(2^{n}) = 10^{n}$, either $x y = 1$ or $x y = 10$. Because $2^{n}$ and $5^{n}$ begin with the same digit, $x$ and $y$ are bounded by the same pair of adjacent integers. It follows that either $x = y = 1$ or $3 \leq x, y < 4$. Because $n$ is positive, neither $2^{n}$ nor $5^{n}$ is a perfect power of $10$, so the former is impossible.