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Question-213482




Question Number 213482 by efronzo1 last updated on 06/Nov/24
Answered by A5T last updated on 06/Nov/24
3∣Y⇒8+c+2+3+d≡0(mod 3)⇒c+d≡2(mod 3)  3∣X⇒9∣Y :   3+1+a+5+b≡0(mod 3)⇒8+c+2+3+d≡0(mod 9)  ⇒a+b≡0(mod 3)⇒c+d≡5(mod 9)  3∣X⇒c+d=5⇒a+b=0,3,6,9 ∧ c+d=5  3∤X⇒a+b∈{1,2,4,5,7,8} ∧ c+d=2,5,8  And the possible values follow from those.
$$\mathrm{3}\mid{Y}\Rightarrow\mathrm{8}+{c}+\mathrm{2}+\mathrm{3}+{d}\equiv\mathrm{0}\left({mod}\:\mathrm{3}\right)\Rightarrow{c}+{d}\equiv\mathrm{2}\left({mod}\:\mathrm{3}\right) \\ $$$$\mathrm{3}\mid{X}\Rightarrow\mathrm{9}\mid{Y}\::\: \\ $$$$\mathrm{3}+\mathrm{1}+{a}+\mathrm{5}+{b}\equiv\mathrm{0}\left({mod}\:\mathrm{3}\right)\Rightarrow\mathrm{8}+{c}+\mathrm{2}+\mathrm{3}+{d}\equiv\mathrm{0}\left({mod}\:\mathrm{9}\right) \\ $$$$\Rightarrow{a}+{b}\equiv\mathrm{0}\left({mod}\:\mathrm{3}\right)\Rightarrow{c}+{d}\equiv\mathrm{5}\left({mod}\:\mathrm{9}\right) \\ $$$$\mathrm{3}\mid{X}\Rightarrow{c}+{d}=\mathrm{5}\Rightarrow{a}+{b}=\mathrm{0},\mathrm{3},\mathrm{6},\mathrm{9}\:\wedge\:{c}+{d}=\mathrm{5} \\ $$$$\mathrm{3}\nmid{X}\Rightarrow{a}+{b}\in\left\{\mathrm{1},\mathrm{2},\mathrm{4},\mathrm{5},\mathrm{7},\mathrm{8}\right\}\:\wedge\:{c}+{d}=\mathrm{2},\mathrm{5},\mathrm{8} \\ $$$${And}\:{the}\:{possible}\:{values}\:{follow}\:{from}\:{those}. \\ $$

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