Find a remainder when $1 !~ + 2 ! ~+ 3 !~ + \cdots ~+ 100 !$ is divided by $12$ .

 Example: Find a remainder when $1 !~ + 2  ! ~+ 3  !~ + \cdots ~+ 100  !$ is divided by $12$ 

Answer: One can observe that $4! \equiv 24 \equiv 0 ~(mod ~12)$; thus, for $k \geq 4,~ k !  \equiv 4! \cdot 5 \cdot 6 ~\cdots ~k \equiv 0 \cdot  5 \cdot 6 ~\cdots~ k \equiv 0 ~(mod ~12)$. Therefore $1!~ + 2!~ + 3!~ + 4! ~+ \cdots ~+ 100! \equiv 1!~ + 2!~ + 3!~ + ~0 +~ \cdots + ~ 0 \equiv 9 ~(mod ~12) $ i.e. $9 $ a remainder when $1 ! + 2  ! + 3  ! + \cdots + 100  !$ divided by 12.

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