Find all natural numbers $k$ for which

$$
1^{k}+9^{k}+10^{k}=5^{k}+6^{k}+11^{k}
$$

holds.
First, we prove that if $k$ is greater than 4, then $11^{k}>10^{k}+9^{k}+1^{k}$. This is equivalent to

$$
1>\left(\frac{10}{11}\right)^{k}+\left(\frac{9}{11}\right)^{k}+\left(\frac{1}{11}\right)^{k}
$$

Using the fact that the function $x \mapsto a^{x}$ is strictly decreasing if $010^{k}+9^{k}+1$. Therefore, the only possible solutions for the problem are $0, 1, 2, 3$, or 4. Checking these five cases, we find that $k=0, 2, 4$ are the solutions.

Based on the work of Márta Jáger (Budapest, Veres Pálné Gymnasium, 4th grade)