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. 
 A: 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 $\ell\le\min\{k,m\}$.
A: $10^{99}=2^{99} \times 5^{99}$
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 $\frac {16}{10000}$.
A: As $10^{99}=2^{99}\cdot 5^{99},$ 
using divisor function 1,2,  the number of divisors is $(1+99)(1+99)$
which are the product of  $$\{1, 2^1,\cdots,2^{98},2^{99}\}\text{ and }\{1, 5^1,\cdots,5^{98},5^{99}\}$$
To be multiple of $10^{96}$ we need to take $$\{ 2^{96},2^{97},2^{98},2^{99}\}\text{ and }\{ 5^{96},5^{97},5^{98},5^{99}\}$$ i.e., there are $4\cdot4=16$ of them
So, the required probability is $$\frac{4^2}{100^2}=\frac1{625}$$
A: Since $10^{99} = (2\times5)^{99} = 2^{99} \times 5^{99}$,
Every factor of $10^{99}$ is of the form $2^p\times 5^q$ 
p and q can each be $0,1,2,...,99$ which is $100$ choices each.
So there are $100\times 100 = 10000$  factors of $10^{99}$ 
That's the denominator of the desired probability.

Next we calculate the numerator of the probability.
They are the positive integers of the form $2^p\times5^q$
which are multiples of $10^{96}$.
So p and q can can each be chosen as $96,97,98,99$  
So there are $4\times 4$ = 16 factors of $10^{99}$ which are multiples of $10^{96}$.
Therefore the desired probability is $16/10000=1/625$.
