2024/25 AHSMC Part 1 Problem 15

We have $\bigl|\frac{2}{5} - \frac{m}{n}\bigr| = \dfrac{|5m-2n|}{5n}$. If $|5m-2n| = 0$ then $n=5$, contradicting $n \ge 6$. If $|5m-2n| = 1$ then the error is $\dfrac{1}{5n}$, smallest when $n$ is largest. This congruence requires $n \equiv \pm 3 \pmod{5}$, and the largest such $n$ is $n = 98$, giving $m = 39$ and error $\dfrac{1}{5\cdot 98}$. If $|5m-2n| \ge 2$ then the error is at least $\dfrac{2}{5\cdot 99} > \dfrac{1}{5\cdot 98}$. Thus $\dfrac{39}{98}$ is closest, and $m+n = 137$. The answer is (C).