## Task 2 - 220812

For a 22-digit number $z$, the following properties are required:

$z$ has the unit digit 7; if this last digit is removed and placed in front of the remaining 21 digits, the result is the same as the multiplication $7 \cdot z$.

Prove that there is exactly one such number $z$! Determine this number!
I. If a number $u z$ has the required properties, then:

There is a 21-digit number $x$ with $z=10 x+7$, and for it, the following holds

$$
7 \cdot 10^{21}+x=7 \cdot(10 x+7)
$$

From this, it follows that

$$
\begin{aligned}
69 x & =7 \cdot 10^{21}-49 \\
x & =\frac{7 \cdot 10^{21}-49}{69}=\frac{6999999999999999999951}{69}=1014492753623188405797
\end{aligned}
$$

The verification confirms the solution.