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CET Karnataka Engineering CET - Karnataka Engineering Solved Paper-2006

  • question_answer
    The remainder obtained when \[{{(1!)}^{2}}+{{(2!)}^{2}}+{{(3!)}^{2}}+....+{{(100!)}^{2}}\] is divided by \[{{10}^{2}}\] is:

    A)  27                                         

    B)  28

    C)  17                                         

    D)  14

    Correct Answer: C

    Solution :

    Terms greater than \[5!\] i.e.,  \[(5!),{{(6!)}^{2}},....,{{(100!)}^{2}}\]   is divisible by 100 \[\therefore \] For terms \[{{(5!)}^{2}},{{(6!)}^{2}},....,{{(100!)}^{2}}\] remainder is 0. Now consider \[{{(1!)}^{2}}+{{(2!)}^{2}}+{{(3!)}^{2}}+{{(4!)}^{2}}\]                                 \[=1+4+36+576\]                                 \[=617\] When  \[617\] is divided by \[100,\] its remainder is 17. \[\therefore \] Required remainder is 17.


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