hanulyeongsam wrote:
Find the number of $ x $ which follows the following :
$ x-\frac{1}{x}=[x]-[\frac{1}{x}] $
$ ( \frac{1}{100} \le x \le {100} ) $
$x=1$ fits while $x=100$ does not
If $x$ is a solution, $\frac 1x$ is a solution too. So let us look only for $x\in(1,100)$
Let $x=n+y$ with $n\in\{1,2,...,99\}$ and $y\in[0,1)$
Equation is $y=\frac 1{n+y}$ and so $y=\frac{-n+\sqrt{n^2+4}}2$ which indeed $\in[0,1)$
And so $99$ solutions in $(1,100)$ : $\frac{n+\sqrt{n^2+4}}2$ for $n=1,2,3,...,99$
And so $\boxed{199\text{ such }x\in[\frac 1{100},100]}$ (the $99$ we just got, plus the $\frac 1x$ for each plus $1$)