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

33

We have,

$[\frac{x}{2}]+[\frac{x}{3}]+[\frac{x}{5}]=\frac{31}{30}x$

$⇒[\frac{x}{2}]+[\frac{x}{3}]+[\frac{x}{5}]=\frac{x}{2}+\frac{x}{3}+\frac{x}{5}$

$⇒[\frac{x}{2}]=\frac{x}{2},[\frac{x}{3}]=\frac{x}{3}$ and $[\frac{x}{5}]=\frac{x}{5}$

$⇒\frac{x}{2},\frac{x}{3},\frac{x}{5}$ are all integers

⇒ x is a multiple of the LCM of 2, 3, 5.

⇒ x is a multiple of 30.

Also, $0 < x <1000$.

$∴x = 30, 60, 90,...., 990$

or, $x = 30 × 1, 30 × 2, 30 × 3,...., 30 × 33$

Hence, there are 33 values of x.