Question Number 149568 by mathdanisur last updated on 06/Aug/21
![solve the equation: 20z[z] - 21{z} = 2021 where {∗} is GIF and {z} = z - [z]](https://www.tinkutara.com/question/Q149568.png)
$${solve}\:{the}\:{equation}: \\ $$$$\mathrm{20}{z}\left[{z}\right]\:-\:\mathrm{21}\left\{{z}\right\}\:=\:\mathrm{2021} \\ $$$${where}\:\left\{\ast\right\}\:{is}\:{GIF}\:\:{and}\:\:\left\{{z}\right\}\:=\:{z}\:-\:\left[{z}\right] \\ $$
Answered by Olaf_Thorendsen last updated on 06/Aug/21
![20z[z]−21{z} = 2021 (1) 20([z]+{z})[z]−21{z} = 2021 Let n = [z] and q = {z} (0≤q<1} 20n^2 +20qn−21q = 2021 q = ((2021−20n^2 )/(20n−21)) (2) By plotting q against n, we verify that 0≤q<1 for n = 10 only • if [z] = 10 (2) : {z} = ((2021−20×100)/(200−21)) = ((21)/(179)) ⇒ z = [x]+{z} = 10+((21)/(179)) = ((1811)/(179)) We verify in (1) that : 20.((1811)/(179)).10−21.((21)/(179)) = 2021 z = ((1811)/(179))](https://www.tinkutara.com/question/Q149728.png)
$$\mathrm{20}{z}\left[{z}\right]−\mathrm{21}\left\{{z}\right\}\:=\:\mathrm{2021}\:\:\:\:\left(\mathrm{1}\right) \\ $$$$\mathrm{20}\left(\left[{z}\right]+\left\{{z}\right\}\right)\left[{z}\right]−\mathrm{21}\left\{{z}\right\}\:=\:\mathrm{2021} \\ $$$$\mathrm{Let}\:{n}\:=\:\left[{z}\right]\:\mathrm{and}\:{q}\:=\:\left\{{z}\right\}\:\left(\mathrm{0}\leqslant{q}<\mathrm{1}\right\} \\ $$$$\mathrm{20}{n}^{\mathrm{2}} +\mathrm{20}{qn}−\mathrm{21}{q}\:=\:\mathrm{2021} \\ $$$${q}\:=\:\frac{\mathrm{2021}−\mathrm{20}{n}^{\mathrm{2}} }{\mathrm{20}{n}−\mathrm{21}}\:\:\:\:\left(\mathrm{2}\right) \\ $$$$\mathrm{By}\:\mathrm{plotting}\:{q}\:\mathrm{against}\:{n},\:\mathrm{we}\:\mathrm{verify} \\ $$$$\mathrm{that}\:\mathrm{0}\leqslant{q}<\mathrm{1}\:\mathrm{for}\:{n}\:=\:\mathrm{10}\:\mathrm{only} \\ $$$$ \\ $$$$\bullet\:\mathrm{if}\:\left[{z}\right]\:=\:\mathrm{10} \\ $$$$\left(\mathrm{2}\right)\::\:\left\{{z}\right\}\:=\:\frac{\mathrm{2021}−\mathrm{20}×\mathrm{100}}{\mathrm{200}−\mathrm{21}}\:=\:\frac{\mathrm{21}}{\mathrm{179}} \\ $$$$\Rightarrow\:{z}\:=\:\left[{x}\right]+\left\{{z}\right\}\:=\:\mathrm{10}+\frac{\mathrm{21}}{\mathrm{179}}\:=\:\frac{\mathrm{1811}}{\mathrm{179}} \\ $$$$ \\ $$$$\mathrm{We}\:\mathrm{verify}\:\mathrm{in}\:\left(\mathrm{1}\right)\:\mathrm{that}\:: \\ $$$$\mathrm{20}.\frac{\mathrm{1811}}{\mathrm{179}}.\mathrm{10}−\mathrm{21}.\frac{\mathrm{21}}{\mathrm{179}}\:=\:\mathrm{2021} \\ $$$$ \\ $$$$\boldsymbol{{z}}\:=\:\frac{\mathrm{1811}}{\mathrm{179}} \\ $$
Commented by mathdanisur last updated on 07/Aug/21
$$\boldsymbol{\mathrm{S}}\mathrm{er},\:\mathrm{thank}\:\mathrm{you}\:\mathrm{cool} \\ $$