Task 10.

Find all natural numbers $n$ such that the sum $S(n)$ of the digits in the decimal representation of the number $2^{n}$ is 5.
Solution. To find the solution, we will construct a table of values for the expression $2^{n}$:

| $n$ | $2^{n}$ | $S(n)$ | $n$ | $2^{n}$ | $S(n)$ |
| :---: | ---: | ---: | :---: | ---: | :---: |
| 1 | 2 | 2 | 16 | 65536 | 25 |
| 2 | 4 | 4 | 17 | 131072 | 14 |
| 3 | 8 | 8 | 18 | 262144 | 19 |
| 4 | 16 | 7 | 19 | 524288 | 29 |
| 5 | 32 | 5 | 20 | 1048576 | 31 |
| 6 | 64 | 10 | 21 | 2097152 | 26 |
| 7 | 128 | 11 | 22 | +194304 | 25 |
| 8 | 256 | 13 | 23 | 8388608 | 41 |
| 9 | 512 | 8 | 24 | 16777216 | 37 |
| 10 | 1024 | 7 | 25 | 33554432 | 29 |
| 11 | 2048 | 14 | 26 | 67108864 | 40 |
| 12 | 4096 | 19 | 27 | 134217728 | 35 |
| 13 | 8192 | 20 | 28 | 268435456 | 43 |
| 14 | 16384 | 22 | 29 | 536870912 | 41 |
| 15 | 32768 | 26 |  |  |  |

Careful examination of this table leads to the conjecture that

$$
S(k+6 n)=S(k)+9 m
$$

where $n, k$ are natural numbers; $m$ is a natural number or zero.

But why, when multiplying the number $2^{n}$ by $2^{6}$, do we get a number whose sum of digits is either equal to the sum of the digits of the number $2^{n}$, or differs from it by $9 m$? What property does the number $2^{6}$ and its sum of digits possess?

We notice that $S(6)=10$ and $2^{6}=64=63+1=9 \cdot 7+1$.

But what follows from this? Let's consider an example:

$2^{14} \cdot 2^{6}=16384(9 \cdot 7+1)=9 \cdot 7 \cdot 16384+16384$

The number $9 \cdot 7 \cdot 16384$ is divisible by 9. Therefore, the sum of its digits is also divisible by 9. Now it is clear why equation (1) is true.

So, $S(5)=5$. But why are there no other solutions?

By equation (1), they should be sought among $S(11), S(17), S(23), S(29), \ldots$. But $S(11), S(23), S(35), \ldots$ end in the digit 8. Therefore, we are only interested in $S(17), S(29), S(41), \ldots$, i.e., $S(5+12 k)$, where $k$ is a natural number.

It is easy to show that all $S(5+(2 k+1) 12)$ end in the digits 7 and 2, i.e., $S(5+(2 k+1) 12) \geqslant 9$. Therefore, we are only interested in $S(5+24 k)$, which end in the digits 1 and 2 for $k \geqslant 1$. But numbers of the form $2^{5}+24 k \quad(k \geqslant 1, \quad k$ - a natural number) are greater than 1000000012. The number 2000000012 (its sum of digits $S=5$) after division by $2^{2}$ becomes odd, but $2^{3}+24 k$ is even.

The numbers $1100000012, \quad 1010000012, \quad 1001000012$, $1000100012,1000010012,1000001012$ also give an odd number when divided by $2^{2}$. The number 1000000112 is not divisible by $2^{5}$.

Thus, we get the answer: $n=5$.