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#1 |
Aug 2020
79*6581e-4;3*2539e-3
23×73 Posts |
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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. |
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#2 | |
Bamboozled!
"๐บ๐๐ท๐ท๐ญ"
May 2003
Down not across
262438 Posts |
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#3 |
Aug 2020
79*6581e-4;3*2539e-3
58410 Posts |
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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.
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#4 | |
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
100110101100012 Posts |
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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:
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#5 |
Aug 2020
79*6581e-4;3*2539e-3
58410 Posts |
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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. |
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#6 |
"Matthew Anderson"
Dec 2010
Oregon, USA
5×233 Posts |
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My project - prime constellations and k-tuples
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#7 |
"99(4^34019)99 palind"
Nov 2016
(P^81993)SZ base 36
72×73 Posts |
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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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