2000 AIME I Problem 11

Since $1000 = 2^3 5^3$ and $a, b$ are coprime, the prime $2$ appears in at most one of them, and likewise for $5$

So $S$ factors as $\left(1 + 2 + 4 + 8 + \frac12 + \frac14 + \frac18\right)$ times the analogous sum with $5$s

Write $a = 2^i 5^j$ and $b = 2^k 5^l$ with exponents between $0$ and $3.$ Coprimality means $\min(i, k) = 0$ and $\min(j, l) = 0,$ and these two constraints are independent. So as $(a, b)$ runs over all coprime pairs, the factor $2^{i - k}$ independently takes each value in $\{2^{-3}, \ldots, 2^{3}\}$ exactly once, and similarly for $5^{j - l}.$ Hence $$\begin{aligned} S &= \scriptsize \left(1 + 2 + 4 + 8 + \tfrac{1}{2} + \tfrac{1}{4} + \tfrac{1}{8}\right) \\ &\quad \scriptsize {}\cdot \left(1 + 5 + 25 + 125 + \tfrac{1}{5} + \tfrac{1}{25} + \tfrac{1}{125}\right) \\ &= \frac{127}{8} \cdot \frac{19531}{125}. \end{aligned}$$

This equals $\frac{2480437}{1000} = 2480.437,$ so $\frac{S}{10} = 248.0437,$ and the greatest integer not exceeding it is $248.$