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CET Karnataka Engineering CET - Karnataka Engineering Solved Paper-2011

  • question_answer
    The digit in the unit?s place of \[{{7}^{171}}+(177)!\]is

    A)  3                                            

    B)  2

    C)  1                                            

    D)  0

    Correct Answer: C

    Solution :

    Given expression \[{{(7)}^{171}}+(177)!\] We know that the cycle of 7 is 4, i.e., \[{{(7)}^{3}}=1\] and in case of more the 4 number in factorial notation gives always 0 at unit place. \[\Rightarrow \]               \[{{({{7}^{3}})}^{57}}+(177)!=1+0=1\]


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