How do you solve $[x]+[2x]+[3x]=4x$ on $\Bbb R$? 
Find the arithmetic average of all solutions $x\in\Bbb R$ of the equation
  $$[x]+[2x]+[3x]=4x,$$
  where $[x]$ denotes the integer part of $x$ (e.g. $[2.5]=2$, $[-2.5]=-3$).

I tried solving this problem by looking at $\{x\}$ and writing for example $[2x]$ as $2[x]$ when $\{x\}<1/2$ and $2[x]+1$ when $\{x\}\ge1/2$. This lead to a lot of cases and after half an hour I literally couldn't go on any longer. I was thinking maybe I can somehow find the arithmetic average without actually knowing the solutions, but couldn't find any way to do that. Any help would be appreciated. :)
Thanks!
 A: As @Jakobian suggests we have that
$$6x-3\lt \lfloor x\rfloor+\lfloor 2x\rfloor+\lfloor 3x\rfloor\le6x$$
Hence any solutions would require that
$$6x-3\lt4x\le6x$$
$$2x-3\lt0\le2x$$
$$x-\frac32\lt0\le x$$
$$0\le x\lt\frac32$$
$$0\le 4x\lt6$$
Also, the only valid solutions $x$ are such that $4x=\lfloor x\rfloor+\lfloor 2x\rfloor+\lfloor 3x\rfloor\in\mathbb{Z}$. Using the above range this means that the only possible solutions are if $4x=0,1,2,3,4,5$ or $x=0,\frac14,\frac12,\frac34,1,\frac54$ respectively. We can test each case separately giving the only solutions
$$x=0,\frac12,\frac34$$
Hence the arithmetic average of the solutions is 
$$\overline{x}=\frac{0+\frac12+\frac34}3=\frac5{12}$$
A: Note that $4x \in \mathbb Z$ and hence $x= k +\frac{r}{4}$ for some $k \in \mathbb Z$ and $r \in \{ 0,1,2,3\}$.
Then, the equation becomes 
$$4k+r= k+[2k+\frac{r}{2}]+[3k+\frac{3r}{4}]=6k+[\frac{r}{2}]+[\frac{3r}{4}]$$
which is equivalent to
$$r=2k+[\frac{r}{2}]+[\frac{3r}{4}].$$
Now, just solve this for each $r \in \{ 0,1,2,3\}$:


*

*if $r=0$ then 
$$0=2k \Rightarrow x=0 $$

*if $r=1$ then 
$$1=2k \Rightarrow \mbox{ no solution}  $$

*if $r=2$ then 
$$2=2k+1+1 \Rightarrow x=\frac{1}{2}  $$

*if $r=3$ then 
$$3=2k+1+2 \Rightarrow x=\frac{3}{4}  $$
A: If $x=I+f$ where $0\le f<1$ and $I$ is an integer
If $3f<1,$
$$4I+4f=I+2I+3I\implies2I=4f, 0\le I<\dfrac23\implies I=0,f=?$$
If $1\le3f<2$ and $2f<1$
$$4I+4f=I+2I+3I+1\iff4f=2I+1\implies\dfrac43\le2I+1<2$$ no integer value available for $I$
If $1\le2f$ and $1\le3f<2$
$$4I+4f=6I+2,2f=I+1\implies 1\le I+1<\dfrac43,I=0,2f=1$$
Check if $2\le3f<3$
