1988 AIME Problems/Problem 5

Problem

Let $m/n$ , in lowest terms, be the probability that a randomly chosen positive 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 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 $m + n = 634$ .