Question

Let $\left[k\right]$ denote the greatest integer less than or equal $k.$ If $N$ be the sum of all the values of $x$ satisfying equation $\left[\frac{x}{2}\right]-\left[\frac{x}{3}\right]=\frac{x}{7},$ then $\frac{N}{3}$ is equal to

• A. $-7$
• B. $3$
• C. $-3$
• D. $7$

Answer 1

The correct option is A $-7$

Put $x=6q+r$ where $0\le r\le 5$
$\left[\frac{x}{2}\right]-\left[\frac{x}{3}\right]=\frac{x}{7}\Rightarrow \left[\frac{r}{2}\right]-\left[\frac{r}{3}\right]=\frac{-q}{7}+\frac{r}{7}$
For $r=0,$ we get $q=0\Rightarrow x=0$
For $r=1,$ we get $q=1\Rightarrow x=7$
For $r=2,$ we get $q=-5\Rightarrow x=-28$
For $r=3,$ we get $q=3\Rightarrow x=21$
For $r=4,$ we get $q=-3\Rightarrow x=-14$
For $r=5,$ we get $q=-2\Rightarrow x=-7$
$\therefore N=$ Sum of all values $=-21$
$\Rightarrow \frac{N}{3}=-7$