ISI B.Math Entrance(8)

$\begin{aligned} |2{\color{#3D99F6}x} - \lfloor x \rfloor| & = 4 & \small \color{#3D99F6} \text{Note that } x = \lfloor x \rfloor + \{ x \} \\ |2{\color{#3D99F6} \lfloor x \rfloor} + 2{\color{#3D99F6}\{ x \}} - \lfloor x \rfloor| & = 4 & \small \color{#3D99F6} \text{where } 0 \le \{ x \} < 1 \\ | \lfloor x \rfloor + 2\{ x \} | & = 4 \end{aligned}$

Since the RHS is integral, the LHS must also be integral and there are two cases:

$\begin{cases} \{x\} = 0 & \implies | \lfloor x \rfloor + 0 | = 4 & \implies \lfloor x \rfloor = \begin{cases} -4 & \implies x = -4 \\ 4 & \implies x = 4 \end{cases} \\ \{x\} = \frac 12 & \implies | \lfloor x \rfloor + 1 | = 4 & \implies \lfloor x \rfloor = \begin{cases} -5 & \implies x = -4.5 \\ 3 & \implies x = 3.5 \end{cases} \end{cases}$

Therefore, there are $\boxed{4}$ solutions.

Given, $\displaystyle |x+\{x\}|=4$ and we need to know $\{x\}\in[0,1)\;\forall\;x\in\mathbb{R}$

For $x>0$ , $\displaystyle 3<x=4-\{x\}\le 4$ of which $x=4$ is clearly a solution. To see if other solutions are present we let $x=3+f$ which gives,

$\displaystyle 3+f+f=4\implies f=0.5$ which gives another solution $x=3.5$

Similar investigation for $x<0$ would lead to two more solutions namely $x=-4,x=-4.5$

There are $\boxed{4}$ solutions