Question Number 166435 by mnjuly1970 last updated on 20/Feb/22
$$ \\ $$$$\:\:\:\:\mathscr{E}{quation}\:: \\ $$$$\:\:\:\:\:{Solve}\:{in}\:\:\mathbb{R}\: \\ $$$$\:\:\:\lfloor{x}\rfloor\:+\lfloor\mathrm{2}{x}\:\rfloor\:+\lfloor\:\mathrm{3}{x}\:\rfloor=\mathrm{1} \\ $$$$\:\:\:\:\:\:−−−−−−−−− \\ $$
Answered by mahdipoor last updated on 20/Feb/22
![I) x=n+f 0≤f<1/3 n∈Z ⇒A=[x]+[2x]+[3x]=n+2n+3n=1 ⇒n=(1/6)∉Z ⇒ without answer II)x=n+f 1/3≤f<1/2 ⇒A=n+2n+(3n+1)=1 ⇒n=0 III)x=n+f 1/2≤f<2 A=n+(2n+1)+(3n+1)=1 ⇒n=−(1/6)∉Z ⇒ without answer IV)x=n+f 2/3≤f<1 A=n+(2n+1)+(3n+2)=1 ⇒ ⇒n=−(1/3)∉Z ⇒ without answer ⇒A=1⇒ x∈[1/3,1/2)](Q166439.png)
$$\left.\mathrm{I}\right)\:{x}={n}+{f}\:\:\:\:\:\mathrm{0}\leqslant{f}<\mathrm{1}/\mathrm{3}\:\:\:\:{n}\in\mathrm{Z} \\ $$$$\Rightarrow{A}=\left[{x}\right]+\left[\mathrm{2}{x}\right]+\left[\mathrm{3}{x}\right]={n}+\mathrm{2}{n}+\mathrm{3}{n}=\mathrm{1} \\ $$$$\Rightarrow{n}=\frac{\mathrm{1}}{\mathrm{6}}\notin\mathrm{Z}\:\Rightarrow\:{without}\:{answer} \\ $$$$\left.\mathrm{II}\right){x}={n}+{f}\:\:\:\:\:\:\:\mathrm{1}/\mathrm{3}\leqslant{f}<\mathrm{1}/\mathrm{2} \\ $$$$\Rightarrow{A}={n}+\mathrm{2}{n}+\left(\mathrm{3}{n}+\mathrm{1}\right)=\mathrm{1}\:\Rightarrow{n}=\mathrm{0} \\ $$$$\left.\mathrm{III}\right){x}={n}+{f}\:\:\:\:\mathrm{1}/\mathrm{2}\leqslant{f}<\mathrm{2} \\ $$$${A}={n}+\left(\mathrm{2}{n}+\mathrm{1}\right)+\left(\mathrm{3}{n}+\mathrm{1}\right)=\mathrm{1} \\ $$$$\Rightarrow{n}=−\frac{\mathrm{1}}{\mathrm{6}}\notin\mathrm{Z}\:\Rightarrow\:{without}\:{answer} \\ $$$$\left.\mathrm{IV}\right){x}={n}+{f}\:\:\:\:\mathrm{2}/\mathrm{3}\leqslant{f}<\mathrm{1} \\ $$$${A}={n}+\left(\mathrm{2}{n}+\mathrm{1}\right)+\left(\mathrm{3}{n}+\mathrm{2}\right)=\mathrm{1}\:\Rightarrow\: \\ $$$$\Rightarrow{n}=−\frac{\mathrm{1}}{\mathrm{3}}\notin\mathrm{Z}\:\Rightarrow\:{without}\:{answer} \\ $$$$\Rightarrow{A}=\mathrm{1}\Rightarrow\:{x}\in\left[\mathrm{1}/\mathrm{3},\mathrm{1}/\mathrm{2}\right) \\ $$
Answered by mr W last updated on 20/Feb/22
$${x}={n}+{f} \\ $$$$\mathrm{1}=\mathrm{6}{n}+\lfloor\mathrm{2}{f}\rfloor+\lfloor\mathrm{3}{f}\rfloor\geqslant\mathrm{6}{n} \\ $$$$\Rightarrow{n}\leqslant\frac{\mathrm{1}}{\mathrm{6}}\:\Rightarrow{n}\leqslant\mathrm{0} \\ $$$$\mathrm{1}=\mathrm{6}{n}+\lfloor\mathrm{2}{f}\rfloor+\lfloor\mathrm{3}{f}\rfloor<\mathrm{6}{n}+\mathrm{5} \\ $$$$\Rightarrow{n}>−\frac{\mathrm{2}}{\mathrm{3}}\:\Rightarrow{n}\geqslant\mathrm{0} \\ $$$$\Rightarrow{n}=\mathrm{0} \\ $$$$\lfloor\mathrm{2}{f}\rfloor+\lfloor\mathrm{3}{f}\rfloor=\mathrm{1} \\ $$$$\Rightarrow\mathrm{3}{f}\geqslant\mathrm{1},\:\mathrm{2}{f}<\mathrm{1} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{3}}\leqslant{f}<\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{3}}\leqslant{x}<\frac{\mathrm{1}}{\mathrm{2}} \\ $$
Commented by mahdipoor last updated on 20/Feb/22
![in line 7 : [2f]+[3f]=1 ⇒[ 0≤f<1 ⇒ 2f≤3f ] ⇒ 2f<1 3f≥1 ⇒ (1/3)≤f<(1/2)](Q166469.png)
$${in}\:{line}\:\mathrm{7}\:: \\ $$$$\left[\mathrm{2}{f}\right]+\left[\mathrm{3}{f}\right]=\mathrm{1}\:\Rightarrow\left[\:\mathrm{0}\leqslant{f}<\mathrm{1}\:\Rightarrow\:\mathrm{2}{f}\leqslant\mathrm{3}{f}\:\right]\:\Rightarrow \\ $$$$\mathrm{2}{f}<\mathrm{1}\:\:\:\:\mathrm{3}{f}\geqslant\mathrm{1}\:\Rightarrow\:\frac{\mathrm{1}}{\mathrm{3}}\leqslant{f}<\frac{\mathrm{1}}{\mathrm{2}} \\ $$
Commented by mr W last updated on 20/Feb/22
$${yes},\:{thanks}!\:{it}'{s}\:{fixed}. \\ $$