What is the remainder when 1! + 2! + 3! + ... + 100! is divided by 7

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.

All the factorials of numbers greater than 6 are divisible by 7

If n is a prime number.
1. (n-1)! = -1 mod n
2. (n-2)! = 1 mod n

6! =6 mod 7.
5! =1 mod 7
4!= 3 mod 7
3! = 6 mod 7
2! = 2 mod 7
1! =1 mod 7

6! + 5! + 4! + 3! + 2! + 1!= (6+1+3+6+2+1) mod 7
6! + 5! + 4! + 3! + 2! + 1!= 5 mod 7

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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Asked: What is the remainder when 1! + 2! + 3! + ... + 100! is divided by 7?

1! = 1mod7
2! = 2mod7
3!= 6mod7
4!= 3mod7
5! = 1mod7
6! = 6mod7
7! = 0mod7
8!= 0 mod7

.....
100! = 0mod7

1! + 2! + 3! + ... + 100! = (1+2+6+3+1+6)mod7 = 19mod7 = 5mod7

IMO C

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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5 has to be the answer.
as the factorial greater than 6 will leave zero remainder.
and upto 6! will leave===> 720+120+24+6+2+1==== 5 remainder

Posted from my mobile device
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Responding to a pm:

Correct. You cannot find the value of 1! + 2! + 3! + ... 100!

But note that 7!, 8! etc all the terms have 7 as a factor so they will be divisible by 7. Hence they will leave a remainder of 0 when divided by 7.
Now we only need to worry about the first 6 terms: 1! + 2! + 3! + 4! + 5! + 6!

These values we should already know but even if we don't, they are easy to calculate: 1 + 2 + 6 + 24 + 120 + 720 = 873
When 873 is divided by 7, remainder is 5.

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