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eb1024@gmail.com science forum beginner
Joined: 22 Jul 2005
Posts: 4

Posted: Fri Jul 22, 2005 10:23 pm Post subject:
Primes and Repeated Results in Decimal Digits of Fractions



Let n be an integer (for this case we'll consider the number 1) and m a
prime number.
For certain values of m the real result of n/m is an infinite result
that repeats itself each m1 digits.
examples:
1/7 = 0, 142857 142857 142857 142857 142857
1/13 = 0, 076923076923 076923076923
1/17 = 0, 0588235294117647 0588235294117647
Some prime numbers like 2, 3, 5 and 11 don't follow this rule, what I
want to know is which values of m obbey to this and why...
Does anyone knows? 

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William Elliot science forum Guru
Joined: 24 Mar 2005
Posts: 1906

Posted: Sat Jul 23, 2005 12:42 am Post subject:
Re: Primes and Repeated Results in Decimal Digits of Fractions



On Fri, 22 Jul 2005, eb1024@gmail.com wrote:
Quote:  Let n be an integer (for this case we'll consider the number 1) and m a
prime number.
For certain values of m the real result of n/m is an infinite result
that repeats itself each m1 digits.
examples:
1/7 = 0, 142857 142857 142857 142857 142857
1/13 = 0, 076923076923 076923076923
1/17 = 0, 0588235294117647 0588235294117647
Some prime numbers like 2, 3, 5 and 11 don't follow this rule, what I
want to know is which values of m obbey to this and why...
No. 1/3 = 0.3 3 3 3 3... 
If 10^j / n = k, then n  10^j.
So the only numbers that don't have nontrivial infinite decimal
expansions are those that divide a power of 10. Of the primes,
only 2 and 5 divide (a power of) 10 

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Odysseus science forum addict
Joined: 05 Jun 2005
Posts: 60

Posted: Sat Jul 23, 2005 8:47 pm Post subject:
Re: Primes and Repeated Results in Decimal Digits of Fractions



William Elliot wrote:
Quote: 
On Fri, 22 Jul 2005, eb1024@gmail.com wrote:
Let n be an integer (for this case we'll consider the number 1) and m a
prime number.
For certain values of m the real result of n/m is an infinite result
that repeats itself each m1 digits.
examples:
1/7 = 0, 142857 142857 142857 142857 142857
1/13 = 0, 076923076923 076923076923
1/17 = 0, 0588235294117647 0588235294117647
Some prime numbers like 2, 3, 5 and 11 don't follow this rule, what I
want to know is which values of m obbey to this and why...
No. 1/3 = 0.3 3 3 3 3...

And 1/11 = 0.09 09 09 09 ...
Quote:  If 10^j / n = k, then n  10^j.
So the only numbers that don't have nontrivial infinite decimal
expansions are those that divide a power of 10. Of the primes,
only 2 and 5 divide (a power of) 10

What I find interesting is the periods of the others. For example,
why do the sevenths and thirteenths both have six digits in their
periodic cycles? Is there any _a prioi_ way to predict them, beyond
that for proper fractions of the form n/m the product of the number
of cycles and their length must equal m  1 ? For example, why do the
thirteenths come in two 6cycles rather than e.g. three 4cycles?

Odysseus 

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Jens Kruse Andersen science forum beginner
Joined: 23 Jul 2005
Posts: 40


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Jens Kruse Andersen science forum beginner
Joined: 23 Jul 2005
Posts: 40

Posted: Sun Jul 24, 2005 12:35 am Post subject:
Re: Primes and Repeated Results in Decimal Digits of Fractions



I wrote:
The first case of 18 starts at 13,147,541,941.
19 starts at 23,053,991,353.
20 starts above 10^11.

Jens Kruse Andersen 

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Henry1 science forum addict
Joined: 15 May 2005
Posts: 58

Posted: Sun Jul 24, 2005 2:16 am Post subject:
Re: Primes and Repeated Results in Decimal Digits of Fractions



On Sat, 23 Jul 2005 22:47:31 GMT, Odysseus
<odysseus1479at@yahoodot.ca> wrote:
Quote:  What I find interesting is the periods of the others. For example,
why do the sevenths and thirteenths both have six digits in their
periodic cycles?
Because 999999=3.3.3.7.11.13.37 while 99999=3.3.41.271, 
9999=3.3.11.101, 999=3.3.3.37, 99=3.3.11, and 9=3.3
Quote:  Is there any _a prioi_ way to predict them, beyond
that for proper fractions of the form n/m the product of the number
of cycles and their length must equal m  1 ?
Only by knowing the factors of 10^k1 for enough k which divide m1. 
Quote:  For example, why do the
thirteenths come in two 6cycles rather than e.g. three 4cycles?
Because 13 does divide 999999 but not 9999. 
Similarly 101 is the only prime producing 4cycles, 37 three cycles,
11 two cycles, and 3 (perhaps also 2 and 5) one cycles. 

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Jens Kruse Andersen science forum beginner
Joined: 23 Jul 2005
Posts: 40

Posted: Sun Jul 24, 2005 10:01 am Post subject:
Re: Primes and Repeated Results in Decimal Digits of Fractions



I wrote:
Quote:  Carlos Rivera found 17 consecutive primes p with period length p1,
starting at 2,245,849,783: http://www.primepuzzles.net/puzzles/puzz_095.htm
The first case of 18 starts at 13,147,541,941.
19 starts at 23,053,991,353.

