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J & K CET Engineering J and K - CET Engineering Solved Paper-2003

  • question_answer
    In   the   group \[G=\{1,3,7,9\}\]under multiplication modulo 10, the inverse of 7 is

    A)  \[7\]              

    B)  \[3\]

    C)  \[9\]               

    D)  \[1\]

    Correct Answer: B

    Solution :

    Now,\[7{{\times }_{10}}3\equiv \,1\,\,\bmod 10\]   (given) which gives the identity element. Hence, the inverse of 7 is 3 under the given operation.


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