998. Find a natural number that is divisible by 9 and 5 and has 14 different divisors.
$\triangle$ Let the desired number be denoted by

$$
a=3^{\alpha_{1}} \cdot 5^{\alpha_{2}} \cdot p_{3}^{\alpha_{3}} \ldots p_{n}^{\alpha_{n}}
$$

where $p_{3}, p_{4}, \ldots, p_{n}$ are distinct prime numbers (different from 3 and 5), $\alpha_{1}, \alpha_{2}$ are natural numbers, and $\alpha_{3}, \alpha_{4}, \ldots, \alpha_{n}$ are non-negative integers.

By the condition,

$$
\left(\alpha_{1}+1\right)\left(\alpha_{2}+1\right)\left(\alpha_{3}+1\right) \ldots\left(\alpha_{n}+1\right)=14
$$

with $\alpha_{1} \geqslant 2, \alpha_{2} \geqslant 1$.

If $\alpha_{1}=2$, then $\alpha_{1}+1=3$. But since 14 is not divisible by 3, this case is impossible.

Let $\alpha_{1}>2$. Then already at $\alpha_{1}=3$ we get:

$$
\left(\alpha_{1}+1\right)\left(\alpha_{2}+1\right)=4\left(\alpha_{2}+1\right) \geqslant 8
$$

Therefore, the product

$$
\left(\alpha_{3}+1\right)\left(\alpha_{4}+1\right) \ldots\left(\alpha_{n}+1\right)
$$

can only be equal to 1, otherwise

$$
\left(\alpha_{1}+1\right)\left(\alpha_{2}+1\right)\left(\alpha_{3}+1\right) \ldots\left(\alpha_{n}+1\right)>14
$$

Hence,

$$
\alpha_{3}=\alpha_{4}=\ldots=\alpha_{n}=0
$$

i.e., the number $a$ has no other prime divisors except 3 and 5. We have:

$$
\left(\alpha_{1}+1\right)\left(\alpha_{2}+1\right)=14
$$

Then

$$
\alpha_{1}+1=7, \quad \alpha_{2}+1=2 ; \quad \alpha_{1}=6, \quad \alpha_{2}=1 ; \quad a=3^{6} \cdot 5=729 \cdot 5=3645
$$

Answer: 3645.