Let's find a positive integer whose last digit is 2, and if we remove this digit from the end of the number and write it at the beginning, we get twice the number.
I. solution: If the last digit of the sought number is 2, then its double can only end in 4. This will also be the second-to-last digit of the sought number. The double of this is 8, which will be the second-to-last digit of the number's double, so the third-to-last digit of the sought number must also be 8. The third-to-last digit of the number's double is 6, since the double of 8 ends in 6. Therefore, the fourth-to-last digit of the sought number will be 6. Thus, we already have part of the number: ... 6842. By doubling this number, the fourth digit will also be the fourth digit from the end of the number formed by omitting the 2 from the sought number and writing it at the beginning. This digit is 3, since we must also add the 1 from the carry-over of $2 \cdot 8$. We can continue determining the digits of the sought number in this manner.

We will obtain the smallest number that satisfies the problem when we first reach 2 in the double of the number. Following this path, we arrive at the number

$$
105263157894736842=A
$$

Applying the procedure again to $2A$, the first digit is 2, which can be considered to have arisen from the prewriting of the last digit, and thus $A$ is a solution to the problem, and indeed the smallest solution.

However, we can continue the procedure by writing a 2 digit in front of the 1; in this case, the digits of $A$ would repeat, and we would obtain another solution in the form of $A \cdot 10^{18} + A$, and similarly, we can repeat the digits of $A$ periodically as many times as we like, always obtaining a solution. The procedure by which we obtained these solutions shows that only these solutions are possible. Therefore, all solutions to the problem are the numbers

$$
\begin{gathered}
A \cdot 10^{18 k} + A \cdot 10^{18(k-1)} + \ldots + A \cdot 10^{18} + A = A \cdot \frac{10^{18(k+1)} - 1}{10^{18} - 1} \\
(k=0,1,2, \ldots)
\end{gathered}
$$

Remarks: 1. The digit 2 has already appeared among the digits (the fourth digit of $A$), but we could not stop there because the 2 would be preceded by a 1 when the number is doubled.

2. We could have started determining the digits from the beginning of the number: the 2 can only be preceded by a 1 when doubled, and 21 can only be followed by 05, etc.
3. Many solvers were satisfied with providing $A$ and did not give all the solutions.

II. solution: Let the number after omitting the 2 be $x$, and let it be an $n$-digit number. The sought number can then be written as $10x + 2$. If we write the 2 in front of the number $x$, we get the number $2 \cdot 10^n + x$. According to the condition, we have:

$$
2(10x + 2) = 2 \cdot 10^n + x
$$

from which:

$$
x = 2 \cdot \frac{10^n - 2}{19}
$$

and the original number:

$$
10x + 2 = 20 \cdot \frac{10^n - 2}{19} + 2 = \frac{20 \cdot 10^n - 2}{19} = 10^n + \frac{10^n - 2}{19}
$$

These are integers if $10^n - 2$ is divisible by 19, or if $10^n$ gives a remainder of 2 when divided by 19. This means that if we start converting the number $\frac{1}{19}$ to a decimal, the remainder 2 will appear among the remainders, and then there is a number that satisfies the conditions of the problem; otherwise, there is not. We can obtain such a number by taking the quotient up to the appearance of the remainder 2, without the decimal point, and if the $n$-th 0 is "removed" and the remainder 2 appears, we add $10^n$ to the integer thus obtained.

Performing the division, the remainder 2 first appears after the 17th 0 is removed, and at this point the quotient is 5263157894736842, so the sought number corresponds to 105263157894736842. If we do not stop at the remainder "2" in the division, we get 1 as the next digit of the quotient and 1 as the remainder, so from here the previously obtained remainders and digits of the quotient repeat, including the 0 digit that appears after the next 0 is removed. Thus, continuing the division indefinitely, every 18th step from the first ",2" leads to a new ",2" remainder. We can stop at any of these.

By this method, we obtain all the numbers obtained in the previous solution, which satisfy the requirements of the problem.

Remark: In the conversion of $\frac{1}{19}$ to a decimal, every number from 1 to 18 appears as a remainder, so 19 is a so-called "cyclic denominator". However, the fact that the problem is solvable - that is, that the remainder 2 appears among the remainders - does not depend on whether all possible remainders actually appear. For example, if we required that the number become five times or seven times its value by moving the last digit to the front, we would end up with the requirement that the remainder 5 appears in the division of $\frac{1}{49}$, or the remainder 7 in the division of $\frac{1}{69}$, and both of these do occur, even though neither 49 nor 69 is a cyclic denominator (a composite number cannot be one).[^0]

[^0]: ${ }^{1}$ A more detailed explanation about such numbers is provided, for example, in the following book: Péter Rózsa: The World of Numbers. (Budapest, 1948. Új Nevelés Könyvtára, 144. p.)