1. Determine all values of the natural number $n$ for which the system

$$
\left\{\begin{array}{l}
x+y=n^{2} \\
10 x+y=n^{3}
\end{array}\right.
$$

has a solution in the set of natural numbers.
Solution. If we subtract the first equation from the second equation, we get

$$
9 x=n^{2}(n-1)
$$

Since we are looking for solutions in the set of natural numbers, we have $9 \mid n^{2}(n-1) \ldots$ According to this, there are two possible cases.

Case 1. 3|n. Then $\quad x=k^{2}(3 k-1)$ and $y=k^{2}(10-3 k)$, where $n=3 k$, for some natural number $k$. Since $y$ is a natural number, possible values for $k$ are 1, 2, and 3. In this case, $n=3,6,9$.

Case 2. Let $9 x=n^{2}(n-1)$ and $9 \mid n-1$. Then there exists a natural number $m$ such that $n=9 m+1$, so $x=m(9 m+1)^{2}$ and $y=(9 m+1)^{2}(1-m)$, meaning for no natural number $m$, $y$ can be a natural number.

Therefore, the only values of $n$ for which the system has solutions in the set of natural numbers are 3, 6, and 9.