4. Find all real numbers $x, y$ that satisfy the system of equations

$$
\begin{aligned}
& x \cdot\left\lfloor\frac{y}{x}\right\rfloor=5, \\
& y \cdot\left\lfloor\frac{x}{y}\right\rfloor=-6 .
\end{aligned}
$$

(The notation $\lfloor a\rfloor$ denotes the floor function of $a$, i.e., the greatest integer less than or equal to $a$.)

The regional round of category B is held

## on Tuesday, April 17, 2012

so that it starts in the morning and contestants have 4 hours of pure time to solve the problems. Each problem can earn a contestant 6 points, and a successful solver is a student who scores 10 points or more. Allowed aids are writing and drawing materials and school MF tables. Calculators, laptops, and any other electronic aids are not allowed. This information will be communicated to the students before the start of the competition.
4. It must be $x y \neq 0$. It is also $x \neq y$; if $x=y$, we would get $\lfloor x / y\rfloor=\lfloor y / x\rfloor=1$, so from the first equation we would get $x=5$ and from the second equation $y=-6$, which contradicts the equality $x=y$. Similarly, it cannot be $x=-y$, so it is even $|x| \neq|y|$.

If both unknowns had the same sign, one of the numbers $x / y, y / x$ would be in the interval $(0,1)$, so its integer part would be 0, which does not lead to a solution. Therefore, one of the unknowns must be positive and the other negative, and both integer parts in the equations are negative. From them, it also follows that $x<0$ and $y>0$.

The signs of the numbers $x, y$ are different and the absolute values of these numbers are also different, so exactly one of the numbers $x / y, y / x$ is in the interval $(-1,0)$, and its integer part is therefore -1.

First, we examine the case $\lfloor x / y\rfloor=-1$. From the second given equation, we get $y=6$. The first given equation then takes the form $\lfloor 6 / x\rfloor=5 / x$. Using the definition of the integer part, we get

$$
\frac{5}{x}=\left\lfloor\frac{6}{x}\right\rfloor \leqq \frac{6}{x}, \quad \text { i.e. } \quad \frac{5}{x} \leqq \frac{6}{x}
$$

The last inequality, however, cannot hold for a negative number $x$, as it is equivalent to the inequality $5 \geqq 6$. Therefore, the given system of equations has no solution in the case $\lfloor x / y\rfloor=-1$.

The remaining case is $\lfloor y / x\rfloor=-1$. From the first given equation, we have $x=-5$. The second given equation then takes the form $\lfloor-5 / y\rfloor=-6 / y$. According to the definition of the integer part, we have

$$
\frac{-5}{y}1$. Using the definition of the integer part also for the equation $\lfloor y /-5\rfloor=-1$, we get

$$
-1 \leqq \frac{y}{-5}1 / y \geqq \frac{1}{5} \text{ and } -5<-5 / y \leqq-1
$$

From the last one, it follows that the expression $\lfloor-5 / y\rfloor$ can only take the values $-5,-4,-3,-2,-1$.

From the second equation rewritten as $y=-6 /\lfloor-5 / y\rfloor$, it follows that the unknown $y$ can only take the values $\frac{6}{5}, \frac{3}{2}, 2,3,6$. The last of these, however (unlike the first four), does not satisfy the derived criterion $1<y \leqq 5$ for the validity of the first equation of the system. Whether the second equation is satisfied for the first four values must be verified at least by checking the value of the expression $\lfloor-5 / y\rfloor$ that led us to them. For $y$ equal to $\frac{6}{5}, \frac{3}{2}, 2$, 3, the fraction $-5 / y$ is successively equal to $-\frac{25}{6},-\frac{10}{3},-\frac{5}{2} \text{ and }-\frac{5}{3}$ with corresponding integer parts indeed $-5,-4,-3,-2$.

Summarizing all considerations, we get that the solutions to the given system are the following pairs of numbers $(x, y):\left(-5, \frac{6}{5}\right),\left(-5, \frac{3}{2}\right),(-5,2),(-5,3)$.

For a complete solution, award 6 points, of which 1 point for determining that both integer parts in the equations are necessarily negative, and another two points for determining that $x=-5$ or $y=6$ must hold. For a procedure based on mere experimentation with selected numbers, award 2 points for the fortuitous finding of all four solutions, if any solution is missing, award 1 point, for the fortuitous finding of a single solution, do not award any points.