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

KCETKCET 2008Permutations 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 place is $6$ which is divisible by $3 !$