Which is the smallest natural number ending in 4, such that when its last digit is written at the beginning of the number, we get four times the original number?
I. Solution: Let the unknown digits of the sought number be denoted by $x_{1}, x_{2}, \ldots, x_{n}$. Thus, the problem seeks a number for which

$$
4 \cdot x_{1} x_{2} \ldots x_{n} 4 = 4 x_{1} x_{2} \ldots x_{n}
$$

Here, the consecutive letters represent the digits of a number in the decimal system, and the multiplication sign is always written. Multiplying the last digit on the left side by 4 $(4 \cdot 4 = 16)$, we find that $x_{n} = 6$; writing this on the left side, we can continue the multiplication and find that $4 \cdot 6 = 24, 24 + 1 = 25$, and thus $x_{n-1} = 5$. Continuing in this manner, we get the digits $2, 0, 1$ in sequence. Multiplying the last one by 4 gives 4, and no further carry is left, so we have

$$
4 \cdot 102564 = 410256
$$

There is no need to complete the procedure at the digit 4, as we would then arrive at numbers that are formed by the repeated sequence of the digits 102564. All these numbers have the desired property, and it is clear from the procedure that only these numbers satisfy the condition.

The result can also be obtained essentially in the same way if we determine the digits $x_{1}$ onwards by dividing the number on the right side.

II. Solution: Let the number formed by the digits before the 4 be $x$ and let it be $n$ digits long. Then the given number is $10 x + 4$. Moving the 4 to the front, it will represent $4 \cdot 10^{n}$, followed by the number $x$. Thus, the problem seeks natural numbers $x$ and $n$ for which

$$
4(10 x + 4) = 4 \cdot 10^{n} + x, \quad 39 x = 4(10^{n} - 4)
$$

The problem, therefore, seeks an integer $n$ for which $4(10^{n} - 4)$ is divisible by $39 = 3 \cdot 13$. Since

$$
10^{n} - 4 = (10^{n} - 1) - 3 = 99 \ldots 9 - 3
$$

is divisible by 3, we only need to ensure divisibility by 13. Since 4 is relatively prime to 13, only $10^{n} - 4$ can be divisible by 13. $10 - 4$ and $10^{2} - 4$ are not divisible by it, so $n > 2$. In this case,

$$
10^{n} - 4 = 100 \cdot 10^{n-2} - 4 = 4 \cdot (25 \cdot 10^{n-2} - 1) = 4[26 \cdot 10^{n-2} - (10^{n-2} + 1)]
$$

Here, 4 is relatively prime to 13, 26 is divisible by it, so $10^{n-2} + 1$ must be divisible by 13. We know that 1001 is divisible by 13, so $n-2 = 3, n = 5$ is the smallest exponent that meets the condition.

$$
x = \frac{4 \cdot (10^{5} - 4)}{39} = 10256
$$

Thus, the sought number is

102564.

Note: 1. The expression $10^{m} + 1$ (we wrote $m$ instead of $n-2$) is always divisible by 13 if $m$ is an odd multiple of 3, because if $m = 3(2k + 1)$, then

$$
\begin{gathered}
10^{m} + 1 = (10^{3})^{2k + 1} + 1 = \\
= (10^{3} + 1)(10^{3 \cdot 2k} - 10^{3(2k-1)} + 10^{3(2k-2)} + \ldots - 10^{3} + 1)
\end{gathered}
$$

and $10^{3} + 1 = 7 \cdot 11 \cdot 13$. If, however, $m = 3(2k + 1) + r, r = 1, 2, 3, 4$ or 5, then

$$
10^{m} + 1 = (10^{3(2k + 1) + r} + 10^{r}) - (10^{r} - 1) = 10^{r}(10^{3(2k + 1)} + 1) - (10^{r} - 1)
$$

Here, the first term is divisible by 13, but the second is not, because $9, 99 = 9 \cdot 11, 999 = 9 \cdot 111 = 9 \cdot 3 \cdot 37, 9999 = 9 \cdot 1001 + 990$ and $99999 = 99 \cdot 1001 + 900$; and here neither factor is divisible by 13, or the first term is divisible by it, but the second is not, so (writing $m = n-2$ back) all solutions of the written equation are

$$
x = \frac{4(10^{6k + 5} - 4)}{39}, \quad k = 0, 1, 2 \ldots
$$

It is easy to see that these are precisely the numbers of the form mentioned in the previous solution.

2. The question arises whether it is just a coincidence that we found a number that satisfies the condition. We only note that the last problem does not always have a solution. From a famous theorem of Fermat in number theory, or rather from its generalization by Euler, it follows that for every odd integer $a$ not divisible by 5, there is an exponent $m$ for which $10^{m} - 1$ is divisible by $a$; in contrast, $10^{m} + 1$ cannot be divisible by 3, because $10^{m} + 1 = (10^{m} - 1) + 2$, which gives a remainder of 2 when divided by 3, because the first term is divisible by 3, regardless of the natural number $m$.