What is the remainder when 1! + 2! + 3! + dot | vedclass

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(C) We know that $7! = 7 	imes 6 	imes 5 	imes 4 	imes 3 	imes 2 	imes 1 = 14 	imes 6 	imes 5 	imes 4 	imes 3 = 14k$,where $k = 360$.So,$7!$

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(C) We know that $7! = 7 	imes 6 	imes 5 	imes 4 	imes 3 	imes 2 	imes 1 = 14 	imes 6 	imes 5 	imes 4 	imes 3 = 14k$,where $k = 360$.So,$7!$ is divisible by $14$.For any $n \geq 7$,$n! = 7! 	imes 8 	imes 9 	imes \dots 	imes n$,which is also divisible by $14$.Therefore,the remainder of $(1! + 2! + 3! + \dots + 200!)$ divided by $14$ is the same as the remainder of $(1! + 2! + 3! + 4! + 5! + 6!)$ divided by $14$.Calculating the sum: $1! + 2! + 3! + 4! + 5! + 6! = 1 + 2 + 6 + 24 + 120 + 720 = 873$.Dividing $873$ by $14$: $873 = 14 	imes 62 + 5$.The remainder is $5$.

What is the remainder when $1! + 2! + 3! + \dots + 200!$ is divided by $14$?

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