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Topic: Expressibility (Read 7022 times) 

ThudnBlunder
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Expressibility
« on: Aug 29^{th}, 2004, 10:54pm » 
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Which positive integers cannot be expressed as a sum of 2 or more consecutive integers?


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mistysakura
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Re: Expressibility
« Reply #1 on: Aug 30^{th}, 2004, 4:15am » 
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none. Let the number to be expressed be x (x+1)+(x+2)...+(1)+0+1...+(x2)+(x1)+x =0+x =x (Exception: 1=0+1 (duh.))

« Last Edit: Aug 30^{th}, 2004, 4:17am by mistysakura » 
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towr
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Re: Expressibility
« Reply #2 on: Aug 30^{th}, 2004, 4:35am » 
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ok then, a 'followup question' Which positive integers cannot be expressed as a sum of 2 or more consecutive positive integers?

« Last Edit: Aug 30^{th}, 2004, 4:36am by towr » 
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EZ_Lonny
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Re: Expressibility
« Reply #3 on: Aug 30^{th}, 2004, 8:19am » 
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Let me see ........... At least all even Integers cannot be displayed, 'coz in two consecutive integers there is one odd and one even integer. Even + odd = odd excuse me if i'm wrong all odd integers can be made. If only i knew how to proof that. 1+2=3 ; 2+3=5; 3+4=7; 4+5=9 etc.


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BNC
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Re: Expressibility
« Reply #4 on: Aug 30^{th}, 2004, 8:22am » 
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on Aug 30^{th}, 2004, 4:35am, towr wrote:ok then, a 'followup question' Which positive integers cannot be expressed as a sum of 2 or more consecutive positive integers? 
 Obviously all odd numbers are ok. The question is what even numbers may not be expressed that way.

« Last Edit: Aug 30^{th}, 2004, 8:23am by BNC » 
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EZ_Lonny
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Re: Expressibility
« Reply #5 on: Aug 30^{th}, 2004, 8:22am » 
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I know, I'm working on that


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EZ_Lonny
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Re: Expressibility
« Reply #6 on: Aug 30^{th}, 2004, 8:24am » 
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I'm only guessing: All integers dividable by 4 or that odd onez


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towr
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Re: Expressibility
« Reply #7 on: Aug 30^{th}, 2004, 9:15am » 
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2+3+4+5=14 So even some even numbers not divisble by 4 can be expressed by 2 or more consecutive numbers..

« Last Edit: Aug 30^{th}, 2004, 9:15am by towr » 
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Sir Col
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Re: Expressibility
« Reply #8 on: Aug 30^{th}, 2004, 4:12pm » 
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In fact, all integers greater than 2 that are not divisible by 4 can be expressed as the sum of at least two positive consecutive integers. However, the proof of that statement requires a slightly different perspective. As towr suggested, instead of trying to work out which integers can be written as such a sum, try and work out which integers cannot... on Aug 30^{th}, 2004, 8:24am, EZ_Lonny wrote:I'm only guessing: All integers dividable by 4 or that odd onez 
 What about 8? Clearly some divisible by 4 work and some don't.

« Last Edit: Aug 30^{th}, 2004, 5:04pm by Sir Col » 
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Hooie
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Re: Expressibility
« Reply #9 on: Aug 30^{th}, 2004, 6:37pm » 
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I think I got it. * It seems to be powers of 2 that don't work. Now I'm trying to find out why. I generalized a number formed by adding consecutive integers to the form of (n+1)x + (n/2)(n+1). I set that equal to 2^k and I'm messing around til I find a contradiction.* Can someone tell me if I'm on the right track? Or am I just making this all up?


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Aryabhatta
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Re: Expressibility
« Reply #10 on: Aug 30^{th}, 2004, 6:41pm » 
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on Aug 30^{th}, 2004, 6:37pm, Hooie wrote:I think I got it. Can someone tell me if I'm on the right track? Or am I just making this all up? 
 You are on the right track. (you have the right answer)


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Grimbal
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Re: Expressibility
« Reply #11 on: Aug 31^{st}, 2004, 3:57am » 
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:: Let's write m = 2^{k}*(2n+1) If 2^{k} > n then m can be written as m = sum 2^{k}n ... 2^{k}+n there are 2n+1 terms averaging 2^{k}. If 2^{k} <= n, then m can be written as m = sum n2^{k}+1 ... n+2^{k} there are 2*2^{k} terms averaging n+1/2 = (2n+1)/2 The first case is a valid solution when n>0, else there is only one term The second case is always valid, there are always at lest 2 terms. This proves that all numbers can be written as such a sum, except when n=0, which means m is a power of 2. It remains to prove that powers of 2 can never be written in such a way. Lets say m = sum a ... b is a power of 2 m = (a+b)*(ba+1)/2, b>a, a>0. 2m = (a+b)*(ba+1) must also be a power of 2. But (a+b) or (ba+1) must be odd. The only way out if if the odd factor is 1. But a>0 => (a+b) >= 1+2 and b>a => (ba+1)>=2, so it cannot be. ::

« Last Edit: Aug 31^{st}, 2004, 3:58am by Grimbal » 
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