repunit prime factors So I am working on this problem... which states:

A number consisting entirely of ones is called a repunit. We shall
  define R(k) to be a repunit of length k; for example, R(6) = 111111.
Let us consider repunits of the form R(10^n).
Although R(10), R(100), or R(1000) are not divisible by 17, R(10000)
  is divisible by 17. Yet there is no value of n for which R(10^n) will
  divide by 19. In fact, it is remarkable that 11, 17, 41, and 73 are
  the only four primes below one-hundred that can be a factor of R(10^n).
Find the sum of all the primes below one-hundred thousand that will
  never be a factor of R(10^n).

My general strategy was to find the prime factorization of the highest repunit below 100,000^2 which I assumed if a factor was less than 100,000 it would become a factor of a repunit by 100,000 squared... which apparently isn't true, as that list is: (3,
7,
11,
13,
37,
41,
73,
101,
137,
239,
271,
4649,
9091)
and going as far as I can in 64 bit ints the list is: (3,
7,
11,
13,
17,
19,
31,
37,
41,
53,
73,
79,
101,
137,
239,
271,
4649,
9091,
9901,
21649,
52579)
So most interesting is the numbers are 17 and 19... which leads me to believe that I need to find a very different way around this problem... but I am not sure where to go next. 
PS I haven't unit tested my algorithms to verify that the lists are complete and correct... but they look right when checking against the factorizations listed on wikipedia.
 A: Not sure that your statement is right.
Did you mean $R(10^n)$ and not R(10n)
Note that
$$R(k)= \frac{10^k-1}{9}$$
So for $p \neq 3$, $p | R(k)$ if and only if $$10^k \equiv 1 \mod p$$
This says that for $p$ to divide $R(k)$, $p-1$ and $k$ must have some non-trivial common factor and $10$ to the power of this factor should be $1 \mod p$.
For example, if you set $p=19$ then $10 n$ and $18$ must have a nontrivial factor. 
Thus repunit with 180 ones is
$$9 \, R(180) = 10^{180}-1 = \left(10^{18}\right)^{10} - 1 \equiv 1^{10} -1 = 0 \mod 19$$
In fact every $p$ divides $R(10n)$ where $n = p-1$.
If what you meant was $R(10^n)$ then the above logic says that $p-1$ must have a common factor with $10^n$, i.e. $p-1$ must be divisible by $2$ or $5$ or both. Now argue that if $p-1$ is a power of 2, or power of 2 times a power of 5 then we can always find a $n$.
If in addition $p-1$ is a multiple of a power of $3$ then use the fact that $10\equiv 1 \mod 3$ to argue the case.
A: Here is a new proof using repunits  that there are infinitely many primes by me  with an interesting  new result on its divisors
Let p be a prime. > 7 and R(p) be the repunit  = (10^p - 1)/9.
If q is a prime divisor of R(p) . Then q is congruent to 1 mod p.
Or q is of the form kp + 1 for some k.
This can be established by the fact that q divides R(q-1)  by Fermat's little theorem and further  applying  Euclidean algorithm.
If q divides  111111 ... p times
also q divides 11111 ( q-1) times
then a divides   11111  ... r times
where r = GCD ( p, q-1) =1 or p
q divies 1  is absurd
hence GCD ( p, q-1) = p
p divides ( q-1)
q = kp +1
In other words all the divisors of R(p)  including  R(p) are of this type  where p is a prime
All the divisors of R(p) are congruent to 1 mod p.
R(p) = (2ap+1) (2bp+1) (2cp+1) ....
All the divisors of R(p) are > p
This  also   proves  a result   that for every prime p there exists at least one prime of the type kp+1.
This completes the proof.
