Question

lf $0<x<1000$ 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

• A. $34$
• B. $32$
• C. $33$
• D. $30$

Answer 1

Answer: (C)

$33$

L.H.S is an integer

$\Rightarrow$ R.H.S also should be an integer

$\Rightarrow x=30k$ for $k\in N$

$\left[\frac{x}{2}\right]+\left[\frac{x}{3}\right]+\left[\frac{x}{5}\right]=15k+10k+6k=31k=$R. H. S

$\therefore$ the equation is always true when $x=30k$ and $k\in N$

From the given condition $0<x<1000$

$\therefore 0<30k<1000$

$\therefore k\in \{1,2,3\cdots \cdots 33\}$

Hence, correct answer is $33$