Find the solutions of equation $\{x\} + \{2x\} + \{3x\} + \{4x\}=3, x\in [0, 2017]$

Just find the solution in $(0,1)$, Consider an $x$ belonging to $(0,1)$

If $ \{ x \}<.25$

Clearly it is not possible as then $ \{ x \}+ \{ 2x \}<1$ hence the sum cannot be $3$

Now if it's between $ 0.25\leq\{ x \}<.5$ then also it's not possible as either $ \{ 3x \}$ or $ \{ 4x \}<0.5$ hence by adding we get that the sum is less than $3$

Now if $ 0.5\leq\{ x \}<.75$ then also we will have a contradiction(Do it similarly $ \{ x \}<0.75$ so either $ \{ 3x \}$ or $ \{ 4x \}<0.75$ and $ \{ 2x \}<0.5$, we can get a contradiction by adding all)

So we have $ \{ x \}\geq0.75$ So $ \{ 2x \}=2x-1$ $ \{ 3x \}=3x-2$ $ \{ 4x \}=4x-3$

So $x+2x-1+3x-2+4x-3=3$ so $x=9/10$