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kwaldmanBest ResponseYou've already chosen the best response.0
LOL 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. ;)
 one year ago

estudierBest ResponseYou've already chosen the best response.0
http://en.wikipedia.org/wiki/Repunit
 one year ago

sauravshakyaBest ResponseYou've already chosen the best response.1
But I see no proof there.
 one year ago

estudierBest ResponseYou've already chosen the best response.0
Me 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).
 one year ago

benzillaBest ResponseYou've already chosen the best response.2
To 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
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.0
In[1]:= PrimeQ[1111111111111111111] Out[1]:= True
 one year ago

sauravshakyaBest ResponseYou've already chosen the best response.1
But that is still a very tough task @benzilla
 one year ago

CliffSedgeBest ResponseYou've already chosen the best response.0
Fermat's Little Theorem and proof by contradiction?
 one year ago

CliffSedgeBest ResponseYou've already chosen the best response.0
Seems unfeasible for a number so large, but if you have some hefty computing power . . .
 one year ago

benzillaBest ResponseYou've already chosen the best response.2
A 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
 one year ago

estudierBest ResponseYou've already chosen the best response.0
I guess the conclusion is that there is only proof by computer (via some algorithm).
 one year ago
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