Problems & Puzzles: Puzzles Puzzle 12.- Period Length of 1/p The decimal expansion of the reciprocal of any prime p (except p=2 and p=5) is an endless sequence of numbers that shows a defined "period". By example : 1/7= .142857142857142857….. For the purpose of saving space, this situation is expressed simply as 1/7 = .142857 and the "period" of 1/7 is defined as the quantity of digits that are "periodically" repeated, in this case 6. So for p=7, we can state that "the period of 1/7 is 6", or in a more algebraic manner : "period (1/7) = 6" … G. L. Honaker, Jr. has proposed several questions concerning
prime period lengths : a) "Find the largest known pal-prime having a palindromic period".
The current record is p = 363818363. McCranie has shown that period(1/ 363818363) = 181909181 Can another be found ?
b)"Find the first prime number p whose reciprocal (1/p) has period 666". Very recently Jud McCranie has found that 902659997773 is the smallest prime whose reciprocal has period is 666. Can another be found ? c) Honaker observes that : 6 + 1 = prime has pal-period 6 [Honaker] 66 + 1 = prime has pal-period 33 [Honaker] 606 + 1 = prime has pal-period 202 [Honaker] 6006 + 1 = prime has pal-period 858 [Honaker] 60606 + 1 = prime has pal-period 60606 [Honaker] 606606 + 1 = prime has pal-period 606606 [Honaker] 60666606 + 1 = prime has pal-period 60666606 [McCranie] and asks for the period of p = 6066666606 + 1, etc. Solution part b Warut Roonguthai solves (5-7/July/98) in a magnificent way this problem !… let’s see it : 1) "The primes whose reciprocals have period length = 666 are those that are primitive factors of 10^666-1. A primitive factor of 10^666-1 is a factor that does not divide any other 10^n-1, n<666. I don't have the latest information on the factorization of 10^666-1, but I know another primitive factor, namely: 7116181512528105949933. The first 666 decimal places of its reciprocal is 000000000000000000000140524802274856310017839523742174848861 2) In response to my following comment : "Thanks for this record. I'll add it next time I update my pages. Any comment about the almost-central collection of "9's" already observed by Honaker in the before record?", Warut answered this : "Let A and B be the first and last 333 digits of that number, respectively. Have you noticed that A+B = 10^333-1? Here is why: A*10^333+B = (10^666-1)/p = (10^333-1)(10^333+1)/p Since p is a primitive factor, p does not divide 10^333-1 (333<666) and must divide 10^333+1. So A*10^333+B = 0 (mod 10^333-1). A*10^333+B = A*(10^333-1)+A+B = A+B = 0 (mod 10^333-1) We can infer that A+B = 10^333-1 because 0 < A,B < 10^333-1. Since the leading digits of A are 0's, the leading digits of B must be 9's." 3) Finally Warut solved completely this puzzle. In his next e-mail he says : " There are only 4 primes with period 666 because the complete factorization of the primitive part of 10^666-1 is 902659997773*7116181512528105949933*62562101977662753776437 *P160, where P160 is the 160-digit prime: 249087269190813757516646608829135564382770107989197857827483 The first factor was rediscovered by Jud McCranie. The second factor was mentioned in my previous letter. I obtained the last two factors from http://www.matematik.su.se/~tege/gmp/repunit.html The first 666 decimal places of 1/62562101977662753776437 is 000000000000000000000015984117674899113398885267968238452812 The first 666 decimal places of 1/P160 is 000000000000000000000000000000000000000000000000000000000000 Solution part c Jud McCranie solves (7-8/July/98) the third question of this puzzle : "On the web page it says (puzzle 12 part c) "...and asks for the period of p=6066666606+1, etc". I calculated the periods of more primes of that form (and)… the pattern doesn't continue… : p=6066666606+1 (six 6s in the middle) is prime and 1/p has period 6066666606 (Warut Roongutai answered this question too) p=606666666606+1 (eight 6s in the middle) is prime and 1/p has period 15,555,555,554 (which is (p-1)/39) p=606666666666666666666666606+1 (twenty-three 6s in the middle) is prime and 1/p as period 86,666,666,666,666,666,666,666,658 (which is (p-1)/7) p=6066666666666666666666666666666666666606+1 (thirty-six 6s in the middle) is prime and 1/p has period 866,666,666,666,666,666,666,666,666,666,666,666,658 (which is (p-1)/7)" *** Solution part a Finally a new solution for this part came - and a real large and complex one! - by J. K. Andersen. He wrote, the good 22/01/03 day: www.lrz-muenchen.de/~hr/numb/period.html has more details. If q is prime and q==a (mod 40) for some a in {-1,1,-3,3,-9,9,-13,13} then the period length of 1/q is (q-1)/K for some even K. In our case, where q==3 (mod 40), the only possibility is K=2, so the period length is p. Strategy to find the solution: Let p be a palindrome with an odd number of digits, where all digits in odd places are <=4 and digits in even places are >=5. Then q=2*p+1 is also a palindrome. This is ensured by carry's, as can best be seen from an example: p=16150605161 q=32301210323 Generate p's of the right form (with right modulo 40 value) and test if both p and 2*p+1 are primes. They are in the example, giving 1/q period length p. I wrote a C program for efficient modular trial factoring of "the added palindrome" in my solution form. Candidates were then prp tested with PrimeForm/GW. Finding a solution with probable primes took around 1 hour. The smaller prp p in the solution was proved with Primo in around 50 minutes. q=2*p+1 could then be proved in 0.25 second with PrimeForm, using the proven prime factor p of q-1. *** |
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