11.3. Find all natural numbers $n$ such that the value of the expression $\sqrt{n \sqrt{n \sqrt{n}}}$ is a natural number less than 2219.
Answer: $n=1, n=2^{8}$ and $n=3^{8}$.

Solution. Transform the expression $\sqrt{n \sqrt{n \sqrt{n}}}=\left(n\left(n(n)^{\frac{1}{2}}\right)^{\frac{1}{2}}\right)^{\frac{1}{2}}=\left(n\left(n^{\frac{3}{2}}\right)^{\frac{1}{2}}\right)^{\frac{1}{2}}=$ $=\left(n(n)^{\frac{3}{4}}\right)^{\frac{1}{2}}=\left(n^{\frac{7}{4}}\right)^{\frac{1}{2}}=n^{\frac{7}{8}}$. Therefore, the number must be the eighth power of some natural number. For $n=4^{8}=2^{16}$, the check gives:

$$
\sqrt{2^{16} \sqrt{2^{16} \sqrt{2^{16}}}}=\sqrt{2^{16} \sqrt{2^{24}}}=\sqrt{2^{28}}=2^{14}=4096>2219
$$

The number $n=1$ is clearly a solution. It remains to check only $n=2^{8}$ and $n=3^{8}$.

$$
\begin{aligned}
& \sqrt{3^{8} \sqrt{3^{8} \sqrt{3^{8}}}}=\sqrt{3^{8} \sqrt{3^{12}}}=\sqrt{3^{14}}=3^{7}=2187<2219 \\
& \sqrt{2^{8} \sqrt{2^{8} \sqrt{2^{8}}}}=\sqrt{2^{8} \sqrt{2^{12}}}=\sqrt{2^{14}}=2^{7}=128<2219
\end{aligned}
$$

## Criteria.

7 points. A complete and justified solution.

3 points. The correct answer is given and the check is done for both cases, but it is not proven that there are no other solutions.

3 points. Only a correct estimate is provided, that is, it is proven that $n=4^{8}$ and larger do not satisfy the condition.

1 point. Only the correct answer.

0 points. Incorrect solution.