Let's find the smallest natural number $n$ whose representation in the decimal system ends with 6, and if we remove this last 6, and simultaneously write a 6 in front of the remaining digits, we get four times $n$.
I. solution. The last digit of $4n$ is 4, because in the usual multiplication starting from $4 \cdot 6 = 24$, we write down 4 units. Thus, we have also obtained the digit in the tens place of $n$, and we can continue the multiplication: in $4n$, we write down 8 tens from the product $4 \cdot 4$ tens and the 2 tens carried over from before - this will be the number of hundreds in $n$ - and we carry over 1 hundred. We must continue this process until we write down a 6 in $4n$ and there is no remainder to carry over. At this point, we have obtained the smallest suitable $n$:

![](https://cdn.mathpix.com/cropped/2024_05_02_36d794f115b52811feabg-1.jpg?height=143&width=593&top_left_y=411&top_left_x=755)

Continuing this procedure, the 6-digit number obtained would repeat, so any number with $6k$ digits that is formed by writing the number $153846$ $k$ times in succession has the property in question.

Tamás Berény (Budapest, Kölcsey F. g. II. o. t.)

Remark. Conversely, we can also obtain $n$ by division. The first digit of $4n$ is 6, so the first digit of $n$ is 1. Thus, the first two digits of $4n$ are 6,1, so the first two digits of $n$ are 1,5, and therefore the first three digits of $4n$ are 6,1,5, and so on, until we get a 6 in $n$:

![](https://cdn.mathpix.com/cropped/2024_05_02_36d794f115b52811feabg-1.jpg?height=141&width=482&top_left_y=867&top_left_x=816)

II. solution. Let the number of digits of the sought $n$ be $k$. By omitting the last 6, the remaining digits move to a place with 10 times their value, thus the number $(n-6)/10$ is formed, and by writing the 6 in front, we get a number that is $6 \cdot 10^{k-1}$ larger. Thus,

$$
6 \cdot 10^{k-1} + \frac{n-6}{10} = 4n, \quad \text{from which} \quad 2(10^k - 1) = 13n
$$

Here, the left side is divisible by 13. The factor 2 is relatively prime to 13, so $10^k - 1$ is divisible by 13. This number is written with $k$ nines, so $k$ is equal to the number of nines used until the remainder 0 first appears in the division of $999 \ldots 9$ by 13, which, by trial, is 6:

$999999: 13 = 76923$,
89
119
29
39
0

$n$ is equal to twice the quotient obtained, $n = 153846$.