AIME 1988 · 第 5 题

Solution

$10^{99} = 2^{99}5^{99}$, 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 $m + n = \boxed{634}$.

Solution 2

Like before there are $(99+1)(99+1) = 100^2 = 10,000$ positive divisors of $10^99$. Now we calculate the cases that satisfy the conditions. Notice that every multiple must be in the form $2^i 5^j$, where $88 \le i, j \le 99$. Then, there are $12$ choices for both $i$ and $j$. We can have any combination of $i$ and $j$ together, giving us $144$ favorable outcomes. Thus our probability is $\frac{144}{10,000}$, or $\frac{9}{625}$, in which $\frac{m}{n} = \frac{9}{625}$. Thus, $m+n = 9+625 = \boxed{634}$

~Pinotation