Let $m/n$, in lowest terms, be the [probability](https://artofproblemsolving.com/wiki/index.php/Probability) that a randomly chosen positive [divisor](https://artofproblemsolving.com/wiki/index.php/Divisor) of $10^{99}$ is an integer multiple of $10^{88}$.  Find $m + n$.
$10^{99} = 2^{99}5^{99}$, so it has $(99 + 1)(99 + 1) = 10000$ factors. Out of these, we only want those factors of $10^{99}$ which are divisible by $10^{88}$; it is easy to draw a [bijection](https://artofproblemsolving.com/wiki/index.php/Bijection) to the number of factors that $10^{11} = 2^{11}5^{11}$ has, which is $(11 + 1)(11 + 1) = 144$. Our probability is $\frac{m}{n} = \frac{144}{10000} = \frac{9}{625}$, and $m + n = \boxed{634}$.