15. Find the maximum natural number which are divisible by 30 and have exactly 30 different positive divisors.
15. Answer: 11250 .
Let $a$ be a natural number divisible by 30 . Thus $a$ can be expressed as
$$
a=2^{n_{2}} 3^{n_{3}} 5^{n_{5}} \prod_{1 \leq i \leq r} p_{i}^{k_{1}},
$$
where $r \geq 0, k_{1} \geq 0$, and $n_{2}, n_{3}, n_{5}$ are positive integers. The number of positive divisors of $a$ is
$$
\left(n_{2}+1\right)\left(n_{3}+1\right)\left(n_{5}+1\right) \prod_{1 \leq i \leq r}\left(k_{i}+1\right) .
$$

Since $30=2 \times 3 \times 5$, we have $r=0$ if
$$
\left(n_{2}+1\right)\left(n_{3}+1\right)\left(n_{5}+1\right) \prod_{1 \leq i \leq r}\left(k_{1}+1\right)=30 .
$$

Hence $30 \mid a$ and $a$ has exactly 30 positive divisors iff $a=2^{n_{2}} 3^{n_{3}} 5^{n_{5}}$ and $\left(n_{2}+1\right)\left(n_{3}+\right.$ 1) $\left(n_{5}+1\right)=30$, where $n_{2}, n_{3}, n_{5}$ are all positive. Further, $\left(n_{2}+1\right)\left(n_{3}+1\right)\left(n_{5}+1\right)=30$, where $n_{2}, n_{3}, n_{5}$ are all positive, iff $\left\{n_{2}, n_{3}, n_{5}\right\}=\{1,2,4\}$.
The maximum value of such $a$ is $a=2 \times 3^{2} \times 5^{4}=11250$.