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Question Number 157231 by mnjuly1970 last updated on 21/Oct/21

$$$$$$\mathcal{SOLVE}:$$$$$$$$\lfloor x\rfloor +\lfloor 2x\rfloor +\lfloor 3x\rfloor =1$$$$----------$$$$$$

Answered by Javokhir last updated on 21/Oct/21

$$x=a+z\Rightarrow \left[z\right]=0\left[x\right]=a$$$$\Rightarrow a+[2a+2z]+[3a+3z]=1=6a+\left[2z\right]+\left[3z\right]$$$$\Rightarrow a=0\iff z\in [0;1)\Rightarrow \left[2z\right]=0or\left[2z\right]=1$$$$Step1$$$$\Rightarrow 0⩽2z<1\Rightarrow \left[2z\right]=0or\left[3z\right]=0or\left[3z\right]=1\swarrow$$$$0⩽z<\frac{1}{2}or\frac{1}{3}⩽z<\frac{2}{3}\iff \frac{1}{3}⩽z<\frac{1}{2}◼$$$$x\in [\frac{1}{3};\frac{1}{2})✓$$$$Solvedby\ast Khamidofff\ast$$$$$$

Commented by mnjuly1970 last updated on 21/Oct/21

$$greatsir...$$

Commented by Javokhir last updated on 21/Oct/21

$$ok$$

Answered by mr W last updated on 21/Oct/21

$$x=n+fwith0⩽f<1$$$$6n+\left[2f\right]+\left[3f\right]=1$$$$1-6n=\left[2f\right]+\left[3f\right]<5\Rightarrow n>-\frac{2}{3}\Rightarrow n⩾0$$$$1-6n=\left[2f\right]+\left[3f\right]⩾0\Rightarrow n⩽\frac{1}{6}\Rightarrow n⩽0$$$$\Rightarrow n=0$$$$\left[2f\right]+\left[3f\right]=1$$$$\Rightarrow 2f<1\wedge 3f⩾1$$$$\Rightarrow \frac{1}{3}⩽f<\frac{1}{2}$$$$\Rightarrow \frac{1}{3}⩽x<\frac{1}{2}$$

Commented by mnjuly1970 last updated on 21/Oct/21

$$thanksalotsirW$$

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