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CET Karnataka Engineering CET - Karnataka Engineering Solved Paper-2009

  • question_answer
    The digit in the unit place of the number \[2009!+{{3}^{7886}}\]is

    A)  \[7\]                                    

    B)  \[3\]

    C)  \[1\]                                    

    D)  \[9\]

    Correct Answer: D

    Solution :

    The digit in the unit place of (2009) is 0. Now, \[{{3}^{1}}=3,\] \[{{3}^{2}}=9,\] \[{{3}^{3}}=27,\] \[{{3}^{4}}=81,\] \[{{3}^{5}}=243\] \[\therefore \]  \[{{3}^{7886}}={{(3)}^{4}}^{1971}{{3}^{2}}\] The digit in the unit place of \[{{3}^{7886}}\] is 9. \[\therefore \] The digit in the unit place of \[\left( 2009 \right)!+{{3}^{7886}}\]is 9.


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