Added solution for Project Euler problem 129. (TheAlgorithms#3113)

* Added solution for Project Euler problem 129. 
* Added doctest for solution() in project_euler/problem_129/sol1.py
* Update formatting. Reference: TheAlgorithms#3256 
* More descriptive function and variable names, more doctests.

Author: fpringle

project_euler/problem_129/__init__.py

@@ -0,0 +1,57 @@
+"""
+Project Euler Problem 129: https://projecteuler.net/problem=129
+
+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.
+
+Given that n is a positive integer and GCD(n, 10) = 1, it can be shown that there
+always exists a value, k, for which R(k) is divisible by n, and let A(n) be the least
+such value of k; for example, A(7) = 6 and A(41) = 5.
+
+The least value of n for which A(n) first exceeds ten is 17.
+
+Find the least value of n for which A(n) first exceeds one-million.
+"""
+
+
+def least_divisible_repunit(divisor: int) -> int:
+    """
+    Return the least value k such that the Repunit of length k is divisible by divisor.
+    >>> least_divisible_repunit(7)
+    6
+    >>> least_divisible_repunit(41)
+    5
+    >>> least_divisible_repunit(1234567)
+    34020
+    """
+    if divisor % 5 == 0 or divisor % 2 == 0:
+        return 0
+    repunit = 1
+    repunit_index = 1
+    while repunit:
+        repunit = (10 * repunit + 1) % divisor
+        repunit_index += 1
+    return repunit_index
+
+
+def solution(limit: int = 1000000) -> int:
+    """
+    Return the least value of n for which least_divisible_repunit(n)
+    first exceeds limit.
+    >>> solution(10)
+    17
+    >>> solution(100)
+    109
+    >>> solution(1000)
+    1017
+    """
+    divisor = limit - 1
+    if divisor % 2 == 0:
+        divisor += 1
+    while least_divisible_repunit(divisor) <= limit:
+        divisor += 2
+    return divisor
+
+
+if __name__ == "__main__":
+    print(f"{solution() = }")