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Topic: Coprimality of Two Randomly Chosen Integers (Read 940 times) 

william wu
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Coprimality of Two Randomly Chosen Integers
« on: Aug 21^{st}, 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 21^{st}, 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 22^{nd}, 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)=1P(not coprime) and the first result comes into play.


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