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kwaldman
 2 years ago
Best ResponseYou've already chosen the best response.0LOL divide it by every number up to that number. There's an faster way though if you don't wanna spend all your time do that. ;)

sauravshakya
 2 years ago
Best ResponseYou've already chosen the best response.1But I see no proof there.

estudier
 2 years ago
Best ResponseYou've already chosen the best response.0Me either, just a lot of commentary about how people spend a lot of computer time trying to show that some repunits are prime. If there is a prime number of digits (19 in this case) then it is a candidate (necessary but not sufficient condition).

benzilla
 2 years ago
Best ResponseYou've already chosen the best response.2To find a prime number you have to test all the prime roots up to the number that would be the square which would be just higher than the number tested. For example if we test 101, we would go up to 11. The square of 11 is 121 so if there are any higher roots there would also need to be corresponding lower roots. Still a daunting task because you will need to test up to 10 digits. Here is a little series of numbers which will help you. See if you see the pattern. 101 prime, 111…….3, 121…….7, 131……prime, 141…….3, 151…….prime, 161…….7, 171…….3, 181…….prime, 191…….prime, 201…….3, 211…….prime, 221…….13, 231…….3, 241…….prime, 251…….prime, 261…….3, 271…….prime, 281…….prime, 291…….3, 301…….7, 311……..prime, 321……..3, 331…….prime Cut and paste these and make them vertical. It is easier to see the pattern. These are the first primes from 101 to 331 that end in 1. Good luck

ParthKohli
 2 years ago
Best ResponseYou've already chosen the best response.0In[1]:= PrimeQ[1111111111111111111] Out[1]:= True

sauravshakya
 2 years ago
Best ResponseYou've already chosen the best response.1But that is still a very tough task @benzilla

CliffSedge
 2 years ago
Best ResponseYou've already chosen the best response.0Fermat's Little Theorem and proof by contradiction?

CliffSedge
 2 years ago
Best ResponseYou've already chosen the best response.0Seems unfeasible for a number so large, but if you have some hefty computing power . . .

benzilla
 2 years ago
Best ResponseYou've already chosen the best response.2A little more help: This chart shows you all the smallest prime roots of series of 1's up to 19. This will help but to prove you need to do a little research. It is not as much number crunching as you may think. 2) 11…prime 3) 111……..3 4) 1111……11 5) 11111…...41 6) 111111…..3 7) 1111111….239 8) 11111111…..11 9) 111111111….3 10) 1111111111…..11 11) 11111111111….21649 12) 111111111111…..3 13) 1111111111111…..53 14) 11111111111111….11 15) 111111111111111…3 16) 1111111111111111….11 17) 11111111111111111….2071723 18) 111111111111111111….3 19) 1111111111111111111…prime 20) 11111111111111111111.....11 21) 111111111111111111111...3

estudier
 2 years ago
Best ResponseYou've already chosen the best response.0I guess the conclusion is that there is only proof by computer (via some algorithm).
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