Levin $M$.

Find all natural numbers divisible by 30 and having exactly 30 distinct divisors.
The number of divisors of the number $n=p_{1}^{\alpha_{1}} \ldots p_{k}^{\alpha_{k}}$ is $\left(\alpha_{1}+1\right) \ldots\left(\alpha_{k}+1\right)$ (see problem $\underline{60537}$ a). $n$ is divisible by $30=$ $2 \cdot 3 \cdot 5$, so $k \geq 3$. But

$\left(\alpha_{1}+1\right) \ldots\left(\alpha_{k}+1\right)=30$, and 30 cannot be factored into more than three factors, different from 1 (and it can be factored into three factors in a unique way). Therefore, $k=3,\left\{\alpha_{1}, \alpha_{2}, \alpha_{3}\right)=\{1,2,4\}$. Thus, $n=p q^{2} r^{4}$, where $(p, q, r)$ is a permutation of the numbers $(2,3,5)$.

## Answer

$720,1200,1620,4050,7500,11250$.