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 $1$)

[x]-[1/x] is an integer. ∴x - (1/x)=(integer). Let x - (1/x)=n. x²-nx-1=0. Thus, x=(n+√n²+4)/2. Since 1/100 <=x<=100, it follows that 1/50<=(n+√n²+4)<=200. The integer values of n that satisfy the above inequality are a total of 199, from -99 to 99. The answer is 199.