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ThudnBlunder
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Expressibility  
« on: Aug 29th, 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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Re: Expressibility  
« Reply #1 on: Aug 30th, 2004, 4:15am »
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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.))
« Last Edit: Aug 30th, 2004, 4:17am by mistysakura » IP Logged
towr
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Re: Expressibility  
« Reply #2 on: Aug 30th, 2004, 4:35am »
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ok then, a 'follow-up question' Wink
 
Which positive integers cannot be expressed as a sum of 2 or more consecutive positive integers?
« Last Edit: Aug 30th, 2004, 4:36am by towr » IP Logged

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EZ_Lonny
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Re: Expressibility  
« Reply #3 on: Aug 30th, 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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Re: Expressibility  
« Reply #4 on: Aug 30th, 2004, 8:22am »
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on Aug 30th, 2004, 4:35am, towr wrote:
ok then, a 'follow-up question' Wink
 
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 30th, 2004, 8:23am by BNC » IP Logged

How about supercalifragilisticexpialidociouspuzzler [Towr, 2007]
EZ_Lonny
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Re: Expressibility  
« Reply #5 on: Aug 30th, 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 30th, 2004, 8:24am »
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I'm only guessing:
 
All integers dividable by 4 or that odd onez
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Re: Expressibility  
« Reply #7 on: Aug 30th, 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 30th, 2004, 9:15am by towr » IP Logged

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Sir Col
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Re: Expressibility  
« Reply #8 on: Aug 30th, 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 30th, 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. Wink
« Last Edit: Aug 30th, 2004, 5:04pm by Sir Col » IP Logged

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Re: Expressibility  
« Reply #9 on: Aug 30th, 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 30th, 2004, 6:41pm »
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on Aug 30th, 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 31st, 2004, 3:57am »
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::
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.
::
« Last Edit: Aug 31st, 2004, 3:58am by Grimbal » IP Logged
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