2. Solve in the set of real numbers the equation $[x+2015]-\left[\frac{5 x-2015}{2}\right]=\frac{x}{2}+2015$, where $[a]$ represents the integer part of the real number a.
## Solution.

If $x \in \mathbb{R}$ is a solution of the equation, then $\frac{x}{2} \in \mathbb{Z}$, so $x=2 p, p \in \mathbb{Z}$. (2p)

Then, $[x+2015]=2 p+2015,(2 p)$

$$
\left[\frac{5 x-2015}{2}\right]=\left[\frac{5 \cdot 2 p-2016+1}{2}\right]=\left[5 p-1008+\frac{1}{2}\right]=5 p-1008
$$

Substituting into the equation, we get $p=252$, so $x=504$. 1p