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number properties: consecutive integers

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Schools: Harvard Business School (HBS) - Class of 2014
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I'm not sure where to post this since I am not looking for [#permalink] New post 13 Jun 2010, 23:18
I'm not sure where to post this since I am not looking for an answer to a specific problem, rather confirmation on a potential number property. If true, I'd use it for data sufficienty so I'm posting here.

I am familiar with the following property: The sum of n consecutive integers is divisible by n if n is odd, but it is not divisible by n if n is even.

This makes logical sense. However, I was curious about scenarios in which n=even, so started messing around with numbers. After experimenting with several sets, it appears that if n=even, the sum of n consecutive integers is divisible by n/2. I've tested this with many number sets and it appears to be true. Can anyone confirm?
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Re: number properties: consecutive integers [#permalink] New post 14 Jun 2010, 02:36
Yes, your assumption must be true. Because, sum of n consecutive integers is n(n+1)/2. Now, if you take N as odd, sum would always be divisible by n. Because, sum would be n*some integer (since n+1 would be even, if n is odd and it would definitely be divisible by 2).

Similarly, if n is even, n(n+1)/2 would covert to n/2 as integer as n is even and n+1 which will be an odd number.

Hence, by algebraically, you can come to this conclusion. Please let me know if you have any queries on the explanation.
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 [#permalink] New post 15 Jun 2010, 00:21
vinay,
I would like to post a correction here regarding your post if you dont mind. You have told that sum of n consecutive integers is n(n+1)/2 which is only correct if your first integer is 1. If you start your series with any other number than 1, this formula is no more vaild.
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 [#permalink] New post 15 Jun 2010, 03:38
it's just pairing the first no. with the last one. Repeat and rinse.

Ex. find the sum of 13 14 15 16 17 18 19 20

20+13 = 14+19 = 15+18 = 16+17 = 33

So the sum is 33*4. This is Gauss's method. The formula is (first+last)*n/2
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 [#permalink] New post 15 Jun 2010, 06:23
amitjash, thank you very much for pointing out the problem in my explanation. I was only thinking about consecutive integers starting from 1. I missed the point there. zestzorb, your explanation holds good for this one.
  [#permalink] 15 Jun 2010, 06:23
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