Project Euler 132: Determining the first forty prime factors of a very large repunit

Problem Description

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(10⁹).

Analysis

Another quick solution using the modulus power function to find the prime factors of a large repunit, r(10⁹).

Project Euler 132 Solution

Afterthoughts

• Function is_prime is listed in Common Functions and Routines for Project Euler
  
• See also, Project Euler 132 Solution:
  
• Here are the first 40 prime factors: [11, 17, 41, 73, 101, 137, 251, 257, 271, 353, 401, 449, 641, 751, 1201, 1409, 1601, 3541, 4001, 4801, 5051, 9091, 10753, 15361, 16001, 19841, 21001, 21401, 24001, 25601, 27961, 37501, 40961, 43201, 60101, 62501, 69857, 76001, 76801, 160001]