10.49 Find all natural numbers $n$ such that the sum of the digits of the decimal number $2^n$ is equal to 5.
(10th All-Russian Mathematical Olympiad, 1984)
[Solution] It is not difficult to verify that the smallest value of $n$ that satisfies the problem's requirements is 5. This is because
$$
2^{5}=32, \quad 3+2=5 .
$$

Below, we prove that when $n \geqslant 6$, no $n$ satisfies the problem's requirements.
Since the last digit of $2^{n}$ is 2, 4, 6, or 8, and the sum of the digits of $2^{n}$ is 5, the cases where the last digit is 6 or 8 should be excluded.
(1) If the last digit of $2^{n}$ is 4, and the sum of the digits is 5, then
$$
2^{n}=\frac{100 \cdots 0}{k \uparrow} 4=4(25 \underbrace{00 \cdots 0}_{k-1 \uparrow}+1) \quad(k \geqslant 1) .
$$

Since
then
$$
32 \text { divides. }
$$

However, when $n \geqslant 6$,
$$
32 \mid 2^{n} \text {. }
$$

This leads to a contradiction.
(2) If the last digit of $2^{n}$ is 2, and the sum of the digits is 5, then the last few digits are
$$
02,22,012,112 \text {. }
$$

When the last few digits are $02,22,012$, $8 \nmid 2^{n}$, which contradicts the fact that when $n \geqslant 6$, $8 \mid 2^{n}$.
When the last few digits are 112,
if it is
$$
\begin{array}{c}
1112=2^{3} \cdot 139 \\
10112=2^{7} \cdot 79
\end{array}
$$

neither is a power of 2.
If it is
$$
1 \underbrace{00 \cdots 0}_{m \uparrow} 112=10^{m+3}+112,
$$

it is clear that $32 \nmid 100 \cdots 0112$, while $32 \mid 2^{n}$.
In summary, only $n=5$.