Sum of inverse cubes over terminating-decimal reciprocals

Let $S$ be the set of positive integers $k$ for which the decimal expansion of $\tfrac1k$ ends after finitely many digits (for instance $2\in S$ because $\tfrac12=0.5$, while $7\notin S$ because $\tfrac17$ repeats forever). Compute $\sum_{k\in S}\frac{1}{k^3}.$

$\tfrac{250}{217}$. The reciprocal $1/k$ terminates exactly when $k=2^a5^b$. The sum therefore factors into two geometric series: $\sum_a 8^{-a}\cdot\sum_b 125^{-b}=\tfrac87\cdot\tfrac{125}{124}=\tfrac{250}{217}$.