Added solution for Project Euler problem 129. Fixes: #2695

Author: fpringle

project_euler/problem_129/__init__.py

@@ -0,0 +1,46 @@
+"""
+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 A(n: int) -> int:
+    """
+    Return the least value k such that the Repunit of length k is divisible by n.
+    >>> A(7)
+    6
+    >>> A(41)
+    5
+    >>> A(1234567)
+    34020
+    """
+    if n % 5 == 0 or n % 2 == 0:
+        return 0
+    R = 1
+    k = 1
+    while R:
+        R = (10 * R + 1) % n
+        k += 1
+    return k
+
+
+def solution(limit: int = 1000000) -> int:
+    """
+    Return the least value of n for which A(n) first exceeds limit.
+    """
+    n = limit - 1
+    while A(n) <= limit:
+        n += 2
+    return n
+
+
+if __name__ == "__main__":
+    print(solution())