Problems & Puzzles: Puzzles

Puzzle 405. R(38)

Perhaps you already know that R(38) has the following curio property:

This repunit has the unique property of having only unique primes (11, (10^19-1)/9 and (10^19+1)/11) as factors. Such a number could only occur for a prime where both (10^p-1)/9 and (10^p+1)/11 were prime. It is unlikely any prime except 19 has that property

Question: Find a prime p>19 such that both (10^p-1)/9 & (10^p+1)/11 are primes.



Contributions came from Shyam Sunder Gupta, Farideh Firoozbakht, J. K. Andersen.


Shyam wrote:

It is clear that (10^p-1)/9 is a  repunit. So for (10^p-1)/9 to be prime,it has to be prime is known that  for currently known possible prime repunits p= 2, 19, 23, 317, 1031, 49081, 86453, and 109297.For both (10^p-1)/9 & (10^p+1)/11 to be  primes, we need to check the primality of (10^p+1)/11 for p=2, 19, 23, 317, 1031, 49081, 86453, and 109297. On checking,it is found that  except for p=19, no other prime p exist upto p= 109297 for which both (10^p-1)/9 & (10^p+1)/11 are  primes.


Farideh wrote:

If there exists such prime, it's greater than 21277.


Andersen wrote:

The known p for which (10^p-1)/9 is a repunit prime or prp are:
p = 2, 19, 23, 317, 1031, 49081, 86453, 109297.
PrimeForm/GW says p = 19 is the only of these for which (10^p+1)/11 is also
prime. says Harvey Dubner
has tested all repunits for p from 86453 to 200000. All smaller p have
probably been tested earlier, so the puzzle has no solution below 200000. I
agree that there is probably no solution at all. gives known p for which
(10^p+1)/11 is prime: p = 5, 7, 19, 31, 53, 67, 293, 641, 2137, 3011. It has
been tested to at least 20000 before, and I have tested to 80000 with


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