Floor Function Multiplied By Fractional Part

$\{ x\} [x]=x\\ \{ x\} [x]=[x]+\{ x\} \\ \{ x\} =\frac { [x] }{ [x]-1 } \\ As\quad \{ x\} \quad \in \quad (0,1)\\ \quad 0<\frac { [x] }{ [x]-1 } <1\\ On\quad Solving\quad The\quad Inequality\quad We\quad Get,\\ [x]\quad \le 0\\ For\quad Each\quad [x]\quad We\quad Have\quad One\quad \{ x\} \\ Therefore\quad Each\quad [x]\quad Gives\quad One\\ Solution\\ \therefore \quad Number\quad Of\quad Possible\quad Solutions\quad In\quad Given\\ Range\quad Is\quad 101\\ \\$

By simply drawing graphs of y=x and y={x}[x]. It will have 101 solutions. I like solving such equations by graphical method. That one is cool.!!! try it , i am sure you will enjoy drawing the graphs of the two expressions on L.H.S and R.H.S.

I am using graphic method as Sandeep Bhardwaj has suggested. I plotted $f(x) = \{ x \} \lfloor x \rfloor$ with a spreadsheet.

It can be seen that for every negative number there is a solution to the equation. There is a trivial solution when $x=0$ . When $x>0$ , $f(x)<0$ and there is no solution. Therefore the number of solutions is $100+1=\boxed{101}$ .