Q94 / 160 · TS EAMCET 2010 · Mathematics

Let $n=1 !+4 !+7 !+\ldots+400 !$. Then ten's digit of $n$ is

  • A
    1
  • Correct answer
    B
    6
  • C
    2
  • D
    7

Solution

Given, $n=1 !+4 !+7 !+\ldots+400 !$ $1 !=1,4 !=24,7 !=5040,10 !=3628800$ and further the terms has last two digits $=00$ So, $1 !+4 !+7 !+\ldots+400 !=\ldots \ldots 65$ Here ten digit is 6 . Hence, 6 is the answer.

Hence, the correct answer is option: 'B'.

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