Repeating Digits of FractionsDate: 04/28/99 at 07:39:22 From: Grabowski Subject: Long numbers Hi, I have been interested in working out fractions, and I would like to find patterns. When I divide one number by another I sometimes get a simple decimal tail, as with 1/8 = 0.125, where the number finishes or turns into an endless line of zeros. Others repeat (boringly as with 1/3). But then there are some more interesting repeats, where a group of numbers keeps repeating (e.g. 1/7 = 0.142857.) I am interested in finding longer repeating groups in these number tails, but my calculator doesn't let me see more than a few digits. What can I do to have divisions like 1/13 read more than just 0.076923076, where I can't tell if the 076923 is really a repeat group? I want to try to see what are the longer tail repetitions. I know that pi is infinite, but I have to have a better kind of calculator for this work because ordinary long division is no fun when going to such lengths. Does such a calculator exist? Also, I want to try to divide terms of the Fibonacci series and other interesting sequences (prime numbers?) to see what the repeating situation is with them. I am sure that buried in them there will be a rule to govern such repeating tails. If so, I will find it and it will be Grabowski's Rule. P. Grabowski Date: 04/28/99 at 13:28:26 From: Doctor Rob Subject: Re: Long numbers You have started on an odyssey that can lead you (eventually) to a fact known as Fermat's Little Theorem. It is a worthy journey, and you can learn much along the way. The length of the repeating part is the smallest number of 9's such that the denominator divides 9999...999000...0000 evenly. For example, 7 divides 999999 evenly, so the repeating part has length 6. Furthermore, the digits in the repeating part of 1/7 are the digits of the quotient 999999/7 = 142857. 1/7 = 142857/999999 = 142857/10^6 + 142857/10^12 + 142857/10^18 + ... = 0.142857 + 0.000000142857 + 0.000000000000142857 + ... You will recognize this as a geometric series. Since 13 is a divisor of 999999, with quotient 76923, you can safely predict that 1/13 = 0.076923 076923 076923 076923 076923 076923 076923 ... without needing a high-precision calculator. Next you probably want to know in advance what number 999...999000...000 is a multiple of some denominator N. This is done by factoring N as a product of prime powers, which can be done only one way. They fall into two classes: ones dividing 10 and ones not dividing 10 (the base of our system of numerals). The first consists of just {2,5}, and the second is all the other primes. The number of zeroes in the number in the first line of this paragraph is just the larger of the two exponents of 2 and 5. That's the easy part. And if there are no other primes, then that larger exponent E is such that N is a divisor of 10^E, and the decimal terminates after E decimal places. 1/N = Q/10^E, where Q = 10^E/N. The next step is to find for the rest of the primes p dividing into N evenly what smallest power of 10 has the property that 10^k - 1 is a multiple of p. That number k is called the order of 10 modulo p. The most helpful fact about k is that it is a divisor of p-1. Recall that for p = 7, we had k = 6, and, sure enough, 6 is a divisor of 7-1 = 6. This limits you to just a few possibilities. There is no easy way to tell which of these divisors of p-1 is actually k, you just have to try them. The next step is to find the highest power of p dividing 10^k - 1. We know that it is at least 1, but it may be more. For example, if p = 3, the divisors of 3-1 = 2 are 1 and 2. It turns out that k = 1, because 3 is a divisor of 10^1 - 1 = 9, but, in fact, 3^2 is also a divisor of 9 as well. Call that highest power p^e. Now if the power of p dividing N is p^n, and n <= e, then the period length of 1/p^n is k. If, on the other hand, n > e, then the period length of 1/p^n is k*p^(n-e). Now to get the length of the period of 1/N, take the least common multiple of the lengths of the periods of 1/p^n, for all primes other than 2 or 5 dividing N. Example: Find the length of the period of 1/9450. Start by factoring 9450 = 2 * 3^3 * 5^2 * 7. The number of zeroes will be the larger of 1 and 2 (the exponents of 2 and 5 in this factorization), namely 2. Now first consider 3^3 = p^n. The order of 10 modulo 3 is 1, and 3^2 is a divisor of 10^1 - 1 = 9, so e = 2. Since n = 3 > 2, the period length of 1/3^3 is 1*3^(3-2) = 3. Next consider 7^1 = p^n. The order of 10 modulo 7 is 6, as above, and 7^1 is the largest power of 7 dividing 10^6 - 1 = 999999 = 3^3 * 7 * 11 * 13 * 37, so e = 1. Thus the period length of 1/7 is 6. Taking the LCM of 3 and 6 gives 6. That tells me that the number of 9's should be 6, and 9450 must divide evenly into 99999900. Sure enough, it does, and the quotient is 10582. Thus 1/9450 = 10582/99999900 = 0.00 010582 010582 010582 010582... Notice the 2 digits to the right of the decimal before the repeating part begins, corresponding to the two zeroes at the end of the number 99999900, corresponding to the larger of the exponents of 2 and 5 in the factorization of 9450. Similarly, 4567/9450 = 48327994/99999900 = 0.48 328042 328042 328042... Example: Find the period length of 1/73. Since 73 is prime, we need to find the order of 10 modulo 73. It must be a divisor of 72, so it could be 72, 36, 24, 18, 12, 9, 8, 6, 4, 3, 2 or 1. It turns out to be 8, so k = 8, and 73 divides evenly into 99999999. The period length must be 8. The quotient is 1369863, so the decimal expansion must be 1/73 = 0.01369863 01369863 01369863 01369863... Get the idea? Proofs of the above facts are a part of a subject called Number Theory, which you can study as an advanced undergraduate at a college or university. - Doctor Rob, The Math Forum http://mathforum.org/dr.math/ Date: 05/21/99 at 15:41:35 From: S.J. Lean Subject: RE: Long numbers On 28 April 1999, Dr. Rob wrote me a good reply, except there is no information on the calculator (I mean, software calculator) I must find. For example I have already looked at sevenths, and find tail 142857 - six digits long. But what is tail length of 1/7 times 1/7, or 1/49? When I work out it is very interesting: 0.020408163265306122448979591836734693877551... THEN digits repeat. I noticed that this is 42 digits long! But 42 is obviously six times seven. Maybe it is only a coincidence, but maybe not. So I have an idea that one over a number (n = 7 in this example) would give a tail length of such-and-such, (6 in this example), but that the tail length of one over the same number SQUARED (n^2 = 49) would turn out to be the first number TIMES the first tail length? This is MY "little theory," but how can I check it out? I tried some examples and found it was true. But I can not easily do trial-and-error without a better calculator. And if I can't, how will I find more patterns for my clues? Do you know about software for this application? I want to work out some examples dozens of digits long (hundreds would be better). P. Grabowski Date: 05/25/99 at 09:55:32 From: Doctor Rob Subject: RE: Long numbers Thanks for writing back to Ask Dr. Math! The problem you are trying to deal with does not require high precision calculators to solve. I described the solution in a previous answer. If you insist on using such calculators, here is some information which may be helpful. Here is a website with links to lots of on-line calculators: Martindale's Reference Desk: Calculators On-Line Center http://www.martindalecenter.com/Calculators.html It sounds to me as if you want a software symbolic computation program such as MAPLE, MATHEMATICA, MATLAB, GP/PARI, or the like, which includes high-accuracy multiple precision arithmetic. Some of these are commercial products, which you have to buy. Others are freeware or shareware. Here is a Web page with links to software for math education that may be of some help: http://mathforum.org/mathed/math.software.archives.html You can use an ordinary 10-digit calculator to do the divisions you want to do, too. You want to find the decimal expansion of 1/N for some values of N less than, say, 10000000 = 10^7. Enter 1000000000 into your calculator. That is your first dividend, D1. Divide by N. The integer part of the quotient is the first part of the decimal expansion. Call that Q1. Then enter D1 and subtract Q1*N. That will give you the remainder R1. Now enter R1 followed by as many 0's as will fit in your calculator. That is your next dividend D2. Divide by N. The integer part of the quotient is the next part of the decimal expansion. Call that Q2. Then enter D2 and subtract Q2*N. That will give you the remainder R2. Continue this process as long as you like. Example: Find the decimal expansion of 1/127. N = 127. D1 = 1000000000. D1/N = 1000000000/127 = 7874015.748... so Q1 = 7874015, and then D1 - Q1*N = R1 = 95. D2 = 9500000000. D2/N = 9500000000/127 = 74803149.61... so Q2 = 74803149, and then D2 - Q2*N = R2 = 77. D3 = 7700000000. D3/N = 7700000000/127 = 60629921.25... so Q3 = 60629921, and then D3 - Q3*N = R3 = 33. D4 = 3300000000. D4/N = 3300000000/127 = 25984251.97... so Q4 = 25984251, and then D4 - Q4*N = R4 = 123. D5 = 1230000000. D5/N = 1230000000/127 = 9685039.370... so Q5 = 9685039, and then D5 - Q5*N = R5 = 47. D6 = 4700000000. D6/N = 4700000000/127 = 37007874.01... so Q6 = 37007874. The decimal expansion we have found so far is 1/127 = 0.007874015748031496062992125984251968503937 007874... and you can see the beginning of the second period. The length of the period is 42. This is because 10^42 - 1 is a multiple of 127, and no smaller power of 10 has that property. Note that 42 = (127-1)/3. 999999999999999999999999999999999999999999/127 = 7874015748031496062992125984251968503937 exactly. If N is larger, you get fewer decimal places in the expansion at each step, so you have to take more steps, but the result is just as valid. This will not work if N is larger than D1. - Doctor Rob, The Math Forum http://mathforum.org/dr.math/ Date: 06/05/99 at 09:02:25 From: Grabowski Subject: RE: Long numbers Dear Dr Rob, Thank you for suggesting that I use MAPLE for my hobby. Mr. Lean (I can use his computer) has found a MAPLE demo that I'm using to work out the tails very smoothly and easily. So thank you for that good tip. I would like to tell you about two new qualities for every number I have discovered. Maybe this can lead to a new way to find prime numbers (like the Sieve of Erastosthenes.) I call these qualities "U" and "F", for "unfriendly" and "friendly". Here's a chart for the first few integers: n U F n U F ---------------- ----------------- 1 1 0 17 16 0 2 1 1 18 1 1 3 1 0 19 18 0 4 1 2 20 1 2 5 1 1 21 6 0 6 1 1 22 2 1 7 6 0 23 22 0 8 1 3 24 1 3 9 1 0 25 1 2 10 1 1 26 7 0 11 2 0 27 3 0 12 1 2 28 6 2 13 6 0 29 28 0 14 6 1 30 1 1 15 1 1 31 30 0 16 1 4 32 1 5 From the chart you can see that the unfriendliest numbers are always prime numbers. Also, the friendliest numbers are powers of two. But how are U and F defined? First, take the reciprocal. Then: F is number of digits AFTER the decimal point and BEFORE the repeat happens (including zeros). U is simply the length of the repeating part. So: 1/n (decimal) n 1/n Before / Repeating part F U ------------------------------------------------------------ 1 1/1 1. / 0... 0 1 2 1/2* 0.5 / 0... 1 1 3 1/3* 0. / 3... 0 1 4 1/4 0.25 / 0... 2 1 5 1/5* 0.2 / 0... 1 1 6 1/6 0.1 / 6... 1 1 7 1/7* 0. / 142857... 0 6 8 1/8 0.125 / 0... 3 1 9 1/9 0. / 1... 0 1 10 1/10 0.1 / 0... 1 1 11 1/11* 0. / 09... 0 2 12 1/12 0.08 / 3... 2 1 13 1/13* 0. / 076923... 0 6 14 1/14 0.0 / 714285... 1 6 15 1/15 0.0 / 6... 1 1 16 1/16 0.0625 / 0... 4 1 17 1/17* 0. / 0588235294117647... 0 16 18 1/18 0.0 / 5... 1 1 19 1/19* 0. / 052631578947368421... 0 18 20 1/20 0.05 / 0... 2 1 21 1/21 0. / 047619... 0 6 22 1/22 0.0 / 45... 1 2 23 1/23* 0. / 0434782608695652173913... 0 22 24 1/24 0.041 / 6... 3 1 25 1/25 0.04 / 0... 2 1 26 1/26 0. / 0384615... 0 7 27 1/27 0. / 037... 0 3 28 1/28 0.03 / 571428... 2 6 29 1/29* 0. / 0344827586206896551724137931... 0 28 30 1/30 0.0 / 3... 1 1 31 1/31* 0. / 032258064516129032258064516129... 0 30 32 1/32* 0.03125 / 0... 5 1 60 1/60 0.01 / 6... 2 1 64 1/64 0.015625/ 0... 6 1 360 1/360 0.002 / 7... 3 1 Now you can see a good rule if you look only at the primes. I have discovered that: 1) Except for two and five all primes have zero friendliness. This is very good for a first check. (I think 2 and 5 are different because 2*5 = 10, and that is the base of our number system. I need to check it in a base seven, for instance, to see if this is true.) 2) Tails can only grow to a maximum length equal to a prime number minus one. If they do, I call them Grabowski Primes - marked with G. Examples: 17, 61, 97, etc. (Only prime numbers have this quality.) 3) Tails sometimes grow to that number minus one DIVIDED BY TWO. If they do, I call them Half-Grabowski Primes. (Marked with H. Examples: 13, 43, etc.) 4) Other tails can grow to a maximum of that number minus one DIVIDED BY FOUR. If they do I, call them Quarter-Grabowski Primes. (Marked with "Q". Examples, 5, 53, 173, etc.) And so on for 8, 16, 32 ... this gives an ORDER of Grabowski prime. BUT ... 5) There is a type of prime that is not accounted for in the Grabowski prime series, which has this following characteristic: It is a Grabowski prime PLUS A REMAINDER that is ALWAYS itself a prime number or a product of primes. Examples: 11, 37, 79 Here is a chart of primes up to 200: n 1/n 1/n (decimal) F U Type -------------------------------------------------------------- 2 1/2 0.5 / 0... 1 1 G 3 1/3 0. / 3... 0 1 H 5 1/5 0.2 / 0... 1 1 Q 7 1/7 0. / 142857... 0 6 H 11 1/11 0. / 09... 0 2 *H'5 13 1/13 0. / 076923... 0 6 H 17 1/17 0. / 0588235294117647... 0 16 G 19 1/19 0. / 052631578947368421... 0 18 G 23 1/23 0. / 0434782608695652173913... 0 22 G 29 1/29 0. / 0344827586206896551724137931... 0 28 G 31 1/31 0. / 032258064516129032258064516129... 0 30 G 37 1/37 0. / 027... 0 3 *Q'9=3*3 41 1/41 0. / 02439.. 0 5 8 43 1/43 0. / 023255813953488372093.. 0 21 H 47 1/47 0. too boring to write in here 0 46 G 53 1/53 0. . 0 13 Q 59 1/59 0. . 0 58 G 61 1/61 0. . 0 60 G 67 1/67 0. . 0 33 H 71 1/71 0. . 0 35 H 73 1/73 0. . 0 8 79 1/79 0. . 0 13 *H'39=3*13 83 1/83 0. . 0 41 H 89 1/89 0. . 0 43 H 97 1/97 0. . 0 96 G 101 1/101 0. . 0 4 *Q'25=5*5 103 1/103 0. . 0 34 *H'51=3*17 107 1/107 0. . 0 53 H 109 1/109 0. . 0 108 G 113 1/113 0. . 0 112 G 127 1/127 0. . 0 42 *H'63=3*3*7 131 1/131 0. . 0 130 G 137 1/137 0. . 0 8 8'17 139 1/139 0. . 0 46 H'69=3*23 149 1/149 0. . 0 148 G 151 1/151 0. . 0 75 H 157 1/157 0. . 