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  4. The remainder obtained when 1! + 2! + 3! + ... + 100! is divided by 12, is

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Q. The remainder obtained when $1! + 2! + 3! + ... + 100!$ is divided by $12$, is

Permutations and Combinations

Solution:

Let $S = 1! + 2! + 3! + 4! + 5! + ... + 100!$
$= 1! + 2! + 3! + (4! + 5! + ... + 100!)$
$= 1 + 2 + 6 + 12\lambda (\because H.C.F.$ of $4! + 5! + ...$
$+ 100!$ is $4!$, which is divisible by $12$)
$= 9 + 12 \,\lambda$
$\therefore $ Remainder is $9$.