Finding the divisors of the number $p^3q^6$ 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..
 A: 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?
A: 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.
A: 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:
$$\begin{matrix}
2^03^0 & 2^13^0 & 2^23^0 & 2^33^0 & 2^43^0 & 2^53^0 & 2^63^0  \\
2^03^1 & 2^13^1 & 2^23^1 & 2^33^1 & 2^43^1 & 2^53^1 & 2^63^1 & \\
2^03^2 & 2^13^2 & 2^23^2 & 2^33^2 & 2^43^2 & 2^53^2 & 2^63^2 & \\
2^03^3 & 2^13^3 & 2^23^3 & 2^33^3 & 2^43^3 & 2^53^3 & 2^63^3
\end{matrix}$$
A: 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.
A: 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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