1. The product of all positive divisors of a natural number $n$ is $20^{15}$. Determine $n$.
1. The number $20^{15}=2^{30} 5^{15}$ is divisible only by the prime numbers 2 and 5, so the sought number $n$ must be of the form $n=2^{a} 5^{b}$, where $a, b$ are natural numbers. Each of its positive divisors is thus of the form $2^{\alpha} 5^{\beta}$, where $\alpha \in\{0,1, \ldots, a\}$ and $\beta \in\{0,1, \ldots, b\}$, and from the theorem of the unique factorization of a natural number into a product of primes, it follows that for different $\alpha, \beta$ we get different divisors.

For each $\beta \in\{0,1, \ldots, b\}$, consider now all divisors of the number $n$ that are divisible by 5 exactly to the power of $\beta$. They are

$$
\underbrace{2^{0} 5^{\beta}, 2^{1} 5^{\beta}, 2^{2} 5^{\beta}, \ldots, 2^{a} 5^{\beta}}_{a+1 \text { numbers }}
$$

and their product is

$$
2^{0+1+2+\ldots+a} 5^{(a+1) \beta}=2^{a(a+1) / 2} 5^{(a+1) \beta} .
$$

If we multiply all these products for $\beta=0,1, \ldots, b$, we get the product of all positive divisors of the number $n$, which is thus equal to

$$
2^{a(a+1)(b+1) / 2} 5^{(a+1) b(b+1) / 2} .
$$

It follows that to find the number $n$, it suffices to solve the system of equations in the set of natural numbers

$$
\begin{aligned}
& \frac{1}{2} a(a+1)(b+1)=30 \\
& \frac{1}{2}(a+1) b(b+1)=15 .
\end{aligned}
$$

The expressions on both sides of these equations are clearly non-zero, and by dividing them, we get $a=2 b$ after rearrangement. Substituting into the first equation of the system, we then have

$$
b(b+1)(2 b+1)=30
$$

Since $30=2 \cdot 3 \cdot 5$, we see that one solution is $b=2$. Since the function $x(x+1)(2 x+1)$ is increasing on the set of positive numbers as the product of three increasing positive functions, this is the only solution. The system (1) therefore has a unique solution $a=4, b=2$, which corresponds to the sought natural number $n=2^{4} 5^{2}=400$.

There is a unique natural number $n$ that satisfies the conditions of the problem, and that is $n=400$.

For a complete solution, award 6 points. Of these, award 1 point for noting that each divisor of the number $n$ is of the form $2^{\alpha} 5^{\beta}$, 2 points for finding their product, 1 point for setting up the system (1) (or an equivalent system), and the remaining 2 points for solving it and finding the number $n$. If it does not follow from the solution method (1) that the found solution is unique and the student does not justify it, deduct 1 point. If the solver only proves that the number $n=400$ has the required property without any indication of its derivation, award 2 points (these points cannot be added to the points according to the previous scheme).