What is the remainder left after dividing 1! + 2! + 3! + 100! by 7?

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Solution

The correct option is B 5
$1! + 2! + 3! + . . .100!$ As after $7!$
each term has '7' $∴$ there Remainder is '0'
hence Reminder of above series is
$(1! + 2! + 3! + 4! + 5! + 6!) / 7 = 873 / 7 \Rightarrow 5$

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