If $0<x<1000$and $\left[\frac{x}{2}\right]+\left[\frac{x}{3}\right]+\left[\frac{x}{5}\right]=\frac{31}{30}x$ where $\left[x\right]$ is the greatest integer less than or equal to x the number of possible values of $x$ is

$\left[\frac{x}{2}\right]+\left[\frac{x}{3}\right]+\left[\frac{x}{5}\right]=\frac{31}{30}x=\frac{x}{2}+\frac{x}{3}+\frac{x}{5}$ are all integers

So, $x=$multiple of the l cm of 2, 3, 5 i.e., multiply of 30

The number of multiplies of 30 between 0, 1000 is 33

The number of possible solutions of x is 33