## Task 1 - 120721

Determine the pairs $(x, y)$ of natural numbers $x$ and $y$ for which the following conditions hold:

(1) The sum of the two numbers $x$ and $y$ is 15390.

(2) If the single-digit number $x$ is placed in front of the number $y$, the resulting number $z$ is four times as large as the number $u$ obtained by placing the number $x$ behind the number $y$.
Assuming two numbers $x$ and $y$ have the required properties, then:
(1) $x + y = 15390$
(2) $z = 4u$

From (1) and because $x$ is a single-digit number, $y$ has an 8 as its second-to-last digit and a non-zero digit as its last digit.

Because of (2), $z$ is divisible by 4. According to the divisibility rules for the number 4, the last two digits of $z$, which are also the last two digits of $y$, form a number divisible by 4. Therefore, $y$ ends in 84 or 88. Hence, $x$ can only be 6 or 2.

If $x = 2$, then $y = 15388$, and one would get $z = 215388$ and $u = 153882$, which contradicts $4 \cdot 153882 = 615528 \neq 215388$. Therefore, only $x = 6$ and $y = 15384$ can have the required properties.

Indeed, these two numbers satisfy (1), and with them, one further obtains $z = 615384$ and $u = 153846$, so that because $4 \cdot 153846 = 615384$, (2) is also satisfied. Thus, there is exactly the possibility $x = 6, y = 15384$ to satisfy conditions (1) and (2).