20220623, 16:53  #1 
Aug 2020
79*6581e4;3*2539e3
7×83 Posts 
Peculiar divisors of k.10^n1
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. 
20220623, 20:58  #2  
Bamboozled!
"๐บ๐๐ท๐ท๐ญ"
May 2003
Down not across
11,423 Posts 
Quote:


20220624, 07:14  #3 
Aug 2020
79*6581e4;3*2539e3
7×83 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.

20220624, 07:54  #4  
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
9902_{10} 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:


20220624, 12:30  #5 
Aug 2020
79*6581e4;3*2539e3
7·83 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. 
20220629, 10:19  #6 
"Matthew Anderson"
Dec 2010
Oregon, USA
5·233 Posts 
hi
My project  prime constellations and ktuples

20220629, 18:11  #7 
"99(4^34019)99 palind"
Nov 2016
(P^81993)SZ base 36
DFA_{16} Posts 
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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