1. In the domain of integers, solve the system of equations

$$
\begin{aligned}
(4 x)_{5}+7 y & =14 \\
(2 y)_{5}-(3 x)_{7} & =74
\end{aligned}
$$

where $(n)_{k}$ denotes the multiple of the number $k$ closest to the number $n$.

(P. Černek)
Solution. From the first equation of the given system, it follows that the number $7 y-14=7(y-2)$ is divisible by five, so $y=5 s+2$ for a suitable integer $s$. Then we have $2 y=10 s+4$, and thus $(2 y)_{5}=10 s+5$. After substitution into the system, we get the pair of equations $(4 x)_{5}+35 s=0$ and $10 s-(3 x)_{7}=$ $=69$. By subtracting seven times the second equation from twice the first equation, we eliminate the unknown $s$ and obtain the equation $2(4 x)_{5}+7(3 x)_{7}=-483$ for the unknown $x$. Since the function $F(t)=2(4 t)_{5}+7(3 t)_{7}$ is non-decreasing in the integer variable $t$ and $F(-18)=$ $=-532, F(-17)=-483$ and $F(-16)=-473$, our equation $F(x)=-483$ has the unique solution $x=-17$. From the equation $(4 x)_{5}+35 s=0$ it then follows that $s=2$, so $y=12$. The check for the pair $(x, y)=(-17,12)$ is easily done by substitution.

The given system has the unique solution $(x, y)=(-17,12)$.