The first case of at least 20 starts at 239,651,440,411.
It has no less than 23 consecutive primes:
239,651,440,000 + 411, 447, 451, 487, 531, 543, 577, 579, 607, 657,
697, 703, 709, 727, 781, 783, 789, 811, 901, 909, 913, 937, 949.
This is the only case of at least 22 primes below 10^12. I stopped there.

Jens Kruse Andersen 

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Odysseus science forum addict
Joined: 05 Jun 2005
Posts: 60

Posted: Sun Jul 24, 2005 6:38 pm Post subject:
Re: Primes and Repeated Results in Decimal Digits of Fractions



Henry wrote:
Quote: 
On Sat, 23 Jul 2005 22:47:31 GMT, Odysseus
odysseus1479at@yahoodot.ca> wrote:
What I find interesting is the periods of the others. For example,
why do the sevenths and thirteenths both have six digits in their
periodic cycles?
Because 999999=3.3.3.7.11.13.37 while 99999=3.3.41.271,
9999=3.3.11.101, 999=3.3.3.37, 99=3.3.11, and 9=3.3
Is there any _a prioi_ way to predict them, beyond
that for proper fractions of the form n/m the product of the number
of cycles and their length must equal m  1 ?
Only by knowing the factors of 10^k1 for enough k which divide m1.

Thanks; that makes sense. Does it then follow that for every prime m
(other than 2 & 5) there exists a divisor k of m  1, such that 10^k
 1 is divisible by m ?

Odysseus 

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Henry1 science forum addict
Joined: 15 May 2005
Posts: 58

Posted: Sun Jul 24, 2005 10:52 pm Post subject:
Re: Primes and Repeated Results in Decimal Digits of Fractions



On Sun, 24 Jul 2005 20:38:00 GMT, Odysseus
<odysseus1479at@yahoodot.ca> wrote:
Quote:  Henry wrote:
On Sat, 23 Jul 2005 22:47:31 GMT, Odysseus
odysseus1479at@yahoodot.ca> wrote:
What I find interesting is the periods of the others. For example,
why do the sevenths and thirteenths both have six digits in their
periodic cycles?
Because 999999=3.3.3.7.11.13.37 while 99999=3.3.41.271,
9999=3.3.11.101, 999=3.3.3.37, 99=3.3.11, and 9=3.3
Is there any _a prioi_ way to predict them, beyond
that for proper fractions of the form n/m the product of the number
of cycles and their length must equal m  1 ?
Only by knowing the factors of 10^k1 for enough k which divide m1.
Thanks; that makes sense. Does it then follow that for every prime m
(other than 2 & 5) there exists a divisor k of m  1, such that
10^k  1 is divisible by m ?

Well 10^(m1)  1 is divisible by m for m prime and not 2 or 5, and
m1 divides m1 (once) so yes, there is always such a k. 

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Odysseus science forum addict
Joined: 05 Jun 2005
Posts: 60

Posted: Mon Jul 25, 2005 10:06 am Post subject:
Re: Primes and Repeated Results in Decimal Digits of Fractions



Henry wrote:
Quote:  [...] Does it then follow that for every prime m
(other than 2 & 5) there exists a divisor k of m  1, such that
10^k  1 is divisible by m ?
Well 10^(m1)  1 is divisible by m for m prime and not 2 or 5, and
m1 divides m1 (once) so yes, there is always such a k.

How do you arrive at the first part of that? Sorry if this should be
obvious, but I'm pretty slow on the uptake when it comes to number
theory. I've got a couple of basic texts; I'll read a chapter
thinking I'm understanding it, then find 90% of the exercises at the
end completely baffling.

Odysseus 

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Jens Kruse Andersen science forum beginner
Joined: 23 Jul 2005
Posts: 40

Posted: Mon Jul 25, 2005 1:29 pm Post subject:
Re: Primes and Repeated Results in Decimal Digits of Fractions



Odysseus wrote:
Quote:  Henry wrote:
Well 10^(m1)  1 is divisible by m for m prime and not 2 or 5, and
m1 divides m1 (once) so yes, there is always such a k.
How do you arrive at the first part of that?

This is Fermat's little Theorem: http://primes.utm.edu/prove/prove2_2.html
For more information about periods, see the link I gave:
http://www.lrzmuenchen.de/~hr/numb/period.html

Jens Kruse Andersen 

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Odysseus science forum addict
Joined: 05 Jun 2005
Posts: 60

Posted: Tue Jul 26, 2005 6:45 am Post subject:
Re: Primes and Repeated Results in Decimal Digits of Fractions



Jens Kruse Andersen wrote:
Quote: 
Odysseus wrote:
Henry wrote:
Well 10^(m1)  1 is divisible by m for m prime and not 2 or 5, and
m1 divides m1 (once) so yes, there is always such a k.
How do you arrive at the first part of that?
This is Fermat's little Theorem: http://primes.utm.edu/prove/prove2_2.html
That's a very interesting site: lots there to chew on! 
Thanks.

Odysseus 

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