Problem 243

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chrisb555
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Joined: Fri Jan 09, 2009 5:27 am

Problem 243

Post by chrisb555 » Mon May 04, 2009 5:55 am

Hi,

Could someone clarify for me what is meant by "ratio of its proper fractions that are resilient"?
What I understand to be the ratio of the resilient fractions for d=12 is :
(1/12):(5/12):(7/12):(11/12)
and I calculate this to be 144/385. So clearly my understanding is different to yours.
Could you clarify?

Thanks,
Chris

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ed_r
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Re: Problem 243

Post by ed_r » Mon May 04, 2009 6:30 am

R(d) is the number of resilient proper fractions, divided by d-1.

"Proportion" would have been a better word than "ratio".
!647 = &8FDF4C

chrisb555
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Re: Problem 243

Post by chrisb555 » Tue May 05, 2009 12:01 am

Thanks Ed - that's very clear :-)

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Oliver1978
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Location: Erfurt, Germany

Re: Problem 243

Post by Oliver1978 » Mon Feb 23, 2015 1:04 am

So primes always give 1?
49.157.5694.1125

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Georg
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Re: Problem 243

Post by Georg » Mon Feb 23, 2015 1:17 am

Yes.

frodenburg
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Re: Problem 243

Post by frodenburg » Sun Nov 25, 2018 10:32 am

How come R(12) is the lowest resilient denominator below 40% and that there is ANOTHER lowest denominator below 16%?
What is meant with minimal and what should I be looking for? Is my english the issue?

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Animus
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Re: Problem 243

Post by Animus » Sun Nov 25, 2018 12:37 pm

You can find a lower resilent fraction for higher denominators, the question is how big the denominator must get in order to get the resilent fraction below the given treshold for the very first time. That's no contradiction.

PS: Your nickname sounds german. If english is an issue and german is fine, you can PM me "auf Deutsch".

frodenburg
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Re: Problem 243

Post by frodenburg » Tue Nov 27, 2018 2:45 pm

So in this case of lowest below 4/10 means that the 4 is fixed and the lowest x of D(x) with numerator 4 is to be found?
I'll try that and see.

frodenburg
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Re: Problem 243

Post by frodenburg » Tue Nov 27, 2018 2:47 pm

By the way, I am Dutch, so German does not really help :-(
Thanks for your explanation though, I give it a try to see if I understood.

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