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Question Number 191343 by mnjuly1970 last updated on 23/Apr/23

        solve  in  R          ⌊ (1/x) ⌋ + ⌊ (2/x) ⌋ + ⌊ (3/x) ⌋ = 1

$$ \\ $$$$\:\:\:\:\:\:\mathrm{solve}\:\:{in}\:\:\mathbb{R} \\ $$$$ \\ $$$$\:\:\:\:\:\:\lfloor\:\frac{\mathrm{1}}{{x}}\:\rfloor\:+\:\lfloor\:\frac{\mathrm{2}}{{x}}\:\rfloor\:+\:\lfloor\:\frac{\mathrm{3}}{{x}}\:\rfloor\:=\:\mathrm{1} \\ $$$$ \\ $$

Answered by mr W last updated on 23/Apr/23

⌊(3/x)⌋=1 and ⌊(2/x)⌋=0  ⇒1≤(3/x)<2 ⇒ (3/2)<x≤3 and  ⇒(2/x)<1 ⇒x>2  solution is 2<x≤3 ✓

$$\lfloor\frac{\mathrm{3}}{{x}}\rfloor=\mathrm{1}\:{and}\:\lfloor\frac{\mathrm{2}}{{x}}\rfloor=\mathrm{0} \\ $$$$\Rightarrow\mathrm{1}\leqslant\frac{\mathrm{3}}{{x}}<\mathrm{2}\:\Rightarrow\:\frac{\mathrm{3}}{\mathrm{2}}<{x}\leqslant\mathrm{3}\:{and} \\ $$$$\Rightarrow\frac{\mathrm{2}}{{x}}<\mathrm{1}\:\Rightarrow{x}>\mathrm{2} \\ $$$${solution}\:{is}\:\mathrm{2}<{x}\leqslant\mathrm{3}\:\checkmark \\ $$

Commented by mnjuly1970 last updated on 23/Apr/23

  thank you so much sir W  ⋛

$$\:\:{thank}\:{you}\:{so}\:{much}\:{sir}\:{W}\:\:\underline{\underbrace{\lesseqgtr}} \\ $$

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