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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 m-1 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?
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 m-1 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 non-trivial infinite decimal
expansions are those that divide a power of 10. Of the primes,
only 2 and 5 divide (a power of) 10
Odysseus

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 m-1 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 non-trivial 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 6-cycles rather than e.g. three 4-cycles?

--
Odysseus
Jens Kruse Andersen
science forum beginner

Joined: 23 Jul 2005
Posts: 40

Posted: Sat Jul 23, 2005 9:06 pm    Post subject: Re: Primes and Repeated Results in Decimal Digits of Fractions

eb1024@gmail.com wrote:

 Quote: For certain values of m the real result of n/m is an infinite result that repeats itself each m-1 digits.

See http://www.lrz-muenchen.de/~hr/numb/period.html

Carlos Rivera found 17 consecutive primes p with period length p-1,
starting at 2,245,849,783: http://www.primepuzzles.net/puzzles/puzz_095.htm

--
Jens Kruse Andersen
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:
 Quote: Carlos Rivera found 17 consecutive primes p with period length p-1, 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.
20 starts above 10^11.

--
Jens Kruse Andersen
Henry1

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
<odysseus1479-at@yahoo-dot.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^k-1 for enough k which divide m-1.

 Quote: For example, why do the thirteenths come in two 6-cycles rather than e.g. three 4-cycles? Because 13 does divide 999999 but not 9999.

Similarly 101 is the only prime producing 4-cycles, 37 three cycles,
11 two cycles, and 3 (perhaps also 2 and 5) one cycles.
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 p-1, 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
Odysseus

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 odysseus1479-at@yahoo-dot.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^k-1 for enough k which divide m-1.

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
Henry1

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
<odysseus1479-at@yahoo-dot.ca> wrote:

 Quote: Henry wrote: On Sat, 23 Jul 2005 22:47:31 GMT, Odysseus odysseus1479-at@yahoo-dot.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^k-1 for enough k which divide m-1. 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^(m-1) - 1 is divisible by m for m prime and not 2 or 5, and
m-1 divides m-1 (once) so yes, there is always such a k.
Odysseus

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: On Sun, 24 Jul 2005 20:38:00 GMT, Odysseus odysseus1479-at@yahoo-dot.ca> wrote: snip

 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^(m-1) - 1 is divisible by m for m prime and not 2 or 5, and m-1 divides m-1 (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
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^(m-1) - 1 is divisible by m for m prime and not 2 or 5, and m-1 divides m-1 (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

http://www.lrz-muenchen.de/~hr/numb/period.html

--
Jens Kruse Andersen
Odysseus

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^(m-1) - 1 is divisible by m for m prime and not 2 or 5, and m-1 divides m-1 (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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