7!, 8!, 9!, and so on are all divisible by 7. Since we want the remainder of a sum when we divide by 7, we can just ignore any multiples of 7, since they will only affect the quotient and not the remainder when we divide them by 7. So the question is just asking:

What is the remainder when 6! + 5! + 4! + 3! + 2! + 1! is divided by 7?

There's no especially useful shortcut here faster than just computing the value of that number - we find it equals 720+120+24+6+2+1 = 873. If we want the remainder dividing this by 7, we can now subtract any multiple of 7 we like, so subtracting 700 and 140, we're left with 33, and the remainder when we divide that by 7 is 5, so that's the answer.
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7!, 8!, 9!...100! will be divisible by 7 because they will contain a 7. Therefore the remainder should be zero for these factorials.

However, for the first six factorials, we need to find the remainder. The easiest way is to add them and then divide it by 7 and find the remainder

(1! + 2! + 3! + 4! + 5! + 6!) = 873
873/7 gives remainder as 5. Hence, the answer should be 5. Option C.
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7! onward everything is divisible by 7.

Hence remainders are as follows:


\frac{(1!)}{7} = 1
\frac{(2!)}{7} = 2
\frac{(3!)}{7} = 6
\frac{(4!)}{7} = 3
\frac{(5!)}{7} = 1
\frac{(6!)}{7} = 6

The remainders sum to 19 which leaves a remainder of 5 when divided by 7.

Answer (C)
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