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CET Karnataka Engineering CET - Karnataka Engineering Solved Paper-2008

  • question_answer
    The         ten's         digit         in \[1!4!+7!+10!+12!+13!+15!+16!+17!\]is divisible by

    A) \[4\]                                     

    B)  \[3!\]

    C)  \[5\]                                    

    D)  \[7\]

    Correct Answer: B

    Solution :

    As we know the last two digits of 10! and above factorials will be zero-zero. \[\therefore \] \[1!+4!+7!+10!+12!+13!+15!+16!+17!\] \[=1+24+5040+10!+12!+13!\]                                 \[+15!+16!+17!\] \[=5065+10!+12!+13!+15!+16!+17!\] In this series, the digit in the ten place is 6 which is divisible by \[3!\].


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