72nd Czech and Slovak Mathematical Olympiad

Since $\lfloor y\rfloor$ and $2022$ are integers, the equation $2x+\lfloor y\rfloor =2022$ implies that $2x$ is also an integer, so $\lfloor 2x\rfloor =2x$. Thus we can eliminate the unknown $x$ by subtracting the first equation of the system from the second one. We get $$3y-\lfloor y\rfloor =1.(1)$$ Thanks to (1), $3y$ is an integer, so it has (according to its remainder after division by three) one of the forms $3k$, $3k+1$ or $3k+2$, 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 $3k-k=1$ with the non-integer solution $k=\frac{1}{2}$.

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

* In the case of $y=k+\frac{2}{3}$, (1) is the equation $(3k+2)-k=1$ with a non-integer solution $k=-\frac{1}{2}$.

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

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Alternative solution.

The verbal definition attached to the problem formulation tells us that $\lfloor a\rfloor$ is an integer for which $\lfloor a\rfloor \le a$ and at the same time $\lfloor a\rfloor +1>a$. So, estimates $a-1<\lfloor a\rfloor \le a$ are valid for every real number $a$. According to them, we get from the first equation of the given system $$2022\le 2x+y<2023.(2)$$ Similarly, from the second equation follows $$2023\le 3y+2x<2024.(3)$$ We can combine these inequalities in two ways. Combining the second part of (2) with the first part of (3) we obtain $2x+y<2023\le 3y+2x$, whence from the comparison of the outermost expressions follows $y>0$. If we modify the first part of (2) to $2024\le 2x+y+2$, then together with the second part of (3) we obtain $3y+2x<2024\le 2x+y+2$. 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 $2x=2022$, which is satisfied only for $x=1011$. By inserting it into the second equation, we get $3y+2022=2023$ with the only solution $y=\frac{1}{3}$ which indeed satisfies the condition $\lfloor y\rfloor =0$ used in the first equation. The pair $(x,y)=(1011,\frac{1}{3})$ is therefore the only solution of the given system.