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Question Number 166257 by mnjuly1970 last updated on 16/Feb/22

     ⌊x⌋⌊2x⌋⌊3x⌋= 6         x=?^

$$ \\ $$$$\:\:\:\lfloor{x}\rfloor\lfloor\mathrm{2}{x}\rfloor\lfloor\mathrm{3}{x}\rfloor=\:\mathrm{6} \\ $$$$\:\:\:\:\:\:\:{x}=\overset{} {?}\: \\ $$

Commented by MJS_new last updated on 16/Feb/22

1≤x<(4/3)

$$\mathrm{1}\leqslant{x}<\frac{\mathrm{4}}{\mathrm{3}} \\ $$

Commented by mnjuly1970 last updated on 17/Feb/22

  ✓✓  thank you

$$\:\:\checkmark\checkmark\:\:{thank}\:{you} \\ $$

Answered by mr W last updated on 17/Feb/22

say x=n+f with 0≤f<1  6=⌊x⌋⌊2x⌋⌊3x⌋≥n×2n×3n=6n^3   ⇒n^3 ≤1 ⇒n≤1  6=⌊x⌋⌊2x⌋⌊3x⌋<n×(2n+2)×(3n+3)=6n^3 +..+6  ⇒n>0  ⇒n=1  1×(2+⌊2f⌋)×(3+⌊3f⌋)=6  6+3⌊2f⌋+2⌊3f⌋+⌊2f⌋⌊3f⌋=6  3⌊2f⌋+2⌊3f⌋+⌊2f⌋⌊3f⌋=0  ⇒⌊2f⌋=0, ⌊3f⌋=0  ⇒0≤3f<1 ⇒0≤f<(1/3)  ⇒solution is 1+0≤x<1+(1/3), i.e.  1≤x<(4/3)

$${say}\:{x}={n}+{f}\:{with}\:\mathrm{0}\leqslant{f}<\mathrm{1} \\ $$$$\mathrm{6}=\lfloor{x}\rfloor\lfloor\mathrm{2}{x}\rfloor\lfloor\mathrm{3}{x}\rfloor\geqslant{n}×\mathrm{2}{n}×\mathrm{3}{n}=\mathrm{6}{n}^{\mathrm{3}} \\ $$$$\Rightarrow{n}^{\mathrm{3}} \leqslant\mathrm{1}\:\Rightarrow{n}\leqslant\mathrm{1} \\ $$$$\mathrm{6}=\lfloor{x}\rfloor\lfloor\mathrm{2}{x}\rfloor\lfloor\mathrm{3}{x}\rfloor<{n}×\left(\mathrm{2}{n}+\mathrm{2}\right)×\left(\mathrm{3}{n}+\mathrm{3}\right)=\mathrm{6}{n}^{\mathrm{3}} +..+\mathrm{6} \\ $$$$\Rightarrow{n}>\mathrm{0} \\ $$$$\Rightarrow{n}=\mathrm{1} \\ $$$$\mathrm{1}×\left(\mathrm{2}+\lfloor\mathrm{2}{f}\rfloor\right)×\left(\mathrm{3}+\lfloor\mathrm{3}{f}\rfloor\right)=\mathrm{6} \\ $$$$\mathrm{6}+\mathrm{3}\lfloor\mathrm{2}{f}\rfloor+\mathrm{2}\lfloor\mathrm{3}{f}\rfloor+\lfloor\mathrm{2}{f}\rfloor\lfloor\mathrm{3}{f}\rfloor=\mathrm{6} \\ $$$$\mathrm{3}\lfloor\mathrm{2}{f}\rfloor+\mathrm{2}\lfloor\mathrm{3}{f}\rfloor+\lfloor\mathrm{2}{f}\rfloor\lfloor\mathrm{3}{f}\rfloor=\mathrm{0} \\ $$$$\Rightarrow\lfloor\mathrm{2}{f}\rfloor=\mathrm{0},\:\lfloor\mathrm{3}{f}\rfloor=\mathrm{0} \\ $$$$\Rightarrow\mathrm{0}\leqslant\mathrm{3}{f}<\mathrm{1}\:\Rightarrow\mathrm{0}\leqslant{f}<\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\Rightarrow{solution}\:{is}\:\mathrm{1}+\mathrm{0}\leqslant{x}<\mathrm{1}+\frac{\mathrm{1}}{\mathrm{3}},\:{i}.{e}. \\ $$$$\mathrm{1}\leqslant{x}<\frac{\mathrm{4}}{\mathrm{3}} \\ $$

Commented by mnjuly1970 last updated on 17/Feb/22

thanks alot sir W

$${thanks}\:{alot}\:{sir}\:{W} \\ $$

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