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Question Number 45291 by naka3546 last updated on 11/Oct/18
How many possible triple of (a,b,c) ∈ integers so that : ∣ a + b ∣ + c = 19 ab + ∣ c ∣ = 97
$${How}\:\:{many}\:\:{possible}\:\:{triple}\:\:{of}\:\:\left({a},{b},{c}\right)\:\:\in\:\:{integers}\:\:{so}\:\:{that}\:\:: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\mid\:{a}\:+\:{b}\:\mid\:+\:{c}\:\:=\:\:\mathrm{19} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{ab}\:+\:\mid\:{c}\:\mid\:\:=\:\:\mathrm{97} \\ $$
Answered by MJS last updated on 12/Oct/18
c=19−∣a+b∣ c=97−ab ∨ c=ab−97 19−∣a+b∣=97−ab ∨ 19−∣a+b∣=ab−97 ∣a+b∣=ab−78 ∨ ∣a+b∣=116−ab a+b=ab−78 ∨ a+b=78−ab ∨ a+b=116−ab ∨ a+b=ab−116 b=((a+78)/(a−1)) ∨ b=((78−a)/(a+1)) ∨ b=((116−a)/(a+1)) ∨ b=((a+116)/(a−1)) case 1 a=0 b=−78 no solution for c a=2 b=80 no solution for c case 2 a=−80 b=−2 no solution for c a=0 b=78 no solution for c case 3 a=−118 b=−2 no solution for c a=−40 b=−4 no solution for c a=−14 b=−10 no solution for c a=0 b=116 c=−97 a=2 b=38 c=−21 a=8 b=12 c=−1 case 4 a=−116 b=0 c=−97 a=−38 b=−2 c=−21 a=−12 b=−8 c=−1 a=2 b=118 no solution for c a=4 b=40 no solution for c a=10 b=14 no solution for c ⇒ because a and b are interchangeable we have 12 solutions
$${c}=\mathrm{19}−\mid{a}+{b}\mid \\ $$$${c}=\mathrm{97}−{ab}\:\vee\:{c}={ab}−\mathrm{97} \\ $$$$ \\ $$$$\mathrm{19}−\mid{a}+{b}\mid=\mathrm{97}−{ab}\:\vee\:\mathrm{19}−\mid{a}+{b}\mid={ab}−\mathrm{97} \\ $$$$\mid{a}+{b}\mid={ab}−\mathrm{78}\:\vee\:\mid{a}+{b}\mid=\mathrm{116}−{ab} \\ $$$${a}+{b}={ab}−\mathrm{78}\:\vee\:{a}+{b}=\mathrm{78}−{ab}\:\vee\:{a}+{b}=\mathrm{116}−{ab}\:\vee\:{a}+{b}={ab}−\mathrm{116} \\ $$$${b}=\frac{{a}+\mathrm{78}}{{a}−\mathrm{1}}\:\vee\:{b}=\frac{\mathrm{78}−{a}}{{a}+\mathrm{1}}\:\vee\:{b}=\frac{\mathrm{116}−{a}}{{a}+\mathrm{1}}\:\vee\:{b}=\frac{{a}+\mathrm{116}}{{a}−\mathrm{1}} \\ $$$$ \\ $$$$\mathrm{case}\:\mathrm{1} \\ $$$${a}=\mathrm{0}\:{b}=−\mathrm{78}\:\mathrm{no}\:\mathrm{solution}\:\mathrm{for}\:{c} \\ $$$${a}=\mathrm{2}\:{b}=\mathrm{80}\:\mathrm{no}\:\mathrm{solution}\:\mathrm{for}\:{c} \\ $$$$ \\ $$$$\mathrm{case}\:\mathrm{2} \\ $$$${a}=−\mathrm{80}\:{b}=−\mathrm{2}\:\mathrm{no}\:\mathrm{solution}\:\mathrm{for}\:{c} \\ $$$${a}=\mathrm{0}\:{b}=\mathrm{78}\:\mathrm{no}\:\mathrm{solution}\:\mathrm{for}\:{c} \\ $$$$ \\ $$$$\mathrm{case}\:\mathrm{3} \\ $$$${a}=−\mathrm{118}\:{b}=−\mathrm{2}\:\mathrm{no}\:\mathrm{solution}\:\mathrm{for}\:{c} \\ $$$${a}=−\mathrm{40}\:{b}=−\mathrm{4}\:\mathrm{no}\:\mathrm{solution}\:\mathrm{for}\:{c} \\ $$$${a}=−\mathrm{14}\:{b}=−\mathrm{10}\:\mathrm{no}\:\mathrm{solution}\:\mathrm{for}\:{c} \\ $$$${a}=\mathrm{0}\:{b}=\mathrm{116}\:{c}=−\mathrm{97} \\ $$$${a}=\mathrm{2}\:{b}=\mathrm{38}\:{c}=−\mathrm{21} \\ $$$${a}=\mathrm{8}\:{b}=\mathrm{12}\:{c}=−\mathrm{1} \\ $$$$ \\ $$$$\mathrm{case}\:\mathrm{4} \\ $$$${a}=−\mathrm{116}\:{b}=\mathrm{0}\:{c}=−\mathrm{97} \\ $$$${a}=−\mathrm{38}\:{b}=−\mathrm{2}\:{c}=−\mathrm{21} \\ $$$${a}=−\mathrm{12}\:{b}=−\mathrm{8}\:{c}=−\mathrm{1} \\ $$$${a}=\mathrm{2}\:{b}=\mathrm{118}\:\mathrm{no}\:\mathrm{solution}\:\mathrm{for}\:{c} \\ $$$${a}=\mathrm{4}\:{b}=\mathrm{40}\:\mathrm{no}\:\mathrm{solution}\:\mathrm{for}\:{c} \\ $$$${a}=\mathrm{10}\:{b}=\mathrm{14}\:\mathrm{no}\:\mathrm{solution}\:\mathrm{for}\:{c} \\ $$$$ \\ $$$$\Rightarrow\:\mathrm{because}\:{a}\:\mathrm{and}\:{b}\:\mathrm{are}\:\mathrm{interchangeable} \\ $$$$\mathrm{we}\:\mathrm{have}\:\mathrm{12}\:\mathrm{solutions} \\ $$

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