4. A natural number $n$ is such that the number $36 n^{2}$ has exactly 51 different natural divisors. How many natural divisors does the number $5 n$ have?
Answer: 16.

Solution. Let $m=p_{1}^{k_{1}} \cdot p_{2}^{k_{2}} \cdot \ldots \cdot p_{l}^{k_{l}}$, where $p_{1}, p_{2}, \ldots, p_{l}$ are pairwise distinct prime numbers. Then the number of natural divisors of the number $m$ is

$$
\tau(m)=\left(k_{1}+1\right)\left(k_{2}+1\right) \ldots\left(k_{l}+1\right)
$$

Indeed, a common divisor of the number $m$ has the form

$$
d=p_{1}^{a_{1}} \cdot p_{2}^{a_{2}} \cdot \ldots \cdot p_{l}^{a_{l}}
$$

where for each $i$ the exponent $a_{i}$ takes values from 0 to $k_{i}-$ a total of $k_{i}+1$ values. By the multiplication rule, we obtain the formula (*).

The number $m=36 n^{2}$ has at least two prime divisors (2 and 3).

Considering the equality $\tau(m)=3 \cdot 17$, we get two factors in the formula (*). Therefore, the number $m$ has the form $m=2^{2} 3^{16}$ or $m=2^{16} 3^{2}$. In this case, $n=p^{7}$, where $p$ is equal to 2 or 3. Then

$$
5 n=5 p^{7}, \quad \tau(5 n)=2 \cdot 8=16
$$

Evaluation. 13 points for a complete solution. Participants in the Olympiad can use the formula (*) without proof.