# Finding the divisors of a number

My text says that $p^3q^6$ has 28 divisors. Could anyone please explain to me how they got 28 here ? Edit: $p$ and $q$ are distinct prime numbers Sorry for the late addition..

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So far there are three answers and only one up-vote: mine. – Michael Hardy Jul 23 '12 at 3:04
@Michael - what's your point? – Gerry Myerson Jul 23 '12 at 3:05
Usually a question worth answering is worth up-voting. It seems as if people often neglect that. – Michael Hardy Jul 23 '12 at 3:12
@Michael: I would be interested to hear your rationale for "Usually a question worth answering is worth up-voting". – MJD Jul 23 '12 at 3:15
An upvote, to me, indicates that the question significantly increases the value of the resource. In my opinion, this question doesn't rise to that level. It's good enough to answer, but not good enough to upvote. – Gerry Myerson Jul 23 '12 at 3:54

There can be 0-3 factors of $p$, so there are 4 ways for that to occur. There are 0-6 factors for $q$, so there are 7 ways for that to occur.

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Let's take $p=3$ and $q=2$ as an example. Then the number is $3^32^6 = 1728$, and the 28 divisors of 1728 are:

$$\begin{matrix} 1&2&4&8&16&32&64 \\ 3&6&12&24&48&96&192 \\ 9 & 18 & 36 & 72 & 144 & 288 & 576 \\ 27 & 54 & 108 & 216 & 432 & 864 & 1728 \end{matrix}$$

These values are, respectively:

$$\begin{matrix} 2^03^0 & 2^13^0 & 2^23^0 & 2^33^0 & 2^43^0 & 2^53^0 & 2^63^0 \\ 2^03^1 & 2^13^1 & 2^23^1 & 2^33^1 & 2^43^1 & 2^53^1 & 2^63^1 & \\ 2^03^2 & 2^13^2 & 2^23^2 & 2^33^2 & 2^43^2 & 2^53^2 & 2^63^2 & \\ 2^03^3 & 2^13^3 & 2^23^3 & 2^33^3 & 2^43^3 & 2^53^3 & 2^63^3 \end{matrix}$$

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In fact, here is the general statement: if $$n=\prod_k p_k^{a_k}$$ is the prime factorization of $n$, then $$\sigma_0(n)=\prod_k (a_k+1)$$ – J. M. Jul 23 '12 at 3:31
+1 Inasmuch as it matters, I like this sort of answer. A concrete example, presented clearly, that tells it all. – copper.hat Jul 23 '12 at 4:09
even those with not-so-good-math can understand this answer. Great! – woliveirajr Jul 23 '12 at 16:19

I assume your text also says $p$ and $q$ are primes, and $p\ne q$ --- otherwise, the statement isn't true.

Do you know that any divisor of $p^3q^6$ must itself be of the form $p^aq^b$ for some $a,b$ with $0\le a\le3$ and $0\le b\le6$? If so, do you see how to get from there to 28?

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That follows from the fundamental theorem of arithmetic right? en.wikipedia.org/wiki/Fundamental_theorem_of_arithmetic – Carry on Smiling Jul 23 '12 at 3:06
@Chuck, yes.${}$ – Gerry Myerson Jul 23 '12 at 3:46
@GerryMyerson Yes they are prime. Sorry I didnt put that up there – MistyD Jul 23 '12 at 10:18
@GerryMyerson I still dont get how they got 28 ? – MistyD Jul 23 '12 at 10:20
@Misty, you haven't seen Mark's answer? – J. M. Jul 23 '12 at 11:14

The number of divisors of n=$\prod(p_i^{a_i})$ is $\sum(a_i+1)$ where $p_i$ are distinct primes.

So, the number of divisors of $p^3q^6$ is(1+3)(1+6) where p,q are distinct primes =>(p,q)=1.

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Any number $A$ is the product of a unique set of primes. if $A=P_1^{k_1}*P_2^{k_2}*....P_n^{k_n}$ then a divisor of A needs to be of the form $P_1^{m_1}*P_2^{m_2}*....P_n^{m_n}$ where $m_i\leq k_i$ for any $i\in \Bbb N \wedge i\leq n$

How many combinations of marbles can you make if you can choose from 3 white marbles and 6 black ones? if you pick 0 white there are 7 combinations. (0 black+0 white = 1). if you pick 1 white there are also 7 you can probably see that there are $4*7$ combinations.

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