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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.2012-07-23
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    @Michael - what's your point?2012-07-23
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    Usually a question worth answering is worth up-voting. It seems as if people often neglect that.2012-07-23
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    @Michael: I would be interested to hear your rationale for "Usually a question worth answering is worth up-voting".2012-07-23
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    After a question is asked and answered, it remains visible forever, so others with the same question might find the question and the answer. Posted questions and answers are a permanent resource. We're building that resource. An up-vote indicates that something is worth including in that resource. The rest of the argument is left as an exercise.2012-07-23
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    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.2012-07-23
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    http://number.subwiki.org/wiki/Divisor_count_function and http://www.artofproblemsolving.com/Wiki/index.php/Divisor_function#Counting_divisors2012-07-23
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    @GerryMyerson : That the same combinatorial problem can appear in various guises is worth having here.2012-07-25

5 Answers 5

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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.

link#1 link#2

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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 $$\sigma_0(n)=\prod_k (a_k+1)$$2012-07-23
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    +1 Inasmuch as it matters, I like this sort of answer. A concrete example, presented clearly, that tells it all.2012-07-23
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    even those with not-so-good-math can understand this answer. Great!2012-07-23
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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? http://en.wikipedia.org/wiki/Fundamental_theorem_of_arithmetic2012-07-23
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    @Chuck, yes.${}$2012-07-23
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    @GerryMyerson Yes they are prime. Sorry I didnt put that up there2012-07-23
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    @GerryMyerson I still dont get how they got 28 ?2012-07-23
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    @Misty, you haven't seen Mark's answer?2012-07-23
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    OK, Misty: how many things can you write down of the form $p^aq^b$ with $0\le a\le3$ and $0\le b\le6$?2012-07-23
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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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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.