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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))

letx=n+fwithn∈Z,0⩽f<1RHS=n+2022=LHS⩾2021n⇒n⩽20222020⇒n⩽1...(i)RHS=n+2022=LHS<2021(n+1)⇒n>12020⇒n⩾1...(ii)⇒n=1⇒x=1+f1+[f]+1+[f+12021]+1+[f+22021]+...+1+[f+20202021]=1+20222021+[f]+[f+12021]+[f+22021]+...+[f+20202021]=1+2022[f]+[f+12021]+[f+22021]+...+[f+20182021]+[f+20192021]+[f+20202021]=2eachofthelasttwotermsshouldbe1andallothertermsbeforethemshouldbezero.1⩽f+20192021∧f+20182021<1⇒22021⩽f<32021⇒solutionis1+22021⩽x<1+32021

Commented by HongKing last updated on 13/Nov/21

Very nice solution thank you so much my dear Ser

VerynicesolutionthankyousomuchmydearSer

Commented by Tawa11 last updated on 14/Nov/21

Great sir.

Greatsir.

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