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In the group $G=\{1,5,7,11\}$ under multiplication modulo 12, the solution of $7^{-1} \underset{12}{\times}(x \underset{12}{12} 11)=5$ is $x=$
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Given that, $G=\{1,5,7,11\}$ is a group under multiplication module 12 .
$\because \quad 7^{-1}=7 \quad\left[\because 7^{-1} \otimes_{12} 7=1\right]$
$\begin{array}{ll}\text { Now, } & 7^{-1} \otimes_{12}\left(x \otimes_{12} 11\right)=5 \\ \Rightarrow & 7 \otimes_{12}\left(11 \otimes_{12} x\right)=5 \\ \Rightarrow & \left(7 \otimes_{12} 11\right) \otimes_{12} x=5 \\ \Rightarrow & \left\{7 \otimes_{12} 11=\text { remainder after dividing } 77 \text { from } 12\right\} \\ \Rightarrow & 5 \otimes_{12} x=5 \\ \Rightarrow & x=1\end{array}$
$\because \quad 7^{-1}=7 \quad\left[\because 7^{-1} \otimes_{12} 7=1\right]$
$\begin{array}{ll}\text { Now, } & 7^{-1} \otimes_{12}\left(x \otimes_{12} 11\right)=5 \\ \Rightarrow & 7 \otimes_{12}\left(11 \otimes_{12} x\right)=5 \\ \Rightarrow & \left(7 \otimes_{12} 11\right) \otimes_{12} x=5 \\ \Rightarrow & \left\{7 \otimes_{12} 11=\text { remainder after dividing } 77 \text { from } 12\right\} \\ \Rightarrow & 5 \otimes_{12} x=5 \\ \Rightarrow & x=1\end{array}$
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