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Re: What is the remainder when 1! + 2! + 3! + ... + 100! is divided by 7 [#permalink]
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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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Re: What is the remainder when 1! + 2! + 3! + ... + 100! is divided by 7 [#permalink]
Hovkial wrote:
What is the remainder when 1! + 2! + 3! + ... + 100! is divided by 7?

(A) 0

(B) 3

(C) 5

(D) 6

(E) 7


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
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What is the remainder when 1! + 2! + 3! + ... + 100! is divided by 7 [#permalink]
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Hovkial wrote:
What is the remainder when 1! + 2! + 3! + ... + 100! is divided by 7?

(A) 0

(B) 3

(C) 5

(D) 6

(E) 7


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)

Originally posted by unraveled on 24 Aug 2019, 09:05.
Last edited by unraveled on 16 Dec 2019, 07:35, edited 1 time in total.
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Re: What is the remainder when 1! + 2! + 3! + ... + 100! is divided by 7 [#permalink]
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

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Re: What is the remainder when 1! + 2! + 3! + ... + 100! is divided by 7 [#permalink]
Expert Reply
Hovkial wrote:
What is the remainder when 1! + 2! + 3! + ... + 100! is divided by 7?

(A) 0

(B) 3

(C) 5

(D) 6

(E) 7



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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Re: What is the remainder when 1! + 2! + 3! + ... + 100! is divided by 7 [#permalink]
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Re: What is the remainder when 1! + 2! + 3! + ... + 100! is divided by 7 [#permalink]
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