# 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!

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Start by finding the prime factorization of $4725$. – David Mitra Jun 17 '13 at 15:58

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 $0$.

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I don't understand fully what you have added, can you please clarify just a little the last portion? – Jeel Shah Jun 17 '13 at 16:16
@gekkostate For a divisor to be divisible by $5$, it needs at least one $5$ in its prime factorization, rather than $0$. – Tim Vermeulen Jun 17 '13 at 16:17
@gekkostate, Absolutely. The prime factorization of $4725$ is $7 \cdot 5^2 \cdot 3^3$. So we have $2$ ways to use the $7$ (one seven or no seven). For $5^2$, we have $2$ ways (one five or two five). We can't use zero fives, because it won't be divisible by $5$. We have $4$ ways to choose the $3$ (either $0, 1, 2$ or $3$ threes could be used). The answer is the product, or $2 \cdot 3 \cdot 4$. – George V. Williams Jun 17 '13 at 16:18
@GeorgeV.Williams Wow, that makes sense :D Thanks a bunch! – Jeel Shah Jun 17 '13 at 16:21

Using Divisor function,

as $4725=7^1\cdot5^2\cdot3^3$

the number of divisors will be $(1+1)(2+1)(3+1)$

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Why do we add $1$ to each of number of exponents? Is it to include the number itself? – Jeel Shah Jun 17 '13 at 16:01
@gekkostate, as the divisors are the cross product of $\{1,7\}\{1,5,5^2\},\{1,3,3^3,3^3\}$ – lab bhattacharjee Jun 17 '13 at 16:03
The next part of the question asks me to find how many of those divisors will be divisble by $5$, how would I do that? – Jeel Shah Jun 17 '13 at 16:04
@gekkostate, the number of element of the cross product $\{1,7\}\{5,5^2\},\{1,3,3^3,3^3\}$ i..e, $2\cdot 2\cdot4$ – lab bhattacharjee Jun 17 '13 at 16:08