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

$$$$$$\lfloor x\rfloor \lfloor 2x\rfloor \lfloor 3x\rfloor =6$$$$x=\stackrel{}{?}$$

Commented by MJS_new last updated on 16/Feb/22

$$1⩽x<\frac{4}{3}$$

Commented by mnjuly1970 last updated on 17/Feb/22

$$✓✓thankyou$$

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

$$sayx=n+fwith0⩽f<1$$$$6=\lfloor x\rfloor \lfloor 2x\rfloor \lfloor 3x\rfloor ⩾n\times 2n\times 3n=6n^3$$$$\Rightarrow n^3⩽1\Rightarrow n⩽1$$$$6=\lfloor x\rfloor \lfloor 2x\rfloor \lfloor 3x\rfloor <n\times (2n+2)\times (3n+3)=6n^3+..+6$$$$\Rightarrow n>0$$$$\Rightarrow n=1$$$$1\times (2+\lfloor 2f\rfloor )\times (3+\lfloor 3f\rfloor )=6$$$$6+3\lfloor 2f\rfloor +2\lfloor 3f\rfloor +\lfloor 2f\rfloor \lfloor 3f\rfloor =6$$$$3\lfloor 2f\rfloor +2\lfloor 3f\rfloor +\lfloor 2f\rfloor \lfloor 3f\rfloor =0$$$$\Rightarrow \lfloor 2f\rfloor =0,\lfloor 3f\rfloor =0$$$$\Rightarrow 0⩽3f<1\Rightarrow 0⩽f<\frac{1}{3}$$$$\Rightarrow solutionis1+0⩽x<1+\frac{1}{3},i.e.$$$$1⩽x<\frac{4}{3}$$

Commented by mnjuly1970 last updated on 17/Feb/22

$$thanksalotsirW$$

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