how to find the roots of the following floor-equation:

Question

How to find the roots of $$\lfloor x\rfloor+\lfloor 2x\rfloor+\lfloor 3x\rfloor=6$$

Answer 1

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

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