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Manager
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Joined: 21 May 2008
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PS-Sum of even integers.jpg [#permalink]

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New post 10 Oct 2008, 05:47
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Intern
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Joined: 22 Aug 2008
Posts: 3
Re: PS-Sum of even integers [#permalink]

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New post 10 Oct 2008, 13:40
7550.
explanation-
2+4+6+...+100 = 2550.
sum of all even integers from 102 to 200 is
102+104+106+...+200
=(100+2)+(100+4)+(100+6)+....+(100+100)
=(2+4+6+....+100)+(50*100)
=2550+5000
Senior Manager
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Re: PS-Sum of even integers [#permalink]

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New post 10 Oct 2008, 13:48
7550

102 + 104 + 106 + .... 200
100 + 2 + 100 + 4 + 100 + 6 ..... 100+100
50*100 + (2+4+6 ...100)

= 5000 + 2550 = 7550
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Schools: Wharton'11 HBS'12
Re: PS-Sum of even integers [#permalink]

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New post 10 Oct 2008, 13:54
first figure how how many integers there are between 102-200

200=102+(n-1)(2) => n=50

sum= 50/2 (2*102+49*2)=7550
Manager
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Joined: 27 Sep 2008
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Re: PS-Sum of even integers [#permalink]

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New post 10 Oct 2008, 14:43
lets see an easy one

2+4+6+8+10 = 30

2(1+2+3+4+5) = 30

we know that 1+2+3+4+5 = 5*6/2 (the last one*last one+1)/2

so 2*(5*6/2) = 30

solve

2*(100*101)/2-2*(50*51)/2 = 7550

:)
Re: PS-Sum of even integers   [#permalink] 10 Oct 2008, 14:43
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