TheAlgorithms-Python

1
"""
2
Project Euler Problem 129: https://projecteuler.net/problem=129
3

4
A number consisting entirely of ones is called a repunit. We shall define R(k) to be
5
a repunit of length k; for example, R(6) = 111111.
6

7
Given that n is a positive integer and GCD(n, 10) = 1, it can be shown that there
8
always exists a value, k, for which R(k) is divisible by n, and let A(n) be the least
9
such value of k; for example, A(7) = 6 and A(41) = 5.
10

11
The least value of n for which A(n) first exceeds ten is 17.
12

13
Find the least value of n for which A(n) first exceeds one-million.
14
"""
15

16

17
def least_divisible_repunit(divisor: int) -> int:
18
    """
19
    Return the least value k such that the Repunit of length k is divisible by divisor.
20
    >>> least_divisible_repunit(7)
21
    6
22
    >>> least_divisible_repunit(41)
23
    5
24
    >>> least_divisible_repunit(1234567)
25
    34020
26
    """
27
    if divisor % 5 == 0 or divisor % 2 == 0:
28
        return 0
29
    repunit = 1
30
    repunit_index = 1
31
    while repunit:
32
        repunit = (10 * repunit + 1) % divisor
33
        repunit_index += 1
34
    return repunit_index
35

36

37
def solution(limit: int = 1000000) -> int:
38
    """
39
    Return the least value of n for which least_divisible_repunit(n)
40
    first exceeds limit.
41
    >>> solution(10)
42
    17
43
    >>> solution(100)
44
    109
45
    >>> solution(1000)
46
    1017
47
    """
48
    divisor = limit - 1
49
    if divisor % 2 == 0:
50
        divisor += 1
51
    while least_divisible_repunit(divisor) <= limit:
52
        divisor += 2
53
    return divisor
54

55

56
if __name__ == "__main__":
57
    print(f"{solution() = }")
58