Question Number 157231 by mnjuly1970 last updated on 21/Oct/21
$$ \\ $$$$\:\:\:\:\:\:\mathcal{SOLVE}\:: \\ $$$$ \\ $$$$\:\:\:\:\:\lfloor\:{x}\:\rfloor\:+\:\lfloor\mathrm{2}{x}\:\rfloor\:+\lfloor\:\mathrm{3}{x}\:\rfloor=\:\mathrm{1} \\ $$$$−−−−−−−−−− \\ $$$$ \\ $$
Answered by Javokhir last updated on 21/Oct/21
![x=a+z ⇒[z]=0 [x]=a ⇒a+[2a+2z]+[3a+3z]=1=6a+[2z]+[3z] ⇒ a=0 ⇔ z∈[0;1) ⇒[2z]=0 or [2z]=1 Step1 ⇒0≤2z<1 ⇒[2z]=0 or [3z]=0 or [3z]=1 ↙ 0≤z<(1/2) or (1/3)≤z<(2/3) ⇔ (1/3)≤z< (1/2) ■ x∈[(1/3) ;(1/2))✓ Solved by ∗Khamidofff ∗](Q157242.png)
$$\:\:\:\:\:\:\:\:\:\:\:\:{x}={a}+{z}\:\:\:\:\:\:\Rightarrow\left[{z}\right]=\mathrm{0}\:\:\:\:\left[{x}\right]={a} \\ $$$$\:\:\:\:\:\:\:\:\:\:\Rightarrow{a}+\left[\mathrm{2}{a}+\mathrm{2}{z}\right]+\left[\mathrm{3}{a}+\mathrm{3}{z}\right]=\mathrm{1}=\mathrm{6}{a}+\left[\mathrm{2}{z}\right]+\left[\mathrm{3}{z}\right] \\ $$$$\:\:\:\:\:\:\:\:\Rightarrow\:{a}=\mathrm{0}\:\Leftrightarrow\:\:\:{z}\in\left[\mathrm{0};\mathrm{1}\right)\:\Rightarrow\left[\mathrm{2}{z}\right]=\mathrm{0}\:{or}\:\left[\mathrm{2}{z}\right]=\mathrm{1}\:\:\:\:\: \\ $$$$\:\:\:\:\:\:{Step}\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\Rightarrow\mathrm{0}\leqslant\mathrm{2}{z}<\mathrm{1}\:\Rightarrow\left[\mathrm{2}{z}\right]=\mathrm{0}\:{or}\:\left[\mathrm{3}{z}\right]=\mathrm{0}\:{or}\:\left[\mathrm{3}{z}\right]=\mathrm{1}\:\:\swarrow \\ $$$$\:\:\:\:\:\:\:\:\:\mathrm{0}\leqslant{z}<\frac{\mathrm{1}}{\mathrm{2}}\:{or}\:\:\frac{\mathrm{1}}{\mathrm{3}}\leqslant{z}<\frac{\mathrm{2}}{\mathrm{3}}\:\:\Leftrightarrow\:\:\frac{\mathrm{1}}{\mathrm{3}}\leqslant{z}<\:\frac{\mathrm{1}}{\mathrm{2}}\:\blacksquare \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{x}\in\left[\frac{\mathrm{1}}{\mathrm{3}}\:;\frac{\mathrm{1}}{\mathrm{2}}\right)\checkmark \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{Solved}\:{by}\:\:\ast{Khamidofff}\:\ast \\ $$$$ \\ $$
Commented by mnjuly1970 last updated on 21/Oct/21
$${great}\:{sir}\:... \\ $$
Commented by Javokhir last updated on 21/Oct/21
$${ok} \\ $$
Answered by mr W last updated on 21/Oct/21
![x=n+f with 0≤f<1 6n+[2f]+[3f]=1 1−6n=[2f]+[3f]<5 ⇒n>−(2/3) ⇒n≥0 1−6n=[2f]+[3f]≥0 ⇒n≤(1/6) ⇒n≤0 ⇒n=0 [2f]+[3f]=1 ⇒2f<1 ∧ 3f≥1 ⇒(1/3)≤f<(1/2) ⇒(1/3)≤x<(1/2)](Q157249.png)
$${x}={n}+{f}\:{with}\:\mathrm{0}\leqslant{f}<\mathrm{1} \\ $$$$\mathrm{6}{n}+\left[\mathrm{2}{f}\right]+\left[\mathrm{3}{f}\right]=\mathrm{1} \\ $$$$\mathrm{1}−\mathrm{6}{n}=\left[\mathrm{2}{f}\right]+\left[\mathrm{3}{f}\right]<\mathrm{5}\:\Rightarrow{n}>−\frac{\mathrm{2}}{\mathrm{3}}\:\Rightarrow{n}\geqslant\mathrm{0} \\ $$$$\mathrm{1}−\mathrm{6}{n}=\left[\mathrm{2}{f}\right]+\left[\mathrm{3}{f}\right]\geqslant\mathrm{0}\:\Rightarrow{n}\leqslant\frac{\mathrm{1}}{\mathrm{6}}\:\Rightarrow{n}\leqslant\mathrm{0} \\ $$$$\Rightarrow{n}=\mathrm{0} \\ $$$$\left[\mathrm{2}{f}\right]+\left[\mathrm{3}{f}\right]=\mathrm{1} \\ $$$$\Rightarrow\mathrm{2}{f}<\mathrm{1}\:\wedge\:\mathrm{3}{f}\geqslant\mathrm{1} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{3}}\leqslant{f}<\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{3}}\leqslant{x}<\frac{\mathrm{1}}{\mathrm{2}} \\ $$
Commented by mnjuly1970 last updated on 21/Oct/21
$$\:{thanks}\:{alot}\:{sir}\:{W} \\ $$