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Question Number 182804 by mnjuly1970 last updated on 14/Dec/22

$$$$$$solve$$$$\lfloor \underset{}{\stackrel{}{x}}\rfloor -\lfloor \frac{x}{3}\rfloor =3$$$$$$

Answered by mahdipoor last updated on 14/Dec/22

$$getx=3k+j0⩽j<3k\in Z$$$$\lfloor x\rfloor -\lfloor x/3\rfloor =(3k+\lfloor j\rfloor )-(k+\lfloor j/3\rfloor )=$$$$2k+\lfloor j\rfloor =3$$$$\{\begin{array}{l}0⩽j<1\Rightarrow 2k+0=3\Rightarrow \nexists \\ 1⩽j<2\Rightarrow 2k+1=3\Rightarrow k=1\\ 2⩽j<3\Rightarrow 2k+2=3\Rightarrow \nexists \end{array}$$$$\Rightarrow \Rightarrow 4⩽x<5$$

Commented by mnjuly1970 last updated on 14/Dec/22

$$verynicesirmahdipoor$$

Answered by mr W last updated on 14/Dec/22

$$lett=\frac{x}{3}$$$$\lfloor 3t\rfloor -\lfloor t\rfloor =3$$$$lett=n+fwith0⩽f<1$$$$3n+\lfloor 3f\rfloor -n=3$$$$\lfloor 3f\rfloor =3-2n$$$$3-2n⩾0\Rightarrow n⩽\frac{3}{2}\Rightarrow n⩽1$$$$3-2n<3\Rightarrow n>0\Rightarrow n⩾1$$$$\Rightarrow n=1$$$$\Rightarrow \lfloor 3f\rfloor =3-2=1$$$$\Rightarrow 1⩽3f<2\Rightarrow \frac{1}{3}⩽f<\frac{2}{3}$$$$\Rightarrow \frac{4}{3}⩽t<\frac{5}{3}$$$$\Rightarrow \frac{4}{3}⩽\frac{x}{3}<\frac{5}{3}$$$$\Rightarrow 4⩽x<5✓$$

Commented by mnjuly1970 last updated on 14/Dec/22

$$thxsirWverynicesolution$$

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