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Question Number 155951 by mnjuly1970 last updated on 06/Oct/21

$$$$$$solve:$$$$$$$$\lfloor \frac{1}{x}\rfloor +\lfloor \frac{3}{x}\rfloor =4$$$$$$

Answered by mr W last updated on 06/Oct/21

$$lett=\frac{1}{x}$$$$\lfloor t\rfloor +\lfloor 3t\rfloor =4$$$$lett=n+fwith0⩽f<1$$$$n+3n+\lfloor 3f\rfloor =4$$$$4n+\lfloor 3f\rfloor =4$$$$case0⩽f<\frac{1}{3}:$$$$4n+0=4\Rightarrow n=1\Rightarrow 1⩽t<\frac{4}{3}$$$$case\frac{1}{3}⩽f<\frac{2}{3}:$$$$4n+1=4\Rightarrow nosolution!$$$$case\frac{2}{3}⩽f<1:$$$$4n+2=4\Rightarrow nosolution!$$$$$$$$onlysolutionis1⩽t<\frac{4}{3}$$$$i.e.1⩽\frac{1}{x}<\frac{4}{3}$$$$\Rightarrow \frac{3}{4}<x⩽1$$

Commented by mnjuly1970 last updated on 06/Oct/21

$$gratefulsir\mathcal{W}$$

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