Python/project_euler/problem_129/sol1.py
fpringle c938e7311f
Added solution for Project Euler problem 129. ()
* Added solution for Project Euler problem 129. 
* Added doctest for solution() in project_euler/problem_129/sol1.py
* Update formatting. Reference:  
* More descriptive function and variable names, more doctests.
2020-11-21 08:22:26 +05:30

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() = }")