Q100 / 180 · COMEDK 2018 · Mathematics

The last digit of $\sum_{\substack{1 < p < 100 \\ p \rightarrow \text { prime }}} p !-\sum_{n=1}^{50}(2 n) !$ is

  • Correct answer
    A
    2
  • B
    4
  • C
    6
  • D
    8

Solution

$\underset{\substack{1 < p < 100 \\ p \rightarrow p r i m e}}{ } P !-\sum_{n=1}^{50}(2 n !)=(2 !+3 !+5 !+7 !+\ldots)$ $-(2 !+4 !+6 !+\ldots)$ After 5 ! the number will have 0 at the one's place. So, they will no effect the last digit of the total sum. Now last digit of $(2 !+3 !)=2+6=8$ Last digit of $(2 !+4 !)=2+4=6$ Last digit of $\sum_{\substack{1 < p < 100 \\ p \rightarrow \text { prime }}} p !-\sum_{n=1}^{50}(2 m !)=8-6=2$

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

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