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Question Number 147585 by mathdanisur last updated on 22/Jul/21

Answered by Rasheed.Sindhi last updated on 22/Jul/21

z=w+f : whole number & fraction<1 w+f−2021f=2021 w=2021f−f+2021 w=2020f+2021 Let f=(p/q) ; p & q coprime p<q w=2020((p/q))+2021 z=w+f=2020((p/q))+2021+(p/q) z=2021((p/q))+2021 z=2021(((p+q)/q)) q ∣ 2020 q=2⇒p=1⇒f=(1/2)⇒w=2020((1/2))+2021 w=3031⇒z=3031(1/2) q=4⇒p=1,3⇒f=(1/4),(3/4)⇒w=2020((1/4))+2021 w=2526⇒z=2526(1/4) w=2020((3/4))+2021=3536⇒z=3536(3/4) q=5⇒p=1,2,3,4⇒f=(1/5),(2/5),(3/5),(4/5) w=2020((1/5))+2021=2425 z=2425(1/5) z=2021(((p+q)/q)) Where q ∣ 2020 ∧ (p,q)=1 ∧ p<q

$$\:\:\:{z}={w}+{f}\::\:{whole}\:{number}\:\&\:{fraction}<\mathrm{1} \\ $$$$\:\:\:\:{w}+{f}−\mathrm{2021}{f}=\mathrm{2021} \\ $$$$\:\:\:\:{w}=\mathrm{2021}{f}−{f}+\mathrm{2021} \\ $$$$\:\:\:\:{w}=\mathrm{2020}{f}+\mathrm{2021} \\ $$$${Let}\:{f}=\frac{{p}}{{q}}\:;\:{p}\:\&\:{q}\:{coprime}\:{p}<{q} \\ $$$$\:\:\:\:{w}=\mathrm{2020}\left(\frac{{p}}{{q}}\right)+\mathrm{2021} \\ $$$$\:\:\:{z}={w}+{f}=\mathrm{2020}\left(\frac{{p}}{{q}}\right)+\mathrm{2021}+\frac{{p}}{{q}} \\ $$$$\:\:\:{z}=\mathrm{2021}\left(\frac{{p}}{{q}}\right)+\mathrm{2021} \\ $$$$\:\:\:{z}=\mathrm{2021}\left(\frac{{p}+{q}}{{q}}\right) \\ $$$${q}\:\mid\:\mathrm{2020}\: \\ $$$${q}=\mathrm{2}\Rightarrow{p}=\mathrm{1}\Rightarrow{f}=\frac{\mathrm{1}}{\mathrm{2}}\Rightarrow{w}=\mathrm{2020}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)+\mathrm{2021} \\ $$$${w}=\mathrm{3031}\Rightarrow{z}=\mathrm{3031}\frac{\mathrm{1}}{\mathrm{2}}\: \\ $$$${q}=\mathrm{4}\Rightarrow{p}=\mathrm{1},\mathrm{3}\Rightarrow{f}=\frac{\mathrm{1}}{\mathrm{4}},\frac{\mathrm{3}}{\mathrm{4}}\Rightarrow{w}=\mathrm{2020}\left(\frac{\mathrm{1}}{\mathrm{4}}\right)+\mathrm{2021} \\ $$$${w}=\mathrm{2526}\Rightarrow{z}=\mathrm{2526}\frac{\mathrm{1}}{\mathrm{4}} \\ $$$${w}=\mathrm{2020}\left(\frac{\mathrm{3}}{\mathrm{4}}\right)+\mathrm{2021}=\mathrm{3536}\Rightarrow{z}=\mathrm{3536}\frac{\mathrm{3}}{\mathrm{4}} \\ $$$${q}=\mathrm{5}\Rightarrow{p}=\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{4}\Rightarrow{f}=\frac{\mathrm{1}}{\mathrm{5}},\frac{\mathrm{2}}{\mathrm{5}},\frac{\mathrm{3}}{\mathrm{5}},\frac{\mathrm{4}}{\mathrm{5}} \\ $$$${w}=\mathrm{2020}\left(\frac{\mathrm{1}}{\mathrm{5}}\right)+\mathrm{2021}=\mathrm{2425} \\ $$$${z}=\mathrm{2425}\frac{\mathrm{1}}{\mathrm{5}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{z}=\mathrm{2021}\left(\frac{{p}+{q}}{{q}}\right) \\ $$$$\:{Where}\:\:{q}\:\mid\:\mathrm{2020}\:\wedge\:\left({p},{q}\right)=\mathrm{1}\:\wedge\:{p}<{q} \\ $$

Commented by mathdanisur last updated on 22/Jul/21

thank you Ser answer z=2021+n+n/2020 n=0;1;2;...2020

$${thank}\:{you}\:{Ser} \\ $$$${answer}\:{z}=\mathrm{2021}+{n}+{n}/\mathrm{2020} \\ $$$${n}=\mathrm{0};\mathrm{1};\mathrm{2};...\mathrm{2020} \\ $$

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