The statement of this problem can be found here.

In order to solve this problem the first important thing to notice is how a repunit can be represented: \[ R(k) = \frac{10^k - 1}{9} \]

Therefore, we can express if a repunit is divisible by p like:

\[  \begin{aligned} \\ \frac{10^k - 1}{9} & \equiv & 0 \mod p \\ {10^k - 1} & \equiv & 0 \mod 9p \\ {10^k} & \equiv & 1 \mod 9p \\ \end{aligned} \]

So if \(10^k \mod 9p = 1 \), then p divides \( R(10^k) \).
The problem now is how to calculate the remainder in an efficient way as it is impossible to calculate the remainder to a number of a thousand million digits.
Here we can use Modular exponentiation as what we need to calculate is the remainder of a number than can be expressed as a power with base 10 and exponent 9.

The solution for this code in Python (problem132.py) is really simple (the CommonFunctions file can be found here):

from CommonFunctions import *
from itertools import *

if __name__ == '__main__':
    primes = find_primes_less_than(10 ** 6)
    base = 10
    exp = 10 ** 9
    result = sum(islice((p for p in primes if mod_pow(base, exp, 9 * p) == 1), 0, 40))
    print("The result is:", result)