Math Question and Answers


Question and Answers Forum

All Questions Topic List
Algebra Questions
Previous in All Question Next in All Question
Previous in Algebra Next in Algebra
Question Number 159092 by HongKing last updated on 12/Nov/21

	Answered by Rasheed.Sindhi last updated on 13/Nov/21
	$(i)a(a−b)=2018 { ((a=1 ∧ a−b=2018⇒b=−2017<0)),((a=2018 ∧ a−b=1⇒b=2017)),((a=2 ∧ a−b=1009⇒b=−1007<0)),((a=1009 ∧ a−b=2⇒b=1007)) :} Possible solutions (a,b)=(2018,2017), (1009,1007) (ii) bc+ac−b^2 −c^2 =1999 •If (a,b)=(2018,2017): c(4035)−2017^2 −c^2 =1999 c^2 −4035c+2017^2 +1999=0 No integer solution. •If(a,b)=(1009,1007): c(2016)−1007^2 −c^2 =1999 (c−1012)(c−1004)=0 c=1012,1004 ∵ a>b>c ⇒1009>1007>c ∴ c=1012 rejected and c=1004 accepted (a,b,c)=(1009,1007,1004)$
	$(i) a (a - b) = 2018$ ${\begin{cases} a = 1 \land a - b = 2018 \Rightarrow b = - 2017 < 0 \\ a = 2018 \land a - b = 1 \Rightarrow b = 2017 \\ a = 2 \land a - b = 1009 \Rightarrow b = - 1007 < 0 \\ a = 1009 \land a - b = 2 \Rightarrow b = 1007 \end{cases}$ $P o s s i b l e s o l u t i o n s (a, b) = (2018, 2017),$ $(1009, 1007)$ $(i i) b c + a c - b^{2} - c^{2} = 1999$ $∙ I f (a, b) = (2018, 2017) :$ $c (4035) - 2017^{2} - c^{2} = 1999$ $c^{2} - 4035 c + 2017^{2} + 1999 = 0$ $N o i n t e g e r s o l u t i o n .$ $∙ I f (a, b) = (1009, 1007) :$ $c (2016) - 1007^{2} - c^{2} = 1999$ $(c - 1012) (c - 1004) = 0$ $c = 1012, 1004$ $∵ a > b > c \Rightarrow 1009 > 1007 > c$ $∴ c = 1012 r e j e c t e d a n d c = 1004 a c c e p t e d$ $(a, b, c) = (1009, 1007, 1004)$
	Commented by HongKing last updated on 13/Nov/21

	$thank you very much dear Ser for your$ $awesome solution$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum