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 2003-06-15, 14:11 #1 epatka   Jun 2003 2016 Posts A strange idea...... Hi, I am new, and I am in no way a mathematician or programmer, that's why I need experts' opinion on this. I have been studying the prime numbers in general, and I have discovered something about them, what - I think - might make looking for primes easier. Because, as I said, I am no expert, I have no idea if what I am suggesting here is new at all. And, of course it is possible that using this method would be even more difficult, or impossible. Anyway.... I was playing with prime numbers, and discovered, that if I divide 1 by a prime number, than in many cases the reminders have a unique character. There are exceptions, and there are a few non-prime numbers that have this character, but in general, it might be a good screening process. Of course I wasn't able to test the very high numbers, but I am sure there are members who could do that for me. I'll make it short. When I divide the number 1 by a prime number, in most cases the repeating residue will be a number where starting the first and the very middle digits, they always add up to 9. To explain, here are some examples: 7/11 0.142857 142857 1+8=9 4+5=9 2+7=9 1/11 0.0909090909090909 0+9=9 1/13 0.076923 076923 0+9=9 7+2=9 9+3=9 etc. Larger numbers: 1/29 0.0344827586206896551724137931 0344827586206896551724137931 When I divide the repeating residue into 2 parts, it is easy to see that here, too it works. 03448275862068 96551724137931 0+9=9 3+6=9 4+5=9 etc. 1/61 0.016393442622950819672131147540983606557377149180327868852459 divided up it works: 016393442622950819672131147540 983606557377149180327868852459 There are exceptions, like 1/49 which works out like this, although it is not a prime, and there are primes that don't have this quality. But I thought, based on this, there could be a pre-screening of primes. What I had in mind was, to divide 1 by the number we want to test, and if there is a patterns like this in the repeating residue, the number is probably a prime. I don't know how difficult it is to divide such huge numbers, but if it can be done, it might be easier, than factoring. And once we have the result of the division, we would have to find the repeating residue (it is not always as long as the number itself), and in the middle of the repeating residue there should be a bunch of 9-s of course, because it would start with a bunch of 0-s. As I said, I don't know if this is feasible. or it might be already part of the screening process....... I am just an amateur, trying to be useful. Eva
 2003-06-15, 18:25 #2 epatka   Jun 2003 2016 Posts To explain a bit more what it looks like, I have posted a few on my website: http://www.newhorizonscience.com/prime1.gif http://www.newhorizonscience.com/prime2.gif http://www.newhorizonscience.com/prime3.gif http://www.newhorizonscience.com/prime4.gif Here you can see the residues compared to the ones of the composite numbers.
2003-06-17, 01:02   #3

"Richard B. Woods"
Aug 2002
Wisconsin USA

22·3·599 Posts
Re: A strange idea......

Quote:
 Originally Posted by epatka I am new
... so you have a good chance to present fresh insight,

Quote:
 and I am in no way a mathematician or programmer,
... so perhaps you are not so likely to miss seeing something that a mathematician or programmer might overlook.

Quote:
 that's why I need experts' opinion on this.
... but you'll have to do with this cheesellama's opinion, apparently.

Quote:
 I have no idea if what I am suggesting here is new at all.
Sometimes everything old is new again ... and "useful", "inspiring", and "insightful" are often more important than either "old" or "new".

Quote:
 And, of course it is possible that using this method would be even more difficult, or impossible.
That would not stop a real mathematician-programmer!

Reminds me of a story (in an inverse sort of way):

A (non-math) professor commented that mathematics was of little value. "Why, mathematicians even claim to work with imaginary numbers! How can anything be more useless than that?"

A math major student in the audience protested, "Hey! Imaginary numbers are just as useful as any other numbers!"

The professor called her to come down to the front next to the blackboard. "Show me", he smugly commanded, "the square root of minus one pieces of chalk", handing her a fresh stick of chalk.

The student frowned. The professor smiled and reached to take back the chalk. The student said, "I'll do it! I'll do it if you first show me one-half piece of chalk."

The professor's smile changed to a grin as he snapped the chalk in two and handed one of the halves to the student. "There you are. One-half piece of chalk."

As she held up the chalk piece, the student carefully explained, "No, this is one piece of chalk. You have one piece of chalk in your hand, and I have one piece of chalk in my hand. But I asked you for one-half piece of chalk, not one piece."

The student said, "See? The square root of minus one is just as real as one-half." and turned to resume her seat.

Quote:
 Anyway....
Quote:
 When I divide the number 1 by a prime number, in most cases the repeating residue will be a number where starting the first and the very middle digits, they always add up to 9. To explain, here are some examples: 7/11
(... 7/11 = 1/7 in base-48, of course ...)

Quote:
 0.142857 142857 1+8=9 4+5=9 2+7=9
Oooooookaaaaaaaaaay ... what we have here is a combination of things (which will also explain the rest of your examples, as we'll see).

First of all: Because of what I'm going to explain below, I doubt that what you've outlined here will lead to a new way of finding primes that is more efficient than other methods of finding primes. Note the qualifier! When we start playing with really, really big numbers (with millions and millions of digits, as GIMPS is doing), efficiency is not a trivial consideration -- it's essential.

Second: For any prime p, other than p=2 or p=5, when you compute its inverse 1/p in base-10, you will get a repeating pattern of decimal digits whose cycle length is p-1 or less.

Why?

Well, I don't have a link to a formal proof right now, but (10^(p-1) - 1) is divisible (and when we say divisible, we mean integer division with a zero remainder) by prime p, except for p=2 or p=5. That is, the number 9999...[total of p-1 nines]...999 is always divisible by p.

3 divides 99. (Yes, 3 also divides 9, but I didn't say it wouldn't.)

7 divides 999999.

11 divides 9999999999. (Also, 99, but I didn't say it wouldn't.)

13 divides 999999999999. (Also, 999999, but ...)

17 divides 9999999999999999.

So, this means that the decimal fraction 0.999...[total of p-1 nines]...999 will be "evenly" divisible by p with no remainder to go to more decimal digits.

0.9999999 / 7 = 0.142857 exactly, because 142,857 * 7 = 999,999. And 0.999999 999999 999999 / 7 = 0.142857 142857 142857 exactly, because 142,857,142,857,142,857 * 7 = 999,999,999,999,999,999.

So the infinitely repeating decimal fraction 0.(9)... (which equals 1.0000...), divided by 7 equals the infinitely repeating decimal fraction 0.(142857)...

Similarly, 0.9999999999999999 / 17 = 0.0588235294117647, because 588,235,294,117,647 * 17 = 9,999,999,999,999,999. And the infinitely repeating decimal fraction 0.(9)... (which equals 1.0000...), divided by 17 equals the infinitely repeating decimal fraction 0.(0588235294117647)...

Third: Why do some cycles have length less than p-1?

Let k be the smallest power of ten such than 10^k-1 is divisible by p (p not 2 or 5). The cycle length will be k. For some p, k = p-1. But for other p, k is a divisor of p-1.

For p = 11, k = 2, not 10. For p = 13, k = 6, not 12.

Fourth: Why do the digits in the cycle add to 9 pairwise first-half-to-second-half?

For instance:

Quote:
 29 0.0344827586206896551724137931 0344827586206896551724137931 When I divide the repeating residue into 2 parts, it is easy to see that here, too it works. 03448275862068 96551724137931 0+9=9 3+6=9 4+5=9 etc.
Here's why the sum of all the digits in the cycle is a multiple of 9: Let k be the smallest power of ten such than 10^k-1 is divisible by p. 10^k-1 is also divisible by 9. Since p is a prime, so is not a multiple of 9, (10^k-1)/p must be divisible by 9. That implies that the sum of the decimal digits of (10^k-1)/p must be a multiple of 9.

Why are the pairwise first-half-to-second-half digit sums all 9? Well, first note that for p > 3, k is even. (Proof: hand-wave. Since llamas don't have hands, someone else can provide this one. :) )

k is equal to either p-1 or some divisor of p-1. Hmmmm...

I seem to be drawing a blank about the rest of the explanation. Can someone else finish up?

