Question Number 77067 by Boyka last updated on 03/Jan/20
Commented by Boyka last updated on 03/Jan/20
![[x]−whole part x=[x]+{x}](https://www.tinkutara.com/question/Q77068.png)
$$\left[\mathrm{x}\right]−\mathrm{whole}\:\mathrm{part}\:\: \\ $$$$\boldsymbol{\mathrm{x}}=\left[\boldsymbol{\mathrm{x}}\right]+\left\{\boldsymbol{\mathrm{x}}\right\} \\ $$
Commented by mr W last updated on 03/Jan/20
$${x}=\mathrm{3}\frac{\mathrm{1}}{\mathrm{7}}=\frac{\mathrm{22}}{\mathrm{7}} \\ $$
Answered by mr W last updated on 03/Jan/20
![if x were positive integer, with x=3 we get 3×3×3×3=81<88 with x=4 we get 4×4×4×4=256>88 ⇒3<x<4 say x=3+d with 0<d<1. [x]=3 x[x]=(3+d)3=9+3d if d<(1/3): [x[x]]=9 x[x[x]]=(3+d)9=27+9d if d<(1/9): [x[x[x]]]=27 x[x[x[x]]]=(3+d)27=88 ⇒d=(7/(27))>(1/9) ⇒wrong if (1/9)≤d<(2/9): [x[x[x]]]=[27+9d]=28 x[x[x[x]]]=(3+d)28=88 ⇒d=(1/7)∈[(1/9),(2/9)) ⇒ok that means the searched number is x=3+(1/7)=((22)/7)=3(1/7)](https://www.tinkutara.com/question/Q77075.png)
$${if}\:{x}\:{were}\:{positive}\:{integer}, \\ $$$${with}\:{x}=\mathrm{3}\:{we}\:{get}\:\mathrm{3}×\mathrm{3}×\mathrm{3}×\mathrm{3}=\mathrm{81}<\mathrm{88} \\ $$$${with}\:{x}=\mathrm{4}\:{we}\:{get}\:\mathrm{4}×\mathrm{4}×\mathrm{4}×\mathrm{4}=\mathrm{256}>\mathrm{88} \\ $$$$\Rightarrow\mathrm{3}<{x}<\mathrm{4} \\ $$$${say}\:{x}=\mathrm{3}+{d}\:{with}\:\mathrm{0}<{d}<\mathrm{1}. \\ $$$$\left[{x}\right]=\mathrm{3} \\ $$$${x}\left[{x}\right]=\left(\mathrm{3}+{d}\right)\mathrm{3}=\mathrm{9}+\mathrm{3}{d} \\ $$$${if}\:{d}<\frac{\mathrm{1}}{\mathrm{3}}: \\ $$$$\left[{x}\left[{x}\right]\right]=\mathrm{9} \\ $$$${x}\left[{x}\left[{x}\right]\right]=\left(\mathrm{3}+{d}\right)\mathrm{9}=\mathrm{27}+\mathrm{9}{d} \\ $$$${if}\:{d}<\frac{\mathrm{1}}{\mathrm{9}}: \\ $$$$\left[{x}\left[{x}\left[{x}\right]\right]\right]=\mathrm{27} \\ $$$${x}\left[{x}\left[{x}\left[{x}\right]\right]\right]=\left(\mathrm{3}+{d}\right)\mathrm{27}=\mathrm{88}\:\Rightarrow{d}=\frac{\mathrm{7}}{\mathrm{27}}>\frac{\mathrm{1}}{\mathrm{9}}\:\Rightarrow{wrong} \\ $$$${if}\:\frac{\mathrm{1}}{\mathrm{9}}\leqslant{d}<\frac{\mathrm{2}}{\mathrm{9}}: \\ $$$$\left[{x}\left[{x}\left[{x}\right]\right]\right]=\left[\mathrm{27}+\mathrm{9}{d}\right]=\mathrm{28} \\ $$$${x}\left[{x}\left[{x}\left[{x}\right]\right]\right]=\left(\mathrm{3}+{d}\right)\mathrm{28}=\mathrm{88}\:\Rightarrow{d}=\frac{\mathrm{1}}{\mathrm{7}}\in\left[\frac{\mathrm{1}}{\mathrm{9}},\frac{\mathrm{2}}{\mathrm{9}}\right)\:\Rightarrow{ok} \\ $$$${that}\:{means}\:{the}\:{searched}\:{number}\:{is} \\ $$$${x}=\mathrm{3}+\frac{\mathrm{1}}{\mathrm{7}}=\frac{\mathrm{22}}{\mathrm{7}}=\mathrm{3}\frac{\mathrm{1}}{\mathrm{7}} \\ $$
Commented by Boyka last updated on 03/Jan/20
$$\mathrm{thanks}\:\mathrm{sir} \\ $$