Problem #132 says:

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(10) = 1111111111 = 11×41×271×9091, and the sum of these prime factors is 9414.

Find the sum of the first forty prime factors of R(10^(9)).

To solve this problem we take the number and rearrange it a bit, then check for primes that fit the formula runs in about 12 seconds.

Solution in Python. Requires the RabinMiller.py file.

Problem132.py

from RabinMiller import *
from time import *

t0 = time()
def GeoSum (a,  n,  c):
    if (n == 1):
        return 1
    if (n==2):
        return (1+a)%c
    ret = ((GeoSum(a, n/2, c)%c)*((1+pow(a, n/2, c))%c))%c
    if (n&1):
        ret = (ret+pow(a, n-1, c))%c
    return ret

sum = 0
count = 0
i = 3
while (count <40):
    if (MillerRabin(i) and not GeoSum(10, 10**9,  i)):
        count = count + 1
        sum = sum + i
        print count, " divids by ", i
    i = i + 2
print sum
print time() - t0

RabinMiller.py

import sys
import random

def toBinary(n):
  r = []
  while (n > 0):
    r.append(n % 2)
    n = n / 2
  return r

def test(a, n):
  """
  test(a, n) -> bool Tests whether n is complex.

  Returns:
    - True, if n is complex.
    - False, if n is probably prime.
  """
  b = toBinary(n - 1)
  d = 1
  for i in xrange(len(b) - 1, -1, -1):
    x = d
    d = (d * d) % n
    if d == 1 and x != 1 and x != n - 1:
      return True # Complex
    if b[i] == 1:
      d = (d * a) % n
  if d != 1:
    return True # Complex
  return False # Prime

def MillerRabin(n, s = 3):
  """
    MillerRabin(n, s = 1000) -> bool Checks whether n is prime or not

    Returns:
      - True, if n is probably prime.
      - False, if n is complex.
  """
  for j in xrange(1, s + 1):
    a = random.randint(1, n - 1)
    if (test(a, n)):
      return False # n is complex
  return True # n is prime

Problem #188 says:

     The hyperexponentiation or tetration of a number
a by a positive integer b, denoted by a↑↑b or ^(b)a,
is recursively defined by:

a↑↑1 = a,
a↑↑(k+1) = a^((a↑↑k)).

Thus we have e.g. 3↑↑2 = 3^(3) = 27,
hence 3↑↑3 = 3^(27) = 7625597484987
and 3↑↑4 is roughly 10^(3.6383346400240996*10^12).

Find the last 8 digits of 1777↑↑1855.

This problem rather basic when you think about it mathematically. And even simpler once you realize that python has a built in modulo parameter to its pow() function. Making this whole problem rather simple. And with a bit of poking around one realizes that you don’t have to go through the whole loop as the value we’re looking for appears in the first 20 loops through the program.
Anyway, Solution is in python (2.6).

def Tetrate(Base,Tetrate,Modulo):
    Value = 1
    while Tetrate:
        Value = pow(Base,Value,10**Modulo)
        Tetrate = Tetrate - 1
    return Value

print "Answer : ", Tetrate(1777,1855,8)