 2003-06-17, 01:22 #4 Complex33     Aug 2002 Texas 5×31 Posts A little off topic, but I very much enjoyed the story of the uninformed professor. I am currently reading Innumeracy Mathematical Illiteracy and Its Consequences by John Allen Paulos. It is a little dated but I feel very representative of our culture. Personally I believe that lack of mathematical and scientific knowledge is right up there with illiteracy. Mathematics is leading innovation and without innovation we are moving backwards. That is why GIMPS is so great, people ask me all the time why I get so excited talking about the project and what is its use. I tell them that it is my contribution, however small, to discovering what is surely out there, plus unlike other projects every exponent tested is unique and vital to overall success. I would love to see every high school graduate have a feeling for the concept of Entropy. Ah we can dream can't we... Yes and a belated Happy Father's day to George father of our beloved GIMPS. Oh and I love this guy not that I want to take away anyone's cycles :)
 2003-06-17, 14:40 #5 apocalypse   Feb 2003 132 Posts Cheesehead - I threw together a quick script to test the assertion that k is even when p > 5, and got the following contradictory results for p < 100. [code:1] p k ----------- 31 15 37 3 41 5 43 21 53 13 67 33 71 35 79 13 83 41 [/code:1] Thanks Joe.
 2003-06-17, 14:55 #6 epatka   Jun 2003 25 Posts Re: A strange idea...... Thank you for responding. I was worried I said someting very dumb or broke some unwritten rules, because there was no reaction to my post. "7/11[/quote] (... 7/11 = 1/7 in base-48, of course ...)" Sorry, this was a mistake of mine. It was supposed to mean 1/7. Of course it doesn't change that you are right in this case. "First of all: Because of what I'm going to explain below, I doubt that what you've outlined here will lead to a new way of finding primes that is more efficient than other methods of finding primes. " I was afraid of this. But in my ignorance, I figured, that here we need one division only, and a division is nothing but substructions. Of course, in this case it woul mean 10 million substractions of 2 10 million digits long numbers, probably 10 million times.... But I am not so sure I am right in this. I have read that the problem really is a space problem. Maybe even entering such a large number can be a problem.... "Second: For any prime p, other than p=2 or p=5, when you compute its inverse 1/p in base-10, you will get a repeating pattern of decimal digits whose cycle length is p-1 or less. 3 divides 99. (Yes, 3 also divides 9, but I didn't say it wouldn't.) 7 divides 999999. 11 divides 9999999999. (Also, 99, but I didn't say it wouldn't.) 13 divides 999999999999. (Also, 999999, but ...) 17 divides 9999999999999999. Thanks for the explanation. This makes it more clear. " I noticed that when I do the division by hand, there is always a point, where the remainder is 1. And then the whole cycle starts all over. Question is, if this whole thing is any help in understanding or recognizing prime numbers??? At this point I have no idea. I might be just going in circles. Thanks anyways, and I am thankful for any further insight, explanation or help. ps. sorry, I coudn't figure out how to use the "quote" function.
2003-06-17, 14:56   #7
Joe O

Aug 2002

3·52·7 Posts

Quote:
 Originally Posted by apocalypse Cheesehead - I threw together a quick script to test the assertion that k is even when p > 5, and got the following contradictory results for p < 100. [code:1] p k ----------- 31 15 37 3 41 5 43 21 53 13 67 33 71 35 79 13 83 41 [/code:1] Hrm. I don't know how to make a cleaner table here.
Use {code} {/code} with [ instead of { and ] instead of }

Where did you see the assertion that k is even?
Quote:
 Originally Posted by Cheesehead k is equal to either p-1 or some divisor of p-1.
This is consistent with your results!

2003-06-17, 15:04   #8
apocalypse

Feb 2003

132 Posts

I was referring to the comment near the bottom of cheesehead's post, one line above the one you quoted.

Quote:
 Originally Posted by cheesehead Why are the pairwise first-half-to-second-half digit sums all 9? Well, first note that for p > 3, k is even. (Proof: hand-wave. Since llamas don't have hands, someone else can provide this one. )

2003-06-17, 15:54   #9

"Richard B. Woods"
Aug 2002
Wisconsin USA

22·3·599 Posts

apocalypse,

You are quite correct. k does not have to be even when p > 5.

I was tiring when I wrote that assertion, and got sloppy in my thinking (and it's hard enough keeping track of even/odd when one has no hands). Thank you for noticing and posting a correction!

Quote:
 Originally Posted by cheesehead Well, first note that for p > 3, k is even. (Proof: hand-wave.
Lesson for all: When someone asserts existence of a "hand-waving" proof (which may not be explicitly labelled as "hand-waving"), LOOK OUT! It lacks rigor, and might well be wrong. CHALLENGE IT!

2003-06-17, 16:00   #10
eepiccolo

Dec 2002
Frederick County, MD

2·5·37 Posts

Quote:
 Originally Posted by cheesehead Lesson for all: When someone asserts existence of a "hand-waving" proof (which may not be explicitly labelled as "hand-waving"), LOOK OUT! It lacks rigor, and might well be wrong. CHALLENGE IT!
Unless you're a physicist, who in general doesn't give a llama's lick about the rigor of a proof. :D ;)

 2003-06-17, 17:28 #11 epatka   Jun 2003 25 Posts In case what I discovered here is something new, and might be helpful for the prime number research, would someone help me to write it down in mathematician's language and present it to a math journal? Unfortunately, I am not familiar with math terms, on top of that, English is my second language. But it would be very nice to see my name in the journals :) Thank you Eva

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