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The last digit of $\sum_{\substack{1 < p < 100 \\ p \rightarrow \text { prime }}} p !-\sum_{n=1}^{50}(2 n) !$ is
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$\underset{\substack{1 < p < 100 \\ p \rightarrow p r i m e}}{ } P !-\sum_{n=1}^{50}(2 n !)=(2 !+3 !+5 !+7 !+\ldots)$ $-(2 !+4 !+6 !+\ldots)$
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 $\sum_{\substack{1 < p < 100 \\ p \rightarrow \text { prime }}} p !-\sum_{n=1}^{50}(2 m !)=8-6=2$
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 $\sum_{\substack{1 < p < 100 \\ p \rightarrow \text { prime }}} p !-\sum_{n=1}^{50}(2 m !)=8-6=2$
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