$k$-multiplicatively perfect numbers If $M(n)$ denotes the product of the divisors of $n$, then $n$ is said to be $k$-multiplicatively perfect if $M(n) = n^k$. 

What are the first eight $3$-multiplicatively perfect numbers?

 A: Joffan almost got the right answer, but missed one option. It's true that $n=p^2q$ gives $k=3$, but so does $n=p^5$.
For $n=p^2q$, the list becomes 12, 18, 20, 28, 44, 45, 50, 52, ... You get this list by, for each prime $p$, making lists of $p^2q$, and then merging these lists.
For $n=p^5$, the list becomes $2^5=32$, $3^5=243$, ...
So the first eight should be: 12, 18, 20, 28, 32, 44, 45, 50, 52, ...
More generally, if $n=\prod_{i=1}^r p_i^{m_i}$, the divisors are the numbers on the form $d=\prod_{i=1}^r p_i^{\delta_i}$ where $0\le\delta_i\le m_i$. So there are $\prod_{i=1}^r (m_i+1)$ different divisors.
The next step can be done in a number of ways. One is to note that if $d$ is a divisor, then $n/d$ is a divisor, so when $M(n)=\prod_{d|n} d$, this makes
$$
M(n)^2 = \prod_{d|n} d\times\prod_{d|n} \frac{n}{d}
= \prod_{d|n} n = n^{\prod_i (m_i+1)}
$$
which makes $M(n)=n^k$ where $k=\frac12\prod_{i=1}^r(m_i+1)$. You then obtain the possible lists $m_1,\ldots,m_r$ from taking different ways of factoring $2k$. Just note that the factors need not be prime, so for $k=3$ you can factor $2k=3\cdot2$ which makes $(m_i)=(2,1)$ or $2k=6$ which makes $(m_i)=(5)$.
A: $p^2q$ with $p,q$ prime seems to be required to fit the condition. So the smallest is $12$, then $18, 20$ etc.
Edit for completeness: $p^5$ is also an option since $\sum_0^5 i = 15 $ so $\prod_0^5a^i = a^{15} = (a^5)^3$
