## Task 10/67

We are looking for all natural numbers $z$ with the following property:

If you add the digit 1 as an additional digit at the end of it, you get seven times the number that results from increasing the highest digit of $z$ by 1.
The number $z$ has $n$ digits $(n=0 ; 1 ; 2 ; \ldots)$. By appending a 1, the number $10 z+1$ is obtained, and by increasing the highest digit by 1, the number $10^{n}+z$ is obtained. Then the equation

$$
7\left(10^{n}+z\right)=10 z+1
$$

has the solution $z=\frac{1}{3}\left(7 \cdot 10^{n}-1\right)$.

For $n=0$, $z=2$ follows, for $n=1$, $z=23$ is obtained, and for $n=3$, $z=233$. Generally, the number $z_{k}$ consists of the highest digit 2 and $k$ subsequent digits, all of which are 3.

If the highest digit is increased by 1, a number is obtained whose $n=k+1$ digits are all 3. Multiplying this number by 7, the lowest digit is 1 due to $3 \cdot 7=21$; the next $k$ digits result from the multiplication $3 \cdot 7=21$ and the carry 2, all being 3, while the highest digit is from the carry alone, being 2.

This product is precisely the number $10 z+1$.