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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
"""
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Project Euler Problem 129: https://projecteuler.net/problem=129
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A number consisting entirely of ones is called a repunit. We shall define R(k) to be
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a repunit of length k; for example, R(6) = 111111.
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Given that n is a positive integer and GCD(n, 10) = 1, it can be shown that there
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always exists a value, k, for which R(k) is divisible by n, and let A(n) be the least
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such value of k; for example, A(7) = 6 and A(41) = 5.
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The least value of n for which A(n) first exceeds ten is 17.
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Find the least value of n for which A(n) first exceeds one-million.
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"""
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def least_divisible_repunit(divisor: int) -> int:
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"""
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Return the least value k such that the Repunit of length k is divisible by divisor.
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>>> least_divisible_repunit(7)
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6
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>>> least_divisible_repunit(41)
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5
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>>> least_divisible_repunit(1234567)
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34020
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"""
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if divisor % 5 == 0 or divisor % 2 == 0:
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return 0
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repunit = 1
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repunit_index = 1
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while repunit:
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repunit = (10 * repunit + 1) % divisor
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repunit_index += 1
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return repunit_index
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def solution(limit: int = 1000000) -> int:
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"""
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Return the least value of n for which least_divisible_repunit(n)
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first exceeds limit.
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>>> solution(10)
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17
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>>> solution(100)
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109
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>>> solution(1000)
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1017
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"""
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divisor = limit - 1
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if divisor % 2 == 0:
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divisor += 1
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while least_divisible_repunit(divisor) <= limit:
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divisor += 2
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return divisor
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if __name__ == "__main__":
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print(f"{solution() = }")
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