How do you solve $[x]+[2x]+[3x]=4x$ on $\Bbb R$?

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 $$\overline{x}=\frac{0+\frac12+\frac34}3=\frac5{12}$$

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 $$3=2k+1+2 \Rightarrow x=\frac{3}{4} $$