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Find-a-b-c-and-d-such-that-abc-1-mod-11-abd-2-mod-11-acd-3-mod-11-bcd-4-mod-11-




Question Number 198519 by cortano12 last updated on 21/Oct/23
  Find a,b,c and d such that        { ((abc = 1 (mod 11))),((abd = 2 (mod 11) )),((acd = 3 (mod 11) )),((bcd = 4 (mod 11) )) :}
$$\:\:\mathrm{Find}\:\mathrm{a},\mathrm{b},\mathrm{c}\:\mathrm{and}\:\mathrm{d}\:\mathrm{such}\:\mathrm{that}\: \\ $$$$\:\:\:\:\begin{cases}{\mathrm{abc}\:=\:\mathrm{1}\:\left(\mathrm{mod}\:\mathrm{11}\right)}\\{\mathrm{abd}\:=\:\mathrm{2}\:\left(\mathrm{mod}\:\mathrm{11}\right)\:}\\{\mathrm{acd}\:=\:\mathrm{3}\:\left(\mathrm{mod}\:\mathrm{11}\right)\:}\\{\mathrm{bcd}\:=\:\mathrm{4}\:\left(\mathrm{mod}\:\mathrm{11}\right)\:}\end{cases} \\ $$
Answered by HomeAlone last updated on 21/Oct/23
ab+ca+dc+ad=a+bd+cd is complicative
$${ab}+{ca}+{dc}+{ad}={a}+{bd}+{cd}\:{is}\:{complicative} \\ $$
Answered by Hridiana last updated on 21/Oct/23
a+b=4  a+c=4
$${a}+{b}=\mathrm{4} \\ $$$${a}+{c}=\mathrm{4} \\ $$
Answered by AST last updated on 21/Oct/23
a^2 b^2 c^2 d^3 =(abc)^2 d^3 ≡^(11) 2×3×4≡^(11) 2⇒d^3 ≡^(11) 2  ⇒d≡7(mod 11)  a^2 bcd^2 =(abc)a(d^2 )≡^(11) 2×3⇒49a≡^(11) 6⇒−6a≡^(11) 6  ⇒a≡10(mod 11)  abd≡^(11) 2⇒4b≡^(11) 2⇒b≡6(mod 11)  abc≡^(11) 1⇒5c≡^(11) 1⇒c≡9(mod 11)
$${a}^{\mathrm{2}} {b}^{\mathrm{2}} {c}^{\mathrm{2}} {d}^{\mathrm{3}} =\left({abc}\right)^{\mathrm{2}} {d}^{\mathrm{3}} \overset{\mathrm{11}} {\equiv}\mathrm{2}×\mathrm{3}×\mathrm{4}\overset{\mathrm{11}} {\equiv}\mathrm{2}\Rightarrow{d}^{\mathrm{3}} \overset{\mathrm{11}} {\equiv}\mathrm{2} \\ $$$$\Rightarrow{d}\equiv\mathrm{7}\left({mod}\:\mathrm{11}\right) \\ $$$${a}^{\mathrm{2}} {bcd}^{\mathrm{2}} =\left({abc}\right){a}\left({d}^{\mathrm{2}} \right)\overset{\mathrm{11}} {\equiv}\mathrm{2}×\mathrm{3}\Rightarrow\mathrm{49}{a}\overset{\mathrm{11}} {\equiv}\mathrm{6}\Rightarrow−\mathrm{6}{a}\overset{\mathrm{11}} {\equiv}\mathrm{6} \\ $$$$\Rightarrow{a}\equiv\mathrm{10}\left({mod}\:\mathrm{11}\right) \\ $$$${abd}\overset{\mathrm{11}} {\equiv}\mathrm{2}\Rightarrow\mathrm{4}{b}\overset{\mathrm{11}} {\equiv}\mathrm{2}\Rightarrow{b}\equiv\mathrm{6}\left({mod}\:\mathrm{11}\right) \\ $$$${abc}\overset{\mathrm{11}} {\equiv}\mathrm{1}\Rightarrow\mathrm{5}{c}\overset{\mathrm{11}} {\equiv}\mathrm{1}\Rightarrow{c}\equiv\mathrm{9}\left({mod}\:\mathrm{11}\right) \\ $$
Commented by HomeAlone last updated on 21/Oct/23
good job
$$\mathrm{good}\:\mathrm{job} \\ $$

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