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Title: Expressibility Post by THUDandBLUNDER on Aug 29th, 2004, 10:54pm Which positive integers cannot be expressed as a sum of 2 or more consecutive integers? |
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Title: Re: Expressibility Post by mistysakura on Aug 30th, 2004, 4:15am [hide]none. Let the number to be expressed be x (-x+1)+(-x+2)...+(-1)+0+1...+(x-2)+(x-1)+x =0+x =x (Exception: 1=0+1 (duh.))[/hide] |
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Title: Re: Expressibility Post by towr on Aug 30th, 2004, 4:35am ok then, a 'follow-up question' ;) Which positive integers cannot be expressed as a sum of 2 or more consecutive positive integers? |
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Title: Re: Expressibility Post by EZ_Lonny on Aug 30th, 2004, 8:19am 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 [hide] excuse me if i'm wrong [/hide] [hide] 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.[/hide] |
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Title: Re: Expressibility Post by BNC on Aug 30th, 2004, 8:22am on 08/30/04 at 04:35:08, towr wrote:
Obviously all odd numbers are ok. The question is what even numbers may not be expressed that way. |
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Title: Re: Expressibility Post by EZ_Lonny on Aug 30th, 2004, 8:22am I know, I'm working on that |
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Title: Re: Expressibility Post by EZ_Lonny on Aug 30th, 2004, 8:24am I'm only guessing: All integers dividable by 4 or that odd onez |
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Title: Re: Expressibility Post by towr on Aug 30th, 2004, 9:15am 2+3+4+5=14 So even some even numbers not divisble by 4 can be expressed by 2 or more consecutive numbers.. |
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Title: Re: Expressibility Post by Sir Col on Aug 30th, 2004, 4:12pm 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 08/30/04 at 08:24:55, EZ_Lonny wrote:
What about 8? Clearly some divisible by 4 work and some don't. ;) |
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Title: Re: Expressibility Post by Hooie on Aug 30th, 2004, 6:37pm I think I got it. *[hide] 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.[/hide]* Can someone tell me if I'm on the right track? Or am I just making this all up? |
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Title: Re: Expressibility Post by Aryabhatta on Aug 30th, 2004, 6:41pm on 08/30/04 at 18:37:56, Hooie wrote:
You are on the right track. (you have the right answer) |
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Title: Re: Expressibility Post by Grimbal on Aug 31st, 2004, 3:57am ::[hide] Let's write m = 2k*(2n+1) If 2k > n then m can be written as m = sum 2k-n ... 2k+n there are 2n+1 terms averaging 2k. If 2k <= n, then m can be written as m = sum n-2k+1 ... n+2k there are 2*2k 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)*(b-a+1)/2, b>a, a>0. 2m = (a+b)*(b-a+1) must also be a power of 2. But (a+b) or (b-a+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 => (b-a+1)>=2, so it cannot be. [/hide]:: |
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