
1. In the domain of non-negative real numbers solve the system of equations

$$
\begin{aligned}
\lfloor 3 x+5 y+7 z\rfloor & =7 z, \\
\lfloor 3 y+5 z+7 x\rfloor & =7 x, \\
\lfloor 3 z+5 x+7 y\rfloor & =7 y .
\end{aligned}
$$

(Tomáš Bárta)


Solution. The first equation of the given system is fulfilled if and only if the following two conditions are satisfied:

$\triangleright$ the number $7 z$ is integer,

$\triangleright 7 z \leq 3 x+5 y+7 z<7 z+1$, i.e. $3 x+5 y \in\langle 0,1)$.

Similarly, the second and third equations are fulfilled if and only if the numbers $7 x$ and $7 y$ are integers and $3 y+5 z, 3 z+5 x \in\langle 0,1)$.

Now consider any triple of non-negative numbers $(x, y, z)$, which is the solution to the problem. The inequalities $z \geq 0$ and $3 z+5 x<1$ imply $5 x<1$, whence $7 x<\frac{7}{5}<2$. This means that non-negative integer $7 x$ is equal to one of the numbers 0 or 1 , i.e., $x \in\left\{0, \frac{1}{7}\right\}$. Similarly, $y, z \in\left\{0, \frac{1}{7}\right\}$.

At this point we have only $2^{3}=8$ triples $(x, y, z)$, which are candidates to solve the problem, so we could test them individually. However, this testing can be avoided by noting that if any two of the numbers $x, y, z$ were equal to $\frac{1}{7}$, one one of the expressions $3 x+5 y, 3 y+5 z, 3 z+5 x$ would be $\frac{8}{7}$, which is greater than 1 , and that is a contradiction. So, at most one of the numbers $x, y, z$ is equal to $\frac{1}{7}$ and the others are equal to zero. But then each of the three (non-negative) expressions $3 x+5 y, 3 y+5 z, 3 z+5 x$ is at most equal to $\frac{5}{7}$, so the conditions stated in the beginning of the solution as equivalence are satisfied and all such triples are solutions.

Conclusion. The problem has exactly 4 solutions

$$
(x, y, z) \in\left\{(0,0,0),\left(\frac{1}{7}, 0,0\right),\left(0,0, \frac{1}{7}, 0\right),\left(0,0, \frac{1}{7}\right)\right\} .
$$
