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