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Question Number 166435 by mnjuly1970 last updated on 20/Feb/22

$$$$$$\mathcal{E}quation:$$$$Solvein\mathbb{R}$$$$\lfloor x\rfloor +\lfloor 2x\rfloor +\lfloor 3x\rfloor =1$$$$---------$$

Answered by mahdipoor last updated on 20/Feb/22

$$\mathrm{I})x=n+f0⩽f<1/3n\in \mathrm{Z}$$$$\Rightarrow A=\left[x\right]+\left[2x\right]+\left[3x\right]=n+2n+3n=1$$$$\Rightarrow n=\frac{1}{6}\notin \mathrm{Z}\Rightarrow withoutanswer$$$$\mathrm{II})x=n+f1/3⩽f<1/2$$$$\Rightarrow A=n+2n+(3n+1)=1\Rightarrow n=0$$$$\mathrm{III})x=n+f1/2⩽f<2$$$$A=n+(2n+1)+(3n+1)=1$$$$\Rightarrow n=-\frac{1}{6}\notin \mathrm{Z}\Rightarrow withoutanswer$$$$\mathrm{IV})x=n+f2/3⩽f<1$$$$A=n+(2n+1)+(3n+2)=1\Rightarrow$$$$\Rightarrow n=-\frac{1}{3}\notin \mathrm{Z}\Rightarrow withoutanswer$$$$\Rightarrow A=1\Rightarrow x\in [1/3,1/2)$$

Answered by mr W last updated on 20/Feb/22

$$x=n+f$$$$1=6n+\lfloor 2f\rfloor +\lfloor 3f\rfloor ⩾6n$$$$\Rightarrow n⩽\frac{1}{6}\Rightarrow n⩽0$$$$1=6n+\lfloor 2f\rfloor +\lfloor 3f\rfloor <6n+5$$$$\Rightarrow n>-\frac{2}{3}\Rightarrow n⩾0$$$$\Rightarrow n=0$$$$\lfloor 2f\rfloor +\lfloor 3f\rfloor =1$$$$\Rightarrow 3f⩾1,2f<1$$$$\Rightarrow \frac{1}{3}⩽f<\frac{1}{2}$$$$\Rightarrow \frac{1}{3}⩽x<\frac{1}{2}$$

Commented by mahdipoor last updated on 20/Feb/22

$$inline7:$$$$\left[2f\right]+\left[3f\right]=1\Rightarrow [0⩽f<1\Rightarrow 2f⩽3f]\Rightarrow$$$$2f<13f⩾1\Rightarrow \frac{1}{3}⩽f<\frac{1}{2}$$

Commented by mr W last updated on 20/Feb/22

$$yes,thanks!{it}^{\prime }sfixed.$$

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