TITLE: Total number of divisors of factorial of a number
QUESTION [5 upvotes]: I came across a problem of how to calculate total number of divisors of factorial of a number. I know that total number of divisor of a number  $n= p_1^a p_2^b p_3^c $ is $(a+1)*(b+1)*(c+1)$ where $a,b,c$ are the powers of a number $n$ and $1<n<100000$. But how to calculate total number of divisiors for $n!$

REPLY [6 votes]: Just an addition to Ross Millikan's answer:
Remember that the highest power of a prime $p$ dividing $n!$ is given by the procedure:

Greatest integer less than or equal to $\frac{n}{p}$
Greatest integer less than or equal to $\frac{n}{p^2}$
Greatest integer less than or equal to $\frac{n}{p^3}$
Repeat until the greatest integer less than or equal to $\frac{n}{p^k}$ is $0$.
Add all of your numbers up!

Example: Power of $3$ in the factorization of $100!$:

Greatest integer less than or equal to $\frac{100}{3}$ is $33$
Greatest integer less than or equal to $\frac{100}{3^2}$ is $11$
Greatest integer less than or equal to $\frac{100}{3^3}$ is $3$
Greatest integer less than or equal to $\frac{100}{3^4}$ is $1$
Greatest integer less than or equal to all the fractions after this is zero: $\frac{100}{3^5} > \frac{100}{3^6} > \cdots$
Add: $33 + 11 + 3 + 1 = 48$.

I would assume you would do this for every prime under $n$ and use the formula in Ross's answer.