mersenneforum.org Peculiar divisors of k.10^n-1
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 2022-06-23, 16:53 #1 bur     Aug 2020 79*6581e-4;3*2539e-3 601 Posts Peculiar divisors of k.10^n-1 I noticed that 3 * 10^272 - 1 / 13 = 230769...23076923. This periodic appearance in base 10 shows for all exponents where the number is divisible by 13, i.e. 3 * 10^32 - 1 / 13 = 23076923076923076923076923076923. It is also not restricted to k= 3, as 9 * 10^16 - 1 / 13 = 6923076923076923 or 12 * 10^13 - 1 / 13 = 923076923076923 or 35 * 10^16 - 1 / 13 = 26923076923076923 etc. It's not actually homework or an exercise, but if someone could give a hint instead of the full explanation, I'd be happy to try and figure the rest out myself.
2022-06-23, 20:58   #2
xilman
Bamboozled!

"๐บ๐๐ท๐ท๐ญ"
May 2003
Down not across

264768 Posts

Quote:
 Originally Posted by bur I noticed that 3 * 10^272 - 1 / 13 = 230769...23076923. This periodic appearance in base 10 shows for all exponents where the number is divisible by 13, i.e. 3 * 10^32 - 1 / 13 = 23076923076923076923076923076923. It is also not restricted to k= 3, as 9 * 10^16 - 1 / 13 = 6923076923076923 or 12 * 10^13 - 1 / 13 = 923076923076923 or 35 * 10^16 - 1 / 13 = 26923076923076923 etc. It's not actually homework or an exercise, but if someone could give a hint instead of the full explanation, I'd be happy to try and figure the rest out myself.
Hint: use the long division algorithm you (should have) learned at school and pay attention to the remainders.

 2022-06-24, 07:14 #3 bur     Aug 2020 79*6581e-4;3*2539e-3 11318 Posts Ah ok, so it's that simple... :D I was spending too much time analyzing the values of k Mod 13 and their order. Thanks.
2022-06-24, 07:54   #4
Batalov

"Serge"
Mar 2008
Phi(4,2^7658614+1)/2

270516 Posts

This is a good organoleptic entrance into the concept of unique primes. You have discovered it by experimentation. That's good.

You have just found that 1/13 is periodic with period of six. And so is 1/7, also periodic with period of six.
As a result both 7 and 13 are not unique primes.

Quote:
 Originally Posted by https://primes.utm.edu/top20/page.php?id=62 The reciprocal of every prime p (other than two and five) has a period, that is the decimal expansion of 1/p repeats in blocks of some set length (see the period of a decimal expansion). This is called the period of the prime p. Samuel Yates defined a unique prime (or unique period prime) to be a prime which has a period that it shares with no other prime. For example: 3, 11, 37, and 101 are the only primes with periods one, two, three, and four respectively--so they are unique primes. But 41 and 271 both have period five, 7 and 13 both have period six, 239 and 4649 both have period seven, and each of 353, 449, 641, 1409, and 69857 have period thirty-two, showing that these primes are not unique primes.

 2022-06-24, 12:30 #5 bur     Aug 2020 79*6581e-4;3*2539e-3 25916 Posts Thanks, actually, I saw your various unique primes at Caldwell's list before, so I knew that concept, but didn't connect it to this phenomenon. In hindsight this obviously occurs for every prime divisor of these numbers. I just didn't notice it because the period is much longer.
 2022-06-29, 10:19 #6 MattcAnderson     "Matthew Anderson" Dec 2010 Oregon, USA 49B16 Posts hi My project - prime constellations and k-tuples
2022-06-29, 18:11   #7
sweety439

"99(4^34019)99 palind"
Nov 2016
(P^81993)SZ base 36

2×52×73 Posts

Quote:
 Originally Posted by Batalov As a result both 7 and 13 are not unique primes.
7 and 13 are unique primes in base 2

7 is the only prime with period length 3 in base 2, and 13 is the only prime with period length 12 in base 2

see factorization of Phi(n,2)

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