Question

For a real number $x,$ let $\left[x\right]$ denote the largest integer less than or equal to $x,$ and let $\left\{x\right\}=x–\left[x\right].$ The number of solutions $x$ to the equation $\left[x\right]\left\{x\right\}=5$ with $0\le x\le 2015$ is

• A. $0$
• B. $3$
• C. $2008$
• D. $2009$

Answer 1

The correct option is D $2009$

$\left[x\right]\left\{x\right\}=5\Rightarrow \left\{x\right\}=\frac{5}{\left[x\right]}$
$\therefore 0\le \frac{5}{\left[x\right]}<1$
$\Rightarrow 1\le \frac{5}{\left\{x\right\}}<\infty$
$\Rightarrow 5<\left[x\right]<\infty$
So, $\left[x\right]=6,7,\mathrm{8.....},2015$
$\because x=\left[x\right]+\left\{x\right\}=\{n+\frac{5}{n}|n\in \{6,7,8,....2015\left\}\right\}$
Hence, number of values of ${}^{\prime }{x}^{\prime }=2015-6=2009$