8 Posts • Page 1 of 1
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Peter
Birch & Swinnerton Dyer


Offline Joined: 05 May 2004 Posts: 5202 Location: Ghent
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A 9 [IHH, pp. 211]
Prove that among any ten consecutive positive integers at least one is relatively prime to the product of the others.
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Posted: Fri May 25, 2007 2:24 am |
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Ilthigore
Riemann Hypothesis


Offline Joined: 31 Dec 2005 Posts: 305 Location: Bristol, UK
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Clearly the only common prime factors amongst 10 consecutive positive integers will be 2,3,5,7.
5 of them will be divisible by 2, at least one of which must also be divisible by 3 and at least one of which must also be divisible by 5, and, if two of the numbers are divisible by 7, one of them will be even.
This leaves two multiples of 3, one multiple of 5, and one multiple of 7 unaccounted for, enough factors to dish out to 4 more of our integers.
But this still leaves one integer not divisible by 2,3,5,7, and therefore co-prime with all other integers of the set, and so coprime with their product.
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Posted: Fri May 25, 2007 2:24 am |
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manjil
Riemann Hypothesis


Offline Joined: 29 Jun 2008 Posts: 254 Location: Tezpur
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Generalisation?
Can we replace by ?
I have found that this fails for , although I am not sure of the other values. Can anyone give anything?
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_________________ Manjil P. Saikia
Posted: Fri Aug 15, 2008 5:40 pm |
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Peter
Birch & Swinnerton Dyer


Offline Joined: 05 May 2004 Posts: 5202 Location: Ghent
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Can you post your example for 25? Maybe that gives us a clue.
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_________________ Boo!
Posted: Fri Aug 15, 2008 8:20 pm |
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MayankM
New Member

Offline Joined: 04 Jul 2005 Posts: 14 Location: New Delhi
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I have read some where that this result is true for n up to 16 consecutive integers. I'll look up the source and post soon.
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_________________ Imposible Nothing
Posted: Tue Aug 26, 2008 6:48 pm |
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manjil
Riemann Hypothesis


Offline Joined: 29 Jun 2008 Posts: 254 Location: Tezpur
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Re:Generalisation
| peter wrote: |
| Can you post your example for 25? Maybe that gives us a clue.
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Well my example is the natural numbers from to .
| MayankM wrote: |
I have read some where that this result is true for n up to 16 consecutive integers. I'll look up the source and post soon.
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The source of the problem said IHH pp.211 and so I took that out and I found what MayankM had siad to be true.
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_________________ Manjil P. Saikia
Posted: Sun Aug 31, 2008 11:03 am |
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Peter
Birch & Swinnerton Dyer


Offline Joined: 05 May 2004 Posts: 5202 Location: Ghent
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Re:Generalisation
| manjil wrote: |
| peter wrote: |
| Can you post your example for 25? Maybe that gives us a clue.
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Well my example is the natural numbers from to .
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Really? I think is relatively prime to the product of the others though... 
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_________________ Boo!
Posted: Sun Aug 31, 2008 11:48 am |
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Jure the frEEEk
Poincare Conjecture


Offline Joined: 12 Aug 2007 Posts: 144 Location: Ljubljana, Slovenija
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| MayankM wrote: |
| I have read some where that this result is true for n up to 16 consecutive integers. I'll look up the source and post soon.
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it's true. 16 can be done and 17 can't. the counterexample for are consecutive numbers of which the first equals .
(in fact the first can be for a natural , mybe there exists a smaller example that i missed but the point is that there are infinitely many)
u can proove it for basicaly same as ilthigore has done for above only a bit more complicated.
let be these consecutive numbers. if we set to be devisable by , devisable by , by and by . it's easy to check thet none of them is than relatively prime to all others. we can calculate with china reminder theorem now.
what is left to do is to show that if there is a counter wxample for than we can also find a counter example for . im working on this now.
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_________________ Archimedes will be remembered when Aeschylus is forgotten, because languages die and mathematical ideas do not!
Posted: Mon Sep 01, 2008 3:32 pm |
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8 Posts • Page 1 of 1
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