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

7

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6

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8

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9

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Solution

The correct option is D $9$
Given $1! + 2! + 3! + 4! + . . . . . . . . . + 100!$
From $4!$ onwards, every factorial expansion has $4 \times 3$ in itself which will be divisible by $12$ .
Hence , in the above sum every thing in $4! + 5! + 6! . . . . . . . + 100!$ will be divisible by $12$
The sum before $4!$ i.e $1! + 2! + 3!$ will be the remainder .
Hence the remainder when $1! + 2! + 3! + 4! + . . . . . . . . . + 100!$ is divisible by $12$ will be $1! + 2! + 3! = 9$

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