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  4. The ten's digit in 1 !+4 !+7 !+10 !+12 !+ 13 !+15 !+16 !+17 ! is divisible by

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Q. The ten's digit in $1 !+4 !+7 !+10 !+12 !+$ $13 !+15 !+16 !+17 !$ is divisible by

Permutations and Combinations

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's place is 6 which is divisible by $3 !$.