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
Answer & explanation
Correct answer: option 3
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.