2. Determine the natural number whose square is equal to the fifth power of the sum of its digits (in decimal notation).
Solution. Let the number $n$ in decimal notation have $k$ digits and let $S$ be the sum of the digits of the number $n$. Then $9^{k-1}776 \geq 6^{5}$
Suppose for some $k \geq 6$ it holds that $9^{2 k-7} \geq k^{5}$. Then

$$
9^{2(k+1)-7}=81 \cdot 9^{2 k-7} \geq 81 k^{5}>(2 k)^{5}>(k+1)^{5}
$$

i.e., the inequality holds for $k+1$ as well.

From the above, it follows that $k \leq 5$ and $S \leq 45$. From the condition $n^{2}=S^{5}$ it follows that $n=S^{2} \sqrt{S}$, so $S$ must be a perfect square. Therefore, $S \in\{1,4,9,16,25,36\}$. We further check:

$$
\begin{array}{ll}
S=1, \quad n=S^{2} \sqrt{S}=1, & 1=1 \\
S=4, \quad n=32, & 3+2 \neq 4 \\
S=9, \quad n=243, & 2+4+3=9 \\
S=16, n=1024, & 1+0+2+4 \neq 16 \\
S=25, n=3125, & 3+1+2+5 \neq 25 \\
S=36, n=46656, & 4+6+6+5+6 \neq 36
\end{array}
$$

Therefore, the required numbers are 1 and 243.