Which is the smallest natural number ending in 4, such that moving its last digit to the beginning of the number results in four times the original number?
The task received two types of solutions. The first one is as follows:

Let the sought natural number with $n$ digits be denoted by $A$, that is,

$$
A=10^{n-1} \cdot a_{n-1}+10^{n-2} \cdot a_{n-2}+\ldots+10 a_{1}+4
$$

According to the condition,

$$
4 A=\frac{A-4}{10}+4 \cdot 10^{n-1}
$$

from which

$$
A=\frac{4\left(10^{n}-1\right)}{39}
$$

The smallest solution for $A$ is obtained when we find the smallest $n$ (with $n>0$) for which $10^{n}$ divided by 39 leaves a remainder of 1. Since $10^{n}$ divided by 3 leaves a remainder of 1, it is sufficient to find the remainder of the division by 13.

The result can be seen in the following table.

| $n$ | 1 | 2 | 3 | 4 | 5 | 6 |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| Remainder of $10^{n}$ divided by 13 | 10 | 9 | 12 | 3 | 4 | 1 |

Due to the cyclic nature of the remainder, it was sufficient to check up to $n=13$ to see if there is such a power of 10. There is, for $n=6$. Thus, the smallest solution is

$$
A=\frac{4\left(10^{6}-1\right)}{39}=102564
$$

The second type of solution:

If there exists a number with the above property, then writing its digits in the form $x_{n} x_{n-1} \ldots x_{1} 4$, we can say:

$$
\frac{x_{n} x_{n-1} \ldots x_{1} 4 \cdot 4}{4 x_{n} \ldots x_{2} x_{1}}
$$

Therefore,

$$
\begin{aligned}
& 4 \cdot 4=\quad 16 \text{, so } x_{1} \text{ can only be } 6, \\
& 64 \cdot 4=\quad 256 \text{, so } x_{2} \text{ can only be } 5, \\
& 564 \cdot 4=2256 \text{, so } x_{3} \text{ can only be } 2, \\
& 2564 \cdot 4=10256 \text{, so } x_{4} \text{ can only be } 0 \text{ and } \\
& x_{5} \text{ can only be } 1, \\
& 102564 \cdot 4=410256, \text{ so } x_{6}=4
\end{aligned}
$$

From here, the digits repeat cyclically, meaning the number can only be of the form $A, A A, \ldots A A \ldots A$, and among these, the smallest is 102564.