how to find the roots of the following floor-equation: How to find the roots of  $$\lfloor x\rfloor+\lfloor 2x\rfloor+\lfloor 3x\rfloor=6$$
 A: As my previous solution was not complete i would like to write a new one.
Let assume $\left\lfloor   x \right\rfloor=n$  so that we know $n\le x<n+1$ and let $\delta$ the fractional part of x, so that $\delta \epsilon \left( 0,1 \right) $ then
$$\left\lfloor x \right\rfloor +\left\lfloor 2x \right\rfloor +\left\lfloor 3x \right\rfloor =6\quad \Longrightarrow \quad \left\lfloor n+\delta  \right\rfloor +\left\lfloor 2\left( n+\delta  \right)  \right\rfloor +\left\lfloor 3\left( n+\delta  \right)  \right\rfloor =6$$
obviosly,it is equal to $$\left\lfloor n+\delta  \right\rfloor +\left\lfloor 2n+2\delta  \right\rfloor +\left\lfloor 3n+3\delta  \right\rfloor =6$$
Due to the fact that $n$,$2n$ and $3n$  is an integer we can write it as follows:
$$n+\left\lfloor \delta  \right\rfloor +2n+\left\lfloor 2\delta  \right\rfloor +3n+\left\lfloor 3\delta  \right\rfloor =6$$
Now,there are three sub-cases,i.e. we will divide $\left( 0,1 \right) $ in to three parts by $\delta $:
$1$.If $0\le \delta <\frac { 1 }{ 3 }   $ then
$$
\left\lfloor n+\delta  \right\rfloor +\left\lfloor 2n+2\delta  \right\rfloor +\left\lfloor 3n+3\delta  \right\rfloor =6
      \Rightarrow  n+2n+3n=6    \Rightarrow  n=1     $$
so 
$$1\le x<\frac { 4 }{ 3 } $$
$2.$ If $\frac { 1 }{ 3 } \le \delta <\frac { 2 }{ 3 } $ then 
$$
n+\left\lfloor \delta  \right\rfloor +2n+\left\lfloor 2\delta  \right\rfloor +3n+\left\lfloor 3\delta  \right\rfloor =6 \Rightarrow 
n+2n+3n+1=6   \Rightarrow  n=\frac { 5 }{ 6 } 
$$ 
but n should be integer 
$3.$ If $\frac { 2 }{ 3 } \le \delta <1$ then 
$$
n+\left\lfloor \delta  \right\rfloor +2n+\left\lfloor 2\delta  \right\rfloor +3n+\left\lfloor 3\delta  \right\rfloor =6 \Rightarrow 
n+2n+1+3n+2=6   \Rightarrow  n=\frac { 1 }{ 2 } 
$$ in that case n is not integer either. 
So our final answer is 

$$1\le x<\frac { 4 }{ 3 } $$

