How many divsors of $4725$ are there? I need to solve the following problem: 

How many divsors of $4725$ are there?

I found the number of divsors between $0-9$ that can divide $4725$ which are: $3,5,7,9$ but how do I find the others? Also, what is a good way to approach such problems?
Thanks!
 A: Here's a proof I made for you which should let you finish the problem:

Theorem: If the prime factorization of $n$ is ${p_1}^{a_1} {p_2}^{a_2}{p_3}^{a_3} \ldots {p_n}^{a_n}$, then the number of factors of $n$ is the product of one plus each of the exponents, that is:
$$ \prod_{k=1}^n (a_k + 1) $$
Proof: For each $p_k$, we have $a_k + 1$ choices, we can include it $0$ times, $1$ time, $2$ times, all the way up until $a_k$ times. Since prime factorizations are unique, this is the only way to form these factors. By the counting principle, the number of choices is the product of the ways to choose each individual factor.


For the second part of the problem, consider this: we no longer have a choice to put $0$ fives. We must now put at least $1$ five. What effect will this have on the number of choices? Well there are still $a_k +1$ choices for all other $p_k$, but for $5$ we will only have $a_k$ choices, since there is no choice of putting $0$.
A: Using Divisor function,
as $4725=7^1\cdot5^2\cdot3^3$
the number of divisors will be $(1+1)(2+1)(3+1)$
