Question-159092

Question Number 159092 by HongKing last updated on 12/Nov/21

Answered by Rasheed.Sindhi last updated on 13/Nov/21

(i)a(a−b)=2018 { ((a=1 ∧ a−b=2018⇒b=−2017<0)),((a=2018 ∧ a−b=1⇒b=2017)),((a=2 ∧ a−b=1009⇒b=−1007<0)),((a=1009 ∧ a−b=2⇒b=1007)) :} Possible solutions (a,b)=(2018,2017), (1009,1007) (ii) bc+ac−b^2 −c^2 =1999 •If (a,b)=(2018,2017): c(4035)−2017^2 −c^2 =1999 c^2 −4035c+2017^2 +1999=0 No integer solution. •If(a,b)=(1009,1007): c(2016)−1007^2 −c^2 =1999 (c−1012)(c−1004)=0 c=1012,1004 ∵ a>b>c ⇒1009>1007>c ∴ c=1012 rejected and c=1004 accepted (a,b,c)=(1009,1007,1004)

$$\left({i}\right){a}\left({a}−{b}\right)=\mathrm{2018} \\ $$$$\begin{cases}{{a}=\mathrm{1}\:\wedge\:{a}−{b}=\mathrm{2018}\Rightarrow{b}=−\mathrm{2017}<\mathrm{0}}\\{{a}=\mathrm{2018}\:\wedge\:{a}−{b}=\mathrm{1}\Rightarrow{b}=\mathrm{2017}}\\{{a}=\mathrm{2}\:\wedge\:{a}−{b}=\mathrm{1009}\Rightarrow{b}=−\mathrm{1007}<\mathrm{0}}\\{{a}=\mathrm{1009}\:\wedge\:{a}−{b}=\mathrm{2}\Rightarrow{b}=\mathrm{1007}}\end{cases} \\ $$$${Possible}\:{solutions}\:\left({a},{b}\right)=\left(\mathrm{2018},\mathrm{2017}\right), \\ $$$$\left(\mathrm{1009},\mathrm{1007}\right) \\ $$$$\left({ii}\right)\:{bc}+{ac}−{b}^{\mathrm{2}} −{c}^{\mathrm{2}} =\mathrm{1999} \\ $$$$\bullet{If}\:\left({a},{b}\right)=\left(\mathrm{2018},\mathrm{2017}\right): \\ $$$$\:\:\:\:\:\:\:\:\:{c}\left(\mathrm{4035}\right)−\mathrm{2017}^{\mathrm{2}} −{c}^{\mathrm{2}} =\mathrm{1999} \\ $$$${c}^{\mathrm{2}} −\mathrm{4035}{c}+\mathrm{2017}^{\mathrm{2}} +\mathrm{1999}=\mathrm{0} \\ $$$${No}\:{integer}\:{solution}. \\ $$$$\bullet{If}\left({a},{b}\right)=\left(\mathrm{1009},\mathrm{1007}\right): \\ $$$$\:\:\:\:\:\:\:{c}\left(\mathrm{2016}\right)−\mathrm{1007}^{\mathrm{2}} −{c}^{\mathrm{2}} =\mathrm{1999} \\ $$$$\:\:\:\:\:\left({c}−\mathrm{1012}\right)\left({c}−\mathrm{1004}\right)=\mathrm{0} \\ $$$$\:\:\:{c}=\mathrm{1012},\mathrm{1004} \\ $$$$\because\:{a}>{b}>{c}\:\Rightarrow\mathrm{1009}>\mathrm{1007}>{c} \\ $$$$\therefore\:{c}=\mathrm{1012}\:{rejected}\:{and}\:{c}=\mathrm{1004}\:{accepted} \\ $$$$\left({a},{b},{c}\right)=\left(\mathrm{1009},\mathrm{1007},\mathrm{1004}\right) \\ $$

Commented by HongKing last updated on 13/Nov/21

thank you very much dear Ser for your awesome solution

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{dear}\:\mathrm{Ser}\:\mathrm{for}\:\mathrm{your} \\ $$$$\mathrm{awesome}\:\mathrm{solution} \\ $$

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