1. In the field of real numbers, solve the system of equations

$$
\begin{aligned}
2 x+\lfloor y\rfloor & =2022, \\
3 y+\lfloor 2 x\rfloor & =2023 .
\end{aligned}
$$

(The symbol $\lfloor a\rfloor$ denotes the floor of the real number $a$, i.e., the greatest integer not greater than $a$. For example, $\lfloor 1.9\rfloor=1$ and $\lfloor-1.1\rfloor=-2$.)

(Jaroslav Švrček)
SOLUTION. Since $\lfloor y\rfloor$ and 2022 are integers, from the equation $2 x+\lfloor y\rfloor=2022$ it follows that the number $2 x$ is also an integer, hence $\lfloor 2 x\rfloor=2 x$. Thus, by subtracting the first equation from the second, we eliminate the unknown $x$ from the given system. We get

$$
3 y-\lfloor y\rfloor=1
$$

Thanks to (1), $3 y$ is an integer, so it has (according to its remainder when divided by three) one of the expressions $3 k, 3 k+1$ or $3 k+2$, where $k$ is an integer. From this (after dividing by three) it follows that for the number $y$ we have either $y=k$, or $y=k+\frac{1}{3}$, or $y=k+\frac{2}{3}$, where $k=\lfloor y\rfloor$. We will now consider these three cases.

- In the case $y=k$, equation (1) is $3 k-k=1$ with a non-integer solution $k=\frac{1}{2}$.
- In the case $y=k+\frac{1}{3}$, equation (1) is $(3 k+1)-k=1$ with the solution $k=0$, which corresponds to $y=\frac{1}{3}$. The original system of equations is then clearly satisfied if and only if $2 x=2022$ or $x=1011$.
- In the case $y=k+\frac{2}{3}$, equation (1) is $(3 k+2)-k=1$ with a non-integer solution $k=-\frac{1}{2}$.

Conclusion. The only solution to the given system of equations is the pair $(x, y)=\left(1011, \frac{1}{3}\right)$.

NOTE. The derived equation (1) can also be solved by expressing a general real number $y$ in the form $y=k+r$, where $k=\lfloor y\rfloor$ and the number $r \in\langle 0,1)$ is the so-called fractional part of the number $y$. Substituting into (1) we get the equation

$$
3(k+r)-k=1 \quad \text { or } \quad 2 k=1-3 r \text {. }
$$

Since $2 k$ is an integer divisible by two and the number $1-3 r$ clearly lies in the interval $(-2,1\rangle$, the equality of these numbers occurs in the only case when $2 k=1-3 r=0$, i.e., $k=0$ and $r=\frac{1}{3}$, i.e., $y=\frac{1}{3}$.