Probability that a divisor of $10^{99}$ is a multiple of $10^{96}$ What is the probability that a divisor of $10^{99}$ is a multiple of $10^{96}$? How to solve this type of question. I know probability but  I'm weak in number theory. 

HINT: $10^n=2^n5^n$, so the divisors of $10^n$ are the numbers of the form $2^k5^m$, where $0\le k,m\le n$. Since there are $n+1$ choices for each of $k$ and $m$, $10^n$ has $(n+1)^2$ divisors. And $10^\ell\mid 2^k5^m$ if and only if $\ell\le\min\{k,m\}$.
