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


It is clear that the equation cannot have irrational solutions (it follows from $\{x\}=x- \lfloor {x} \rfloor$).
Also, if $x$ is solution then $x+n, n \in \mathbb{Z}$ is also solution.
I have found a solution in the required interval $[0,2017], x=\frac 9 {10}$ therefore $x=\frac 9 {10} + n, n \in \mathbb{N}, n\le 2016$ are also solutions.
The question is: how can I prove these are all possible solutions?
 A: 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 $x=9/10$
A: (Expanding my comments to an answer with a different approach)
Fact. For all $x\in\Bbb{R}$ and all $n\in\Bbb{Z}$ the difference 
$n\{x\}-\{nx\}$ is an integer.
Proof.
If $x=m+\{x\}$ where $m\in\Bbb{Z}$, then we have
$$
nx=nm+n\{x\}.
$$
On the other hand $nx=\ell+\{nx\}$ for some integer $\ell$. Therefore
$$
n\{x\}-\{nx\}=(nx-nm)-(nx-\ell)=\ell-nm\in\Bbb{Z}.
$$
QED.
Let us first look for numbers $x$ such that $\{x\}+\{2x\}+\{3x\}+\{4x\}$ is an integer. Assume this is the case. By the above fact
$$
\begin{aligned}
&\{10x\}-\{x\}-\{2x\}-\{3x\}-\{4x\}\\
=&\bigg(\{10x\}-\{x\}-\{2x\}-\{3x\}-\{4x\}\bigg)+(-10+1+2+3+4)\{x\}\\
=&(\{10x\}-10\{x\})+(\{x\}-\{x\})+(2\{x\}-\{2x\})+(3\{x\}-\{3x\})+(4\{x\}-\{4x\})
\end{aligned}
$$
is always an integer. So if $\{x\}+\{2x\}+\{3x\}+\{4x\}$ is an integer we can conclude that $\{10x\}$ must also be an integer as a sum of two integers.
This means that if $x$ is a solution of your equation, then the number $10x$ must be an integer. Together with your observations this implies that you only need to check the choices
$x=0,1/10,2/10,\ldots,9/10$. The conclusion is that $x=9/10$ is the only solution in the range $[0,1)$.
A: Since $\{u\}=u-\text{an integer}$ for any $u$, we have
$$\{x\}+\{2x\}+\{3x\}+\{4x\}=10x-k$$
for some integer $k$.  Therfore, if $\{x\}+\{2x\}+\{3x\}+\{4x\}=3$, we must have
$$x={k+3\over10}$$
for some $k$.  If we restrict $x$ to the interval $(0,1)$, then the two displayed equations imply $0\le k\le6$, which at worst leaves $7$ numbers to check.  There might be some slick way to narrow things down further, but I don't see one offhand.
A: hint
since $x+n $ is a solution if $x $ is,
you just need to find solutions in $[0,1) $.
consider the cases


*

*$0 <x<\frac 14$

*$\frac 14 \leq x <\frac 13$

*$\frac 13 \leq x <\frac 12$

*$\frac 12\leq x <1$


For example, in $(0,\frac 14) $,
the equation becomes
$$10x=3$$.
$\frac{3}{10}\notin (0,\frac 14) $
there is no solution in $ [0,\frac 14) $.
...
