The remainder obtained when 1! + 2! + 3! + . . . + 27! is divided by 15 is

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Solution

The correct option is B 3
$1! + 2! + 3! + 4! + 5! + \dots + 27!$
Divisible by 15
$∴ 1 + 2 + 6 + 24 = \frac{33}{15} \Rightarrow Remainder = 3$

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