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Question Number 96329 by  M±th+et+s last updated on 31/May/20

x⌊x⌊x⌊x⌋⌋⌋=88 x>0

$${x}\lfloor{x}\lfloor{x}\lfloor{x}\rfloor\rfloor\rfloor=\mathrm{88} \\ $$ $${x}>\mathrm{0} \\ $$

Commented byprakash jain last updated on 31/May/20

88=2^3 ×11 x=n+f⇒n^4 ≤88 3^4 =81 ⇒3≤n<4 x=3+f (f>0) (3+f)⌊(3+f)⌊9+3f⌋⌋=88 (3+f)⌊(3+f)(9+⌊3f⌋)⌋=88 (3+f)⌊27+9f+3⌊3f⌋+f⌊3f⌋⌋=88 81+27f+...=88 f<(1/3) ∵27×(1/3)=9>7 ⇒⌊3f⌋=0 (3+f)⌊27+9f⌋=88 81+27f+f⌊9f⌋+3⌊9f⌋=88 27f+f⌊9f⌋+3⌊9f⌋=7 x<(1/3)⇒⌊9f⌋=2,1,0 ⌊9f⌋=0⇒f=(7/(27)) ⇒⌊9f⌋>0 (invalid) ⌊9f⌋=1⇒f=(1/7) ⇒⌊9f⌋=1 (valid) ⌊9f⌋=2⇒f=(1/(29))⇒⌊9f⌋=0 (invalid) x=3+f=((22)/7)

$$\mathrm{88}=\mathrm{2}^{\mathrm{3}} ×\mathrm{11} \\ $$ $${x}={n}+{f}\Rightarrow{n}^{\mathrm{4}} \leqslant\mathrm{88} \\ $$ $$\mathrm{3}^{\mathrm{4}} =\mathrm{81} \\ $$ $$\Rightarrow\mathrm{3}\leqslant{n}<\mathrm{4} \\ $$ $${x}=\mathrm{3}+{f}\:\left({f}>\mathrm{0}\right) \\ $$ $$\left(\mathrm{3}+{f}\right)\lfloor\left(\mathrm{3}+{f}\right)\lfloor\mathrm{9}+\mathrm{3}{f}\rfloor\rfloor=\mathrm{88} \\ $$ $$\left(\mathrm{3}+{f}\right)\lfloor\left(\mathrm{3}+{f}\right)\left(\mathrm{9}+\lfloor\mathrm{3}{f}\rfloor\right)\rfloor=\mathrm{88} \\ $$ $$\left(\mathrm{3}+{f}\right)\lfloor\mathrm{27}+\mathrm{9}{f}+\mathrm{3}\lfloor\mathrm{3}{f}\rfloor+{f}\lfloor\mathrm{3}{f}\rfloor\rfloor=\mathrm{88} \\ $$ $$\mathrm{81}+\mathrm{27}{f}+...=\mathrm{88} \\ $$ $${f}<\frac{\mathrm{1}}{\mathrm{3}}\:\because\mathrm{27}×\frac{\mathrm{1}}{\mathrm{3}}=\mathrm{9}>\mathrm{7}\:\Rightarrow\lfloor\mathrm{3}{f}\rfloor=\mathrm{0} \\ $$ $$\left(\mathrm{3}+{f}\right)\lfloor\mathrm{27}+\mathrm{9}{f}\rfloor=\mathrm{88} \\ $$ $$\mathrm{81}+\mathrm{27}{f}+{f}\lfloor\mathrm{9}{f}\rfloor+\mathrm{3}\lfloor\mathrm{9}{f}\rfloor=\mathrm{88} \\ $$ $$\mathrm{27}{f}+{f}\lfloor\mathrm{9}{f}\rfloor+\mathrm{3}\lfloor\mathrm{9}{f}\rfloor=\mathrm{7} \\ $$ $${x}<\frac{\mathrm{1}}{\mathrm{3}}\Rightarrow\lfloor\mathrm{9}{f}\rfloor=\mathrm{2},\mathrm{1},\mathrm{0} \\ $$ $$\lfloor\mathrm{9}{f}\rfloor=\mathrm{0}\Rightarrow{f}=\frac{\mathrm{7}}{\mathrm{27}}\:\:\Rightarrow\lfloor\mathrm{9}{f}\rfloor>\mathrm{0}\:\left({invalid}\right) \\ $$ $$\lfloor\mathrm{9}{f}\rfloor=\mathrm{1}\Rightarrow{f}=\frac{\mathrm{1}}{\mathrm{7}}\:\Rightarrow\lfloor\mathrm{9}{f}\rfloor=\mathrm{1}\:\left({valid}\right) \\ $$ $$\lfloor\mathrm{9}{f}\rfloor=\mathrm{2}\Rightarrow{f}=\frac{\mathrm{1}}{\mathrm{29}}\Rightarrow\lfloor\mathrm{9}{f}\rfloor=\mathrm{0}\:\left({invalid}\right) \\ $$ $${x}=\mathrm{3}+{f}=\frac{\mathrm{22}}{\mathrm{7}} \\ $$

Commented by M±th+et+s last updated on 31/May/20

correct solution thank you

$${correct}\:{solution}\:{thank}\:{you} \\ $$

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