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# The factorial (!) of a positive integer n denotes the product of all

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Manager
Joined: 13 May 2017
Posts: 122
Location: Finland
Concentration: Accounting, Entrepreneurship
GMAT 1: 530 Q42 V22
GMAT 2: 570 Q36 V31
GMAT 3: 600 Q42 V28
GPA: 3.14
WE: Account Management (Entertainment and Sports)
The factorial (!) of a positive integer n denotes the product of all  [#permalink]

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29 Oct 2018, 06:37
3
00:00

Difficulty:

45% (medium)

Question Stats:

51% (01:03) correct 49% (01:25) wrong based on 92 sessions

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The factorial (!) of a positive integer n denotes the product of all integers from 1 to n, inclusive. If k = 1! + 2! + 3! + . . . + p! , where p is a prime number greater than 10, what is the remainder when k is divided by 4?

A. 0
B. 1
C. 2
D. 3
E. 9
Manager
Joined: 14 Jun 2018
Posts: 210
Re: The factorial (!) of a positive integer n denotes the product of all  [#permalink]

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29 Oct 2018, 07:20
1
Apart from 1! , 2! and 3! , everything else will be divisible by 4.
1+2+6=9
Remainder = 1
Manager
Joined: 29 May 2017
Posts: 158
Location: Pakistan
Concentration: Social Entrepreneurship, Sustainability
Re: The factorial (!) of a positive integer n denotes the product of all  [#permalink]

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08 Nov 2018, 19:23
pandeyashwin wrote:
Apart from 1! , 2! and 3! , everything else will be divisible by 4.
1+2+6=9
Remainder = 1

3!=6
r(6/4)=2
r(2!/4)=2
r(1!/4)=1

1+2+2=5

r(5/4)=1

is my method correct?
Manager
Joined: 14 Jun 2018
Posts: 210
Re: The factorial (!) of a positive integer n denotes the product of all  [#permalink]

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08 Nov 2018, 19:40
Mansoor50 wrote:
pandeyashwin wrote:
Apart from 1! , 2! and 3! , everything else will be divisible by 4.
1+2+6=9
Remainder = 1

3!=6
r(6/4)=2
r(2!/4)=2
r(1!/4)=1

1+2+2=5

r(5/4)=1

is my method correct?

Manager
Joined: 29 May 2017
Posts: 158
Location: Pakistan
Concentration: Social Entrepreneurship, Sustainability
Re: The factorial (!) of a positive integer n denotes the product of all  [#permalink]

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08 Nov 2018, 19:46
pandeyashwin wrote:
Mansoor50 wrote:
pandeyashwin wrote:
Apart from 1! , 2! and 3! , everything else will be divisible by 4.
1+2+6=9
Remainder = 1

3!=6
r(6/4)=2
r(2!/4)=2
r(1!/4)=1

1+2+2=5

r(5/4)=1

is my method correct?

thanks
Intern
Joined: 13 Mar 2018
Posts: 14
Re: The factorial (!) of a positive integer n denotes the product of all  [#permalink]

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10 Nov 2018, 00:48
Can someone kindly explain why everything apart from 1!, 2! and 3! is divisble by 4. May someone please share formula if any
Manager
Joined: 14 Jun 2018
Posts: 210
Re: The factorial (!) of a positive integer n denotes the product of all  [#permalink]

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10 Nov 2018, 00:52
2
SWAPNILP wrote:
Can someone kindly explain why everything apart from 1!, 2! and 3! is divisble by 4. May someone please share formula if any

2! = 2x1
3! = 3x2x1

4! = 4x3x2x1
5! = 5x4x3x2x1
6! = 6x5x4x3x2x1
and so on
Re: The factorial (!) of a positive integer n denotes the product of all   [#permalink] 10 Nov 2018, 00:52
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