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Topic: relatively prime probability (Read 283 times) |
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JocK
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relatively prime probability
« on: Nov 27th, 2005, 11:26am » |
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Randomly select two positive integers.
What is the chance these two integers have no common divisors other than 1?
PS. In case you need things a bit more specific: randomly select twice an element from the set {1, 2, .. , N}. Let PN be the probability that both selected elements are relatively prime. Calculate lim N --> [Infty] PN .
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| « Last Edit: Nov 27th, 2005, 11:30am by JocK » |
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solving abstract problems is like sex: it may occasionally have some practical use, but that is not why we do it.
xy - y = x5 - y4 - y3 = 20; x>0, y>0.
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wizardofozz
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Re: relatively prime probability
« Reply #1 on: Nov 28th, 2005, 11:48am » |
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N pN
1 1.0000
2 0.7500
4 0.6875
8 0.6719
16 0.6211
32 0.6318
64 0.6150
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JohanC
Senior Riddler
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Re: relatively prime probability
« Reply #2 on: Nov 28th, 2005, 1:08pm » |
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6 / Pi2 = 0.607927101854...
But don't ask me why ...
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Barukh
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Re: relatively prime probability
« Reply #3 on: Nov 29th, 2005, 9:43am » |
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JohanC, if you want to know why, try to calculate the probability that two random numbers do not have a common specific prime divisor.
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| « Last Edit: Nov 29th, 2005, 9:44am by Barukh » |
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