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. 

$10^{99}=2^{99} \times 5^{99}$
There are $100$ ways to choose the exponent for $2$ and $100$ ways to choose the exponent for $5$, so that $10^{99}$ has $10000$ distinct factors.
To be divisible  by $10^{96}$ both exponents must be at least $96$, which leaves the four cases $96, 97, 98, 99$ for each exponent. This amounts to $4 \times 4= 16$ cases.
Assuming the distribution is such that any factor is equally likely to be chosen, this gives $\frac {16}{10000}$.
