Question

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 [x] is the greatest integer less than of or equal to x, the number of possible values of x is

• A. 34
• B. 33
• C. 32
• D. none of these

Answer 1

The correct option is B 33

$[\frac{x}{2}],[\frac{x}{3}],[\frac{x}{5}]$ are integers $⟹\frac{31x}{30}$must be a integer

$⟹x$ is a multiple of $30⟹x$ is multiple of $2,3,5$

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

$\frac{x}{2}+\frac{x}{3}+\frac{x}{5}=\frac{31x}{30}$ is always true

So our conditions will be $0<x<1000,x$ is integral multiple of $30$

Number of possible $x$ is $33$