0 78 H 163 1/163 0. . 0 81 H 167 1/167 0. . 0 166 G 173 1/173 0. . 0 43 Q 179 1/179 0. . 0 178 G 181 1/181 0. . 0 180 G 191 1/191 0. . 0 95 H 193 1/193 0. . 0 192 G 197 1/197 0. . 0 98 H 199 1/199 0. . 0 99 H I broke the chart down into separate sequences to show how they climb up each time, from lower Grabowski orders (sometimes non-Grabowski) up to full Grabowski, then start again. This seems to be typical behavior. The non-Grabowski primes (I call them "lean primes") have least "unfriendliness" and even within this type, there are likely to be some different classifications. Such as: those that come from a single prime (11 and 137, etc.) as opposed to those that come from a product of primes. like number tail length type remainder ------------------------------------------- 11 2 H 5 37 3 Q 9=3*3 79 13 H 39=3*13 101 4 Q 25=5*5 103 34 H 51=3*17 127 42 H 63=3*3*7 137 8 8 17 139 46 H 69=3*23 My question is: Can my system be recognized? It is my conjecture that examination of reciprocal tails and classification can be used as a way to work out primes. Thank you for your help, Dr Rob. Grabowski Date: 06/07/99 at 11:46:52 From: Doctor Rob Subject: Re: Long numbers >1) Except for two and five all primes have zero friendliness. >This is very good for a first check. (I think 2 and 5 are >different because 2*5 = 10, and that is the base of our number >system. I need to check it in a base seven, for instance, to >see if this is true.) The friendliness number F is just the larger of the exponents of 2 and 5 in the prime factorization of the denominator n. n = 25 = 5^2 => F = 2. n = 64 = 2^6 => F = 6. n = 200 = 2^3*5^2 => F = max(3,5) = 5. >2) Tails can only grow to a maximum length equal to a prime >number minus one. If they do, I call them Grabowski Primes - >marked with G. Examples: 17, 61, 97, etc. (Only prime numbers >have this quality.) These are the primes for which 10 is a primitive root. Fermat's Little Theorem says that if p doesn't divide a, then a^(p-1) = 1 (mod p). The smallest exponent k such that a^k = 1 (mod p) is called the order of a (mod p), and is written ord_p(a). If that order is p - 1, then a is called a primitive root (mod p). A corollary says that ord_p(a) must always be a divisor of p - 1. >3) Tails sometimes grow to that number minus one DIVIDED BY >TWO. If they do, I call them Half-Grabowski Primes. (Marked >with H. Examples: 13, 43, etc.) These are the primes for which the (p-1)/ord_p(10) = 2. >4) Other tails can grow to a maximum of that number minus one >DIVIDED BY FOUR. If they do I, call them Quater-Grabowski >Primes. (Marked with "Q". Examples, 5, 53, 173, etc.) > >And so on for 8, 16, 32 ... this gives an ORDER of Grabowski >prime. BUT ... These are the primes for which the (p-1)/ord_p(10) = 4. >5) There is a type of prime that is not accounted for in the >Grabowski prime series, which has this following >characteristic: It is a Grabowski prime PLUS A REMAINDER that >is ALWAYS itself a prime number or a product of primes. >Examples: 11, 37, 79 These are primes for which (p-1)/ord_p(10) is not a power of 2. >My question is: Can my system be recognized? It is my >conjecture that examination of reciprocal tails and >classification can be used as a way to work out primes. There is very little known about how to determine ord_p(10) other than trying divisors of p - 1 and doing the arithmetic to see which one is the right one. There are efficient ways to do that, but no theory of which I am aware to predict what will happen without computing. Keep up your good questions! - Doctor Rob, The Math Forum http://mathforum.org/dr.math/ |
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