Question Number 159152 by HongKing last updated on 13/Nov/21
Answered by mr W last updated on 13/Nov/21
![let x=n+f with n∈Z, 0≤f<1 RHS=n+2022=LHS≥2021n ⇒n≤((2022)/(2020)) ⇒n≤1 ...(i) RHS=n+2022=LHS<2021(n+1) ⇒n>(1/(2020)) ⇒n≥1 ...(ii) ⇒n=1 ⇒x=1+f 1+[f]+1+[f+(1/(2021))]+1+[f+(2/(2021))]+...+1+[f+((2020)/(2021))]=1+2022 2021+[f]+[f+(1/(2021))]+[f+(2/(2021))]+...+[f+((2020)/(2021))]=1+2022 [f]+[f+(1/(2021))]+[f+(2/(2021))]+...+[f+((2018)/(2021))]+[f+((2019)/(2021))]+[f+((2020)/(2021))]=2 each of the last two terms should be 1 and all other terms before them should be zero. 1≤f+((2019)/(2021)) ∧ f+((2018)/(2021))<1 ⇒(2/(2021))≤f<(3/(2021)) ⇒solution is 1+(2/(2021))≤x<1+(3/(2021))](https://www.tinkutara.com/question/Q159166.png)
$${let}\:{x}={n}+{f}\:{with}\:{n}\in\mathbb{Z},\:\mathrm{0}\leqslant{f}<\mathrm{1} \\ $$$${RHS}={n}+\mathrm{2022}={LHS}\geqslant\mathrm{2021}{n}\: \\ $$$$\Rightarrow{n}\leqslant\frac{\mathrm{2022}}{\mathrm{2020}}\:\Rightarrow{n}\leqslant\mathrm{1}\:\:\:…\left({i}\right) \\ $$$${RHS}={n}+\mathrm{2022}={LHS}<\mathrm{2021}\left({n}+\mathrm{1}\right)\: \\ $$$$\Rightarrow{n}>\frac{\mathrm{1}}{\mathrm{2020}}\:\Rightarrow{n}\geqslant\mathrm{1}\:\:\:…\left({ii}\right) \\ $$$$\Rightarrow{n}=\mathrm{1} \\ $$$$\Rightarrow{x}=\mathrm{1}+{f} \\ $$$$\mathrm{1}+\left[{f}\right]+\mathrm{1}+\left[{f}+\frac{\mathrm{1}}{\mathrm{2021}}\right]+\mathrm{1}+\left[{f}+\frac{\mathrm{2}}{\mathrm{2021}}\right]+…+\mathrm{1}+\left[{f}+\frac{\mathrm{2020}}{\mathrm{2021}}\right]=\mathrm{1}+\mathrm{2022} \\ $$$$\mathrm{2021}+\left[{f}\right]+\left[{f}+\frac{\mathrm{1}}{\mathrm{2021}}\right]+\left[{f}+\frac{\mathrm{2}}{\mathrm{2021}}\right]+…+\left[{f}+\frac{\mathrm{2020}}{\mathrm{2021}}\right]=\mathrm{1}+\mathrm{2022} \\ $$$$\left[{f}\right]+\left[{f}+\frac{\mathrm{1}}{\mathrm{2021}}\right]+\left[{f}+\frac{\mathrm{2}}{\mathrm{2021}}\right]+…+\left[{f}+\frac{\mathrm{2018}}{\mathrm{2021}}\right]+\left[{f}+\frac{\mathrm{2019}}{\mathrm{2021}}\right]+\left[{f}+\frac{\mathrm{2020}}{\mathrm{2021}}\right]=\mathrm{2} \\ $$$${each}\:{of}\:{the}\:{last}\:{two}\:{terms}\:{should}\:{be}\:\mathrm{1} \\ $$$${and}\:{all}\:{other}\:{terms}\:{before}\:{them}\: \\ $$$${should}\:{be}\:{zero}. \\ $$$$\mathrm{1}\leqslant{f}+\frac{\mathrm{2019}}{\mathrm{2021}}\:\wedge\:{f}+\frac{\mathrm{2018}}{\mathrm{2021}}<\mathrm{1}\:\Rightarrow\frac{\mathrm{2}}{\mathrm{2021}}\leqslant{f}<\frac{\mathrm{3}}{\mathrm{2021}} \\ $$$$\Rightarrow{solution}\:{is}\:\mathrm{1}+\frac{\mathrm{2}}{\mathrm{2021}}\leqslant{x}<\mathrm{1}+\frac{\mathrm{3}}{\mathrm{2021}} \\ $$
Commented by HongKing last updated on 13/Nov/21
$$\mathrm{Very}\:\mathrm{nice}\:\mathrm{solution}\:\mathrm{thank}\:\mathrm{you}\:\mathrm{so}\:\mathrm{much} \\ $$$$\mathrm{my}\:\mathrm{dear}\:\mathrm{Ser} \\ $$
Commented by Tawa11 last updated on 14/Nov/21
$$\mathrm{Great}\:\mathrm{sir}. \\ $$