
1. Solve the following system of equations in the domain of real numbers

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

(The symbol $\lfloor a\rfloor$ denotes the lower integer part of a real number a, i.e. the greatest integer not greater than a. E.g. $\lfloor 1,9\rfloor=1$ and $\lfloor-1,1\rfloor=-2$.) (Jaroslav Švrček)


Solution. Since $\lfloor y\rfloor$ and 2022 are integers, the equation $2 x+\lfloor y\rfloor=2022$ implies that $2 x$ is also an integer, so $\lfloor 2 x\rfloor=2 x$. Thus we can eliminate the unknown $x$ by subtracting the first equation of the system from the second one. We get

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

Thanks to (1), $3 y$ is an integer, so it has (according to its remainder after division by three) one of the forms $3 k, 3 k+1$ or $3 k+2$, where $k$ is an integer. From this it follows that either $y=k$, or $y=k+\frac{1}{3}$, or $y=k+\frac{2}{3}$, where $k=\lfloor y\rfloor$. We now discuss these three cases.

- In the case of $y=k,(1)$ becomes $3 k-k=1$ with the non-integer solution $k=\frac{1}{2}$.
- In the case of $y=k+\frac{1}{3}$, (1) is the equation $(3 k+1)-k=1$ with a solution $k=0$, which corresponds to $y=\frac{1}{3}$. The original system of equations is then apparently fulfilled, precisely when $2 x=2022$, i. e. $x=1011$.
- In the case of $y=k+\frac{2}{3}$, (1) is the equation $(3 k+2)-k=1$ with a non-integer solution $k=-\frac{1}{2}$.

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

REMARK. The equation (1) can also be solved in such a way that we write $y$ 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 $y$. Substituting into (1) we get the equation

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

Since $2 k$ is an integer divisible by two and the number $1-3 r$ apparently lies in the interval $(-2,1)$, 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. $y=\frac{1}{3}$.

Another SOLUtion. The verbal definition attached to the problem formulation tells us that $\lfloor a\rfloor$ is an integer for which $\lfloor a\rfloor \leq a$ and at the same time $\lfloor a\rfloor+1>a$. So, estimates $a-10$. If we modify the first part of (2) to $2024 \leq 2 x+y+2$, then together with the second part of (3) we obtain $3 y+2 x<2024 \leq 2 x+y+2$. This time $y<1$ follows from the comparison of the outermost expressions.

Together, we got $0<y<1$, so $\lfloor y\rfloor=0$ holds. Thanks to this, the first equation of the original system is reduced to the form $2 x=2022$, which is satisfied only for $x=1011$. By inserting it into the second equation, we get $3 y+2022=2023$ with the only solution $y=\frac{1}{3}$ which indeed satisfies the condition $\lfloor y\rfloor=0$ used in the first equation. The pair $(x, y)=\left(1011, \frac{1}{3}\right)$ is therefore the only solution of the given system.

REmark. Derivation of the equality $\lfloor y\rfloor=0$ can be accelerated by the following approach. We subtract the first given equation from the second one and write the result in the form

$$
2 y=1+(2 x-\lfloor 2 x\rfloor)-(y-\lfloor y\rfloor) \text {. }
$$

Since the two expressions in round brackets on the right-hand side belong to the interval $\langle 0,1)$, apparently the entire right-hand side is in the interval $(0,2)$. Thus, $0<2 y<2$, i.e. $0<y<1$ holds, from which $\lfloor y\rfloor=0$ follows.
