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What is the remainder when 1! + 2! + 3! … 100! is divided by 18?

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What is the remainder when 1! + 2! + 3! … 100! is divided by 18?  [#permalink]

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New post 12 Aug 2018, 22:20
1
6
00:00
A
B
C
D
E

Difficulty:

  65% (hard)

Question Stats:

49% (01:28) correct 51% (01:20) wrong based on 78 sessions

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What is the remainder when 1! + 2! + 3! … 100! is divided by 18?

1) 0
2) 1
3) 5
4) 8
5) 9

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What is the remainder when 1! + 2! + 3! … 100! is divided by 18?  [#permalink]

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New post 12 Aug 2018, 22:37
1
Afc0892 wrote:
What is the remainder when 1! + 2! + 3! … 100! is divided by 18?

1) 0
2) 1
3) 5
4) 8
5) 9


Factorial of the positive integers greater than 5 is divisible by 18.

So, we have to find out the remainder when 1!+2!+3!+4!+5!(= 153)is divided by 18. (All other terms yield zero remainder)

So, Remainder =9.

Ans. (E)
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Re: What is the remainder when 1! + 2! + 3! … 100! is divided by 18?  [#permalink]

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New post 21 Aug 2018, 02:39
PKN : Though I understood your explanation, would like to know how to apply this technique in similar questions. Should the divisor be prime factorized and then to be checked with the factorials in the numerator?
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Re: What is the remainder when 1! + 2! + 3! … 100! is divided by 18?  [#permalink]

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New post 21 Aug 2018, 02:47
1
Prime Factorization of 18 gives (3^2)*2. Therefore, we need at least 2 3s and 1 2 to ensure that the factorial is divisible by 18.

1! = 1
2! = 2
3! = 3*2
4! = 4*3*2*1
5! = 5*4*3*2*1
6! = 6*5*4*3*2*1 = (3*2)*5*4*3*2*1. Thus, after 6! (inclusive), every other factorial in the above sum will be divisible by 18.

Adding 1! to 5!, we get 153. When we divide, 153 by 18, the remainder is 9.

Answer: E
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Re: What is the remainder when 1! + 2! + 3! … 100! is divided by 18?  [#permalink]

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New post 21 Aug 2018, 02:54
narendran1990 wrote:
PKN : Though I understood your explanation, would like to know how to apply this technique in similar questions. Should the divisor be prime factorized and then to be checked with the factorials in the numerator?


We have to inspect the terms in the numerator and check the term that is divisible by numerator. Moreover, the pattern needs to be observed.

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Re: What is the remainder when 1! + 2! + 3! … 100! is divided by 18?  [#permalink]

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New post 01 Oct 2018, 02:32
Do we get the same answer if we divide 1!, 2!, 3!, 4!, 5! individually by 18?

i get 1 + 1 +1 +1 +2-->6

not sure where i am going wrong
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Re: What is the remainder when 1! + 2! + 3! … 100! is divided by 18?  [#permalink]

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New post 01 Oct 2018, 02:44
1
Mansoor50 wrote:
Do we get the same answer if we divide 1!, 2!, 3!, 4!, 5! individually by 18?

i get 1 + 1 +1 +1 +2-->6

not sure where i am going wrong



Yes you can divide individually and then add all the remainders and again divide it by 18 to get the final remainder.

1!/18 = 1
2!/18 = 2
3! 18 =6
4!/18 = 6
5!/18 = 12

Now adding all the remainders 1+2+6+6+12 = 27/18. final remainder is 9.

Hope its clear.
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Re: What is the remainder when 1! + 2! + 3! … 100! is divided by 18? &nbs [#permalink] 01 Oct 2018, 02:44
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