## Task 3 - 100933

If $x$ is a real number, let $[x]$ denote the greatest integer not greater than $\mathrm{x}$. (For example, $[3.7]=3,[-3.7]=-4,[4]=4$.)

Determine all real numbers $\mathrm{x}$ for which:

$$
\left[\frac{10+3 x}{6}\right]=\frac{5 x+3}{7}
$$
Assuming there is a real number $x$ that satisfies the given equation. Then $\frac{5 x+3}{7}$ is an integer, and there is a real number $a$ with $0 \leq a<1$, for which

$$
\frac{10+3 x}{6}=\frac{5 x+3}{7}+a
$$

holds. This implies $70+21 x=30 x+18+42 a$, from which we get

$$
x=\frac{52-42 a}{9}
$$

Due to $0 \leq a<1$, this results in

$$
\frac{10}{9}<x \leq \frac{52}{9}
$$

and further

$$
\frac{\frac{50}{9}+3}{7}<\frac{5 x+3}{7} \leq \frac{\frac{260}{9}+3}{7} \quad \text { or } \quad \frac{11}{9}<\frac{5 x+3}{7} \leq \frac{41}{9}
$$

Thus, the expression $\frac{5 x+3}{7}$ (since it is an integer) can only be equal to one of the numbers 2, 3, 4.

From $\frac{5 x+3}{7}=2$ we get $x=\frac{11}{5}$,

from $\frac{5 x+3}{7}=3$ we get $x=\frac{18}{5}$,

from $\frac{5 x+3}{7}=4$ we get $x=5$.

Therefore, $x=\frac{11}{5}, x=\frac{18}{5}, x=5$ can be at most solutions of $\left(^{*}\right)$. Indeed, these are solutions, as

$$
\begin{array}{lll}
{\left[\frac{10+\frac{33}{5}}{6}\right]=2} & , & \frac{\frac{55}{5}+3}{7}=2 \\
{\left[\frac{10+\frac{54}{5}}{6}\right]=3} & , & \frac{\frac{90}{5}+3}{7}=3 \\
{\left[\frac{10+15}{6}\right]=4} & , & \frac{25+3}{7}=4
\end{array}
$$

Taken from [2]