Let $S$ be the sum of all numbers of the form $a/b,$ where $a$ and $b$ are [relatively prime](https://artofproblemsolving.com/wiki/index.php/Relatively_prime) positive [divisors](https://artofproblemsolving.com/wiki/index.php/Divisor) of $1000.$ What is the [greatest integer](https://artofproblemsolving.com/wiki/index.php/Floor_function) that does not exceed $S/10$?
Since all divisors of $1000 = 2^35^3$ can be written in the form of $2^{m}5^{n}$, it follows that $\frac{a}{b}$ can also be expressed in the form of $2^{x}5^{y}$, where $-3 \le x,y \le 3$. Thus every number in the form of $a/b$ will be expressed one time in the product
\[(2^{-3} + 2^{-2} + 2^{-1} + 2^{0} + 2^{1} + 2^2 + 2^3)(5^{-3} + 5^{-2} +5^{-1} + 5^{0} + 5^{1} + 5^2 + 5^3)\]
Using the formula for a [geometric series](https://artofproblemsolving.com/wiki/index.php/Geometric_series), this reduces to $S = \frac{2^{-3}(2^7 - 1)}{2-1} \cdot \frac{5^{-3}(5^{7} - 1)}{5-1} = \frac{127 \cdot 78124}{4000} = 2480 + \frac{437}{1000}$, and $\left\lfloor \frac{S}{10} \right\rfloor = \boxed{248}$.