<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] $$\lfloor{\frac{2x-1}{3}}\rfloor + \lfloor{\frac{4x+1}{6}}\rfloor=5x-4.$$
As you have been mentioned; 
$5x-4$ must be an integer, 
so $x=\frac{a}{5},$ 
where $a$ is from $\mathbb Z,$ 
then replace in the original equation, to get
$$\lfloor\frac{4a-10}{30}\rfloor + \lfloor\frac{4a+5}{30}\rfloor=a-4.$$ 

Notice that by Euclid's algorithm
there are integers $n,t \in \mathbb{Z}$; 
such that $a=15n+t$, with $0 \leq t \leq 14$,
by replacing we get that: 
$$ 
\lfloor\frac{60n+4t-10}{30}] + \lfloor\frac{60n+4t+5}{30}]=15n+t-4 
\Longrightarrow 
\\ 
2n+\lfloor\frac{4t-10}{30}] + 2n+\lfloor\frac{4t+5}{30}]=15n+t-4 
\Longrightarrow 
\\ 
\lfloor\frac{4t-10}{30}] + \lfloor\frac{4t+5}{30}] -t +4 =11n 
\ 
; 
\ \ \ \ \ 
\color{Blue}{\star} 
$$ 

but notice that $0 \leq t \leq 14$; implies the following inequalities: 
$$ 
-10 \leq 4t-10 \leq 46 
\ \ 
%%\text{and} 
; 
\ \ \ 
  5 \leq 4t+5  \leq 61 
\ \ 
%%\text{and} 
; 
\ \ \ 
-10 \leq -t+4 \leq 4 
\ \ 
\Longrightarrow 
\\ 
 \color{Red}{-1} \leq \lfloor\frac{4t-10}{30}\rfloor \leq  \color{Red}{1} 
\ \ 
%%\text{and} 
; 
\ \ \ 
  \color{Red}{0} \leq \lfloor\frac{4t+5}{30}\rfloor  \leq  \color{Red}{2} 
\ \ 
%%\text{and} 
; 
\ \ \ 
-10 \leq -t+4 \leq 4 
\ \ 
\Longrightarrow 
\\ 
 -1+0-10 
\leq 
\lfloor\frac{4t-10}{30}\rfloor + 
\lfloor\frac{4t+ 5}{30}\rfloor 
-t +4 
\leq  
1 + 2 + 4
\ \ 
\color{Green}{\star} 
; 
\overset{\color{Blue}{\star}}{\Longrightarrow} 
\\ 
-11 
\leq 
11n 
\leq  
7; 
$$

so we have only two choices for $n$ ; 


*

*$n=-1$, gives: 
$$ -1+0-10 = -11 = 
\lfloor\frac{4t-10}{30}\rfloor + 
\lfloor\frac{4t+ 5}{30}\rfloor 
-t +4 
; 
$$ 
which implies that: 
$$ 
4t-10 < 0 
\ \ 
; 
\ \ \ 
4t +5 < 30 
\ \ 
; 
\ \ \ 
-t+4 = -10 
; 
$$ 
so there is no possibilities for $t$ in this case. 

*$n=0$, gives: 
$$ 0 = 
\lfloor\frac{4t-10}{30}\rfloor + 
\lfloor\frac{4t+ 5}{30}\rfloor 
-t +4 
; 
$$ 
which implies that: 
$$ 
-3=-(\color{Red}{1+2}) 
\leq 
-t+4 
\leq 
-(\color{Red}{-1+0})=1 
; 
$$ 
so the only possibilities for $t$ is: 
$t=3, 4, 5, 6, 7$.
Now you can check by-hand that only $\color{Green}{t=4}$ gives an answer.
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