Find a prime divisor of 1111 (13 1's)

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SUMMARY

The discussion centers on finding a prime divisor of the repunit number 1111111111111 (R_13), which consists of 13 ones. Participants confirm that the known prime divisors include 53 and 79, with additional larger primes also being factors. The method suggested for finding these divisors involves trial and error, testing each odd number up to the square root of the number. The mathematical property of repunits indicates that if a repunit R_n is divisible by a, then R_n is also divisible by R_a, although this does not simplify the search for R_13 due to its primality.

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mathmadx
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Hello friends,
The problem I am trying to solve sounds simple, but I still haven't been able to find the solution:
Find a prime divisor of 1111111111111 (13 ones), also known as a repunit.
I know the answer(53, 79 and some big prime), but I have no idea how Mathematica calculated those values.
Can anyone help me out? :)

I am afraid I should've posted this question in the number theory forums, sorry for that.
 
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Hi mathmadx! :smile:

As far as I know we're still stuck with trial and error to find prime divisors.
Try each odd number up to the square root and see if it's a divisor.
If you've found one, divide the number and repeat.
 
An interesting, but probably useless fact from http://en.wikipedia.org/wiki/Repunit" :
It is easy to show that if n is divisible by a, then R_n is divisible by R_a:
where R_i is the repunit with i ones. Unfortunately, you're dealing with R_{13} and 13 is prime...
 
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