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Topic: Coprimality of Two Randomly Chosen Integers (Read 979 times) |
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william wu
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Coprimality of Two Randomly Chosen Integers
« on: Aug 21st, 2003, 2:17pm » |
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Show that
[prod]p in primes(1 - p-2) = 6[pi]-2
Conclude that the probability two randomly chosen integers are coprime is 6[pi]-2.
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SWF
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Re: Coprimality of Two Randomly Chosen Integers
« Reply #1 on: Aug 21st, 2003, 6:23pm » |
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What a coincidence! Just a few minutes ago I used that in solution to Random Line Segment in Square riddle. However I left out the details to keep my post from being too long.
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TenaliRaman
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I am no special. I am only passionately curious.
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Re: Coprimality of Two Randomly Chosen Integers
« Reply #2 on: Aug 22nd, 2003, 11:56am » |
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hey the second question's pretty neat!!!
it had me hooked up for the last 5 hours before it dawned on me that P(coprime)=1-P(not coprime) and the first result comes into play.
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Self discovery comes when a man measures himself against an obstacle - Antoine de Saint Exupery
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