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Question Number 182922 by mnjuly1970 last updated on 16/Dec/22

     solve       ⌊ x ⌋ + ⌊ 2x ⌋ + ⌊ 3x ⌋ =1

$$ \\ $$$$\:\:\:{solve} \\ $$$$ \\ $$$$\:\:\:\lfloor\:{x}\:\rfloor\:+\:\lfloor\:\mathrm{2}{x}\:\rfloor\:+\:\lfloor\:\mathrm{3}{x}\:\rfloor\:=\mathrm{1} \\ $$$$ \\ $$

Answered by mr W last updated on 16/Dec/22

if x<(1/3),  ⌊x⌋+⌊2x⌋+⌊3x⌋≤0 <1  ⇒x≥(1/3)  if x≥(1/2),  ⌊x⌋+⌊2x⌋+⌊3x⌋≥2 >1  ⇒x<(1/2)  ⇒solution (1/3)≤x<(1/2)

$${if}\:{x}<\frac{\mathrm{1}}{\mathrm{3}},\:\:\lfloor{x}\rfloor+\lfloor\mathrm{2}{x}\rfloor+\lfloor\mathrm{3}{x}\rfloor\leqslant\mathrm{0}\:<\mathrm{1} \\ $$$$\Rightarrow{x}\geqslant\frac{\mathrm{1}}{\mathrm{3}} \\ $$$${if}\:{x}\geqslant\frac{\mathrm{1}}{\mathrm{2}},\:\:\lfloor{x}\rfloor+\lfloor\mathrm{2}{x}\rfloor+\lfloor\mathrm{3}{x}\rfloor\geqslant\mathrm{2}\:>\mathrm{1} \\ $$$$\Rightarrow{x}<\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow{solution}\:\frac{\mathrm{1}}{\mathrm{3}}\leqslant{x}<\frac{\mathrm{1}}{\mathrm{2}} \\ $$

Commented by mnjuly1970 last updated on 17/Dec/22

 very nice and short

$$\:{very}\:{nice}\:{and}\:{short}\: \\ $$

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