Find all natural $n$ and $k$, satisfying the equation $k^{5}+5 n^{4}=81 k$.

#
$5 n^{4}=k\left(9+k^{2}\right)(3+k)(3-k)$. The left side of this equation is positive for any natural value of $n$, so the right side must also be positive. Therefore, it is sufficient to check two natural values of $k: k=1$ and $k=2$.

1) If $k=1$, then $5 n^{4}=80$, which means $n=2$.
2) If $k=2$, then $5 n^{4}=130$. There is no such natural number $n$.

## Answer

$n=2, k=1$.

Author: Raskin M.A.

There are four cards with digits: $2,0,1,6$. For each of the numbers from 1 to 9, a four-digit number can be formed from these cards that is divisible by the chosen single-digit number. In what year will this happen again?

## Solution

For a number to be divisible by 9, the sum of its digits must be divisible by 9. Therefore, the next such year will not be earlier than 9 years later, that is, in 2025. This year meets the condition of the problem, as from the cards 2, 0, 2, and 5, the number 2520 can be formed. It is divisible by any single-digit number from 1 to 9, since $2520=$ $2^{3} \cdot 3^{2} \cdot 5 \cdot 7$.

## Answer

In the year 2025.