Let $n \in \mathbb{N}$. If $2n$ has $30$ positive factors and $3n$ has $32$ positive factors, how many positive factors does $6n$ have? Question: Let $n \in \mathbb{N}$. If $2n$ has $30$ positive factors and $3n$ has $32$ positive factors, how many positive factors does $6n$ have?
the answer is given: https://artofproblemsolving.com/wiki/index.php/1996_AHSME_Problems/Problem_29
What I have tried so far: I searched for similar question online and I can't understand what the result is saying, here's some of my question: 
(1) what's the relationship between the number of factors and prime factor? Is it just because the prime factor cannot further decompose? 
(2) Is there a way to do this question without testing all possible outcome?
(3) What are the restrictions of this method? Is it applicable in any situation? 
I am just curious when I saw this math competition problem online, and it really tells me I know nothing about number theory... I am in chaos now.
 A: If $n=p^aq^br^c$ with $p,q,r$ prime the number of divisors of $n$ is $(a+1)(b+1)(c+1)$.  This extends to any number of prime factors.  This is because to get a divisor you can choose $0$ to $a$ factors of $p$, $0$ to $b$ factors of $q$, and $0$ to $c$ factors of $r$.
A: If $2$ and $3$ are coprimes to $n$ then $2n$ and $3n$ have the same number of positive factors. It follows without difficulty that $n$ should be divisible by $2$ and also by $3$. Consider 
$n=2^a3^bp^c$ (we prove first with one prime $p\gt3$ which is plausible because $2n$ has only $30$ positive divisors).  We have
$$2n=2^{a+1}3^bp^c\\3n=2^a3^{b+1}p^c\\6n=2^{a+1}3^{b+1}p^c$$  Then
$$(a+2)(b+1)(c+1).=30\\(a+1)(b+2)(c+1).=32$$ 
Putting $a+1=X$, $b+1=Y$ one has
$$XY+Y=\frac{30}{c+1}\\XY+X=\frac{32}{c+1}$$ Then $c+1=2^h$ where easily $h=1$ so 
$$XY+Y=15\Rightarrow X+1=5, Y=3\Rightarrow (a+2,b+1)=(5,3)$$ This is compatible with $XY+X=16=X(Y+1)=4\cdot4$.
Consequently $\color{red}{n=2^33^2p}$ where $p$ is an arbitrary prime greater than $3$ and $6n=2^43^3p$ has $\color{red}{40}$ positive divisors.
A: Write $n=2^a 3^b m$, with $\gcd(m,6)=1$. Let $c=\tau(m)$. Then
$$
30=\tau(2n)=(a+2)(b+1)c
\\
32=\tau(3n)=(a+1)(b+2)c
$$
Subtracting these equations gives $2=(a-b)c$. Thus, there are two cases:


*

*$c=1$ and $a-b=2$: Then $b^2 + 5 b - 26=0$, but the solutions of this equation are not integers.

*$c=2$ and $a-b=1$: Then $16=(a+1)(b+2)$ implies $(b + 2)^2 = 16$, and so $b=2$ and $a=3$.
Therefore,
$$
\tau(6n)=(a+2)(b+2)c=40
$$
Note that $c=2$ implies that $m$ is prime greater than $3$. So, $n=72m$ are the only possible solutions (although this is not required as an answer).
