Thai Mathematical Olympiad

It can be seen from the given equation that $x$ must be a positive integer. Let $x=15k+r$ where $0\le r\le 14$ is an integer and $k$ is a nonnegative integer. Then $$15k+r+\left[5k+\frac{r}{3}\right]=\left[10k+\frac{2r}{3}\right]+\left[9k+\frac{3r}{5}\right]$$ which simplifies to $k+r+\left[\frac{r}{3}\right]=\left[\frac{2r}{3}\right]+\left[\frac{3r}{5}\right]$. Thus, $$k+r+\left(\frac{r}{3}-1\right)<k+r+\left[\frac{r}{3}\right]=\left[\frac{2r}{3}\right]+\left[\frac{3r}{5}\right]\le \frac{2r}{3}+\frac{3r}{5}=\frac{19r}{15}$$ which follows that $k<1-\frac{r}{15}\le 1$. But $k\ge 0$; so, we have $k=0$. It can now be verified that the only solutions are $x=2,5$.