mirror of
https://github.com/TheAlgorithms/Python.git
synced 2024-10-06 21:59:30 +00:00
c938e7311f
* Added solution for Project Euler problem 129. * Added doctest for solution() in project_euler/problem_129/sol1.py * Update formatting. Reference: #3256 * More descriptive function and variable names, more doctests.
58 lines
1.5 KiB
Python
58 lines
1.5 KiB
Python
"""
|
|
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() = }")
|