<s>[INST] Solve $\lfloor \frac{2x-1}{3} \rfloor + \lfloor \frac{4x+1}{6} \rfloor=5x-4$ I came across this problem and it doesn't seem so tricky, but I didn't do it.

Solve the equation
  $$\lfloor {\frac{2x-1}{3}}\rfloor + \lfloor {\frac{4x+1}{6}}\rfloor=5x-4.$$

My thoughts so far:
Trying to use the inequality $k-1 < \lfloor k\rfloor \leq k$
$5x-4$ is integer, so $x=\frac{a}{5},$ where $a$ is from $\mathbb Z,$ then replace in the original equation, to get
$$\lfloor \frac{4a-10}{30}\rfloor + \lfloor \frac{4a+5}{30}\rfloor =a-4$$
Let $k= \frac{4a-10}{30}$
so $\lfloor k\rfloor +\lfloor k+\frac{1}{2}\rfloor =a-4$
but $\lfloor k\rfloor =\lfloor k+\frac{1}{2}\rfloor $ or $\lfloor k\rfloor +1=\lfloor k+\frac{1}{2}\rfloor $, so we have 2 cases
I strongly think that a shorter solution exists, I would appreciate a hint! 
 [/INST] Let $\lfloor{\frac{2x-1}{3}}\rfloor + \lfloor{\frac{4x+1}{6}}\rfloor=n$, where $n\in\mathbb Z$.
Hence, $5x-4=n$, which gives $x=\frac{n+4}{5}$.
Thus,
$$\lfloor{\frac{2n+13}{15}}\rfloor + \lfloor{\frac{4x+21}{30}}\rfloor=n,\tag{1}$$
which gives
$$n\leq\frac{2n+13}{15}+ \frac{4x+21}{30}<n+2,\tag{2}$$ which gives
$0\leq n\leq2$.
Now, for $n=0$ we get $x=\frac{4}{5}$ and a substitution in the original equation gives $1=1$, 
which says that $1$ is the root.
$n=1$ gives $0=1$, $n=2$ gives $0=2$, 
which says that the answer is
$$\left\{\frac{4}{5}\right\}.$$
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