Question Number 155951 by mnjuly1970 last updated on 06/Oct/21
$$ \\ $$$$\:\:\:\:\:\:\:{solve}\:\:: \\ $$$$ \\ $$$$\:\:\:\:\:\lfloor\:\frac{\mathrm{1}}{{x}}\:\rfloor\:+\:\lfloor\:\frac{\mathrm{3}}{{x}}\:\rfloor=\:\mathrm{4}\: \\ $$$$ \\ $$
Answered by mr W last updated on 06/Oct/21
$${let}\:{t}=\frac{\mathrm{1}}{{x}} \\ $$$$\lfloor{t}\rfloor+\lfloor\mathrm{3}{t}\rfloor=\mathrm{4} \\ $$$${let}\:{t}={n}+{f}\:{with}\:\mathrm{0}\leqslant{f}<\mathrm{1} \\ $$$${n}+\mathrm{3}{n}+\lfloor\mathrm{3}{f}\rfloor=\mathrm{4} \\ $$$$\mathrm{4}{n}+\lfloor\mathrm{3}{f}\rfloor=\mathrm{4} \\ $$$${case}\:\mathrm{0}\leqslant{f}<\frac{\mathrm{1}}{\mathrm{3}}: \\ $$$$\mathrm{4}{n}+\mathrm{0}=\mathrm{4}\:\Rightarrow{n}=\mathrm{1}\:\Rightarrow\mathrm{1}\leqslant{t}<\frac{\mathrm{4}}{\mathrm{3}} \\ $$$${case}\:\frac{\mathrm{1}}{\mathrm{3}}\leqslant{f}<\frac{\mathrm{2}}{\mathrm{3}}: \\ $$$$\mathrm{4}{n}+\mathrm{1}=\mathrm{4}\:\Rightarrow{no}\:{solution}! \\ $$$${case}\:\frac{\mathrm{2}}{\mathrm{3}}\leqslant{f}<\mathrm{1}: \\ $$$$\mathrm{4}{n}+\mathrm{2}=\mathrm{4}\:\Rightarrow{no}\:{solution}! \\ $$$$ \\ $$$${only}\:{solution}\:{is}\:\mathrm{1}\leqslant{t}<\frac{\mathrm{4}}{\mathrm{3}} \\ $$$${i}.{e}.\:\mathrm{1}\leqslant\frac{\mathrm{1}}{{x}}<\frac{\mathrm{4}}{\mathrm{3}} \\ $$$$\Rightarrow\frac{\mathrm{3}}{\mathrm{4}}<{x}\leqslant\mathrm{1} \\ $$
Commented by mnjuly1970 last updated on 06/Oct/21
$${grateful}\:\:{sir}\:\mathcal{W} \\ $$