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# Problem # 157 in the OG 12th Ed

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Joined: 01 Nov 2009
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Problem # 157 in the OG 12th Ed [#permalink]  24 Nov 2009, 20:11
00:00

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(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions
For any positive integer n the sum of the first n positive integers equals (n(n+1))/2. What is the sum of all the even integers between 99 and 301?

Could someone give me a better explanation than what the OG provides?
CEO
Joined: 17 Nov 2007
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Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
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Re: Problem # 157 in the OG 12th Ed [#permalink]  24 Nov 2009, 20:51
Expert's post
sum = 100 + 102 + 104 + ... + 296 + 298 + 300.

N = (300-100)/2 +1 = 101 - the number of integers in the sum

Now, you can add first and last integers (100+300=400), then second and 100th integers (102+298=400) and so on.
Finally, we will have 50 times 400 and one integers at the middle (200) as we have odd number of integers.

So, sum = 50 * 400 + 200 = 20200
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Manager
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Re: Problem # 157 in the OG 12th Ed [#permalink]  24 Nov 2009, 21:48
Sum = 100 + 102 + ... + 298 + 300
= 2*(50 + 51 + ... + 149 + 150)
= 2*[(1 + 2 + ... + 149 + 150) - (1 + 2 + ... + 48 + 49)

Using the given formula: Sum of Integers from 1 to n = n(n+1)/2

= 2*[150*151/2 - 49*50/2]
= 2*[22650 - 2450]/2
= 20200
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Location: Tbilisi, Georgia
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Re: Problem # 157 in the OG 12th Ed [#permalink]  25 Nov 2009, 01:28
That's quite easy problem... Consecutive integer concepts are well explained in MGMAT Number properties.
Re: Problem # 157 in the OG 12th Ed   [#permalink] 25 Nov 2009, 01:28
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