## Problem 2

Solve the equation: $\left[x+\frac{1}{2}\right]+\left[x+\frac{1}{3}\right]=1$, where $[a]$ represents the integer part of the real number $a$.
## Problem 2

We denote $\left[x+\frac{1}{2}\right]=k \in \mathbb{Z} \Rightarrow\left[x+\frac{1}{3}\right]=1-k$

We write the inequalities for the two integer parts:

$$
\begin{aligned}
& k-\frac{1}{2} \leq x<k+\frac{1}{2} \\
& \frac{2}{3}-k \leq x<\frac{5}{3}-k
\end{aligned}
$$

From these, by eliminating $x$, we get $k \in\left(\frac{1}{12}, \frac{13}{12}\right)$ and since $k \in \mathbb{Z} \Rightarrow k=1$

Thus, $x \in\left[\frac{1}{2}, \frac{3}{2}\right) \cap\left[-\frac{1}{3}, \frac{2}{3}\right)=\left[\frac{1}{2}, \frac{2}{3}\right)$