For what real numbers $x$ is it true that $[5 x]=[3 x]+2[x]+1$? ( $[x]$ denotes the integer part of $x$.)
Let's transform the equation as follows:

$$
[5 x]=[3 x]+2[x]+1 ;[5 x]-5[x]=[3 x]-3[x]+1 ;[5(x-[x])]=[3(x-[x])]+1 ;[5\{x\}]=[3\{x\}]+1
$$

( $\{x\}$ denotes the fractional part of $x$.)

Let $a=[3\{x\}]$. Considering that $0 \leq 3\{x\}<3$, the possible values of $a$ are 0, 1, and 2. For a certain $a$, the $x$ values that are valid are those for which

$$
a \leq 3\{x\}<a+1 \text { and } a+1 \leq 5\{x\}<a+2
$$

which means

$$
\frac{a}{3} \leq\{x\}<\frac{a+1}{3} \quad \text { and } \quad \frac{a+1}{5} \leq\{x\}<\frac{a+2}{5}
$$

Writing these for the possible values of $a$, the set of possible values of $\{x\}$ is

$$
\begin{gathered}
\left(\left[0 ; \frac{1}{3}\right) \cap\left[\frac{1}{5} ; \frac{2}{5}\right)\right) \cup\left(\left[\frac{1}{3} ; \frac{2}{3}\right) \cap\left[\frac{2}{5} ; \frac{3}{5}\right)\right) \cup\left(\left[\frac{2}{3} ; 1\right) \cap\left[\frac{3}{5} ; \frac{4}{5}\right)\right)= \\
=\left[\frac{1}{5} ; \frac{1}{3}\right) \cup\left[\frac{2}{5} ; \frac{3}{5}\right) \cup\left[\frac{2}{3} ; \frac{4}{5}\right) .
\end{gathered}
$$

([p,q) denotes the half-open interval from $p$ to $q$, closed on the left and open on the right.)

Finally, the solution set of the equation is:

$$
\bigcup_{k \in \mathbf{Z}}\left(\left[k+\frac{1}{5} ; k+\frac{1}{3}\right) \cup\left[k+\frac{2}{5} ; k+\frac{3}{5}\right) \cup\left[k+\frac{2}{3} ; k+\frac{4}{5}\right)\right)
$$

Gábor Braun (Budapest, Szent István Gymnasium, 3rd grade)