Question:
If $n$ is a positive integer such that $2n$ has 28 positive divisors and $3n$ has 30 positive divisors, then how many positive divisors does $6n$ have?

Answer:
Let $\, 2^{e_1} 3^{e_2} 5^{e_3} \cdots \,$ be the prime factorization of $\, n$.  Then the  number of positive divisors of $\, n \,$ is $\, (e_1 + 1)(e_2 + 1)(e_3 + 1) \cdots \; $. In view of the given information, we have \[
28 = (e_1 + 2)(e_2 + 1)P
\]and \[
30 = (e_1 + 1)(e_2 + 2)P,
\]where $\, P = (e_3 + 1)(e_4 + 1) \cdots \; $. Subtracting the first equation from the second, we obtain $\, 2 = (e_1 - e_2)P,
\,$ so either $\, e_1 - e_2 = 1 \,$ and $\, P = 2, \,$ or $\, e_1
- e_2 = 2 \,$ and $\, P = 1$.  The first case yields $\, 14 = (e_1
+ 2)e_1 \,$ and  $\, (e_1 + 1)^2 = 15$; since $\, e_1 \,$ is a nonnegative integer, this is impossible. In the second case, $\,
e_2 = e_1 - 2 \,$ and $\, 30 = (e_1 + 1)e_1, \,$ from which we find $\, e_1 = 5 \,$ and $\, e_2 = 3$.  Thus $\, n = 2^5 3^3, \,$ so $\, 6n = 2^6 3^4 \,$ has $\, (6+1)(4+1) = \boxed{35} \,$ positive divisors.