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501a2ff430
* Fix code style for Project Euler problems: - 13, 17, 21 - Default args - Type hints - File path * Fix code style for multiple problems * Made suggested changes
69 lines
1.9 KiB
Python
69 lines
1.9 KiB
Python
"""
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Problem 33: https://projecteuler.net/problem=33
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The fraction 49/98 is a curious fraction, as an inexperienced
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mathematician in attempting to simplify it may incorrectly believe
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that 49/98 = 4/8, which is correct, is obtained by cancelling the 9s.
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We shall consider fractions like, 30/50 = 3/5, to be trivial examples.
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There are exactly four non-trivial examples of this type of fraction,
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less than one in value, and containing two digits in the numerator
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and denominator.
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If the product of these four fractions is given in its lowest common
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terms, find the value of the denominator.
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"""
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from fractions import Fraction
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from typing import List
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def is_digit_cancelling(num: int, den: int) -> bool:
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if num != den:
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if num % 10 == den // 10:
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if (num // 10) / (den % 10) == num / den:
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return True
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return False
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def fraction_list(digit_len: int) -> List[str]:
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"""
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>>> fraction_list(2)
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['16/64', '19/95', '26/65', '49/98']
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>>> fraction_list(3)
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['16/64', '19/95', '26/65', '49/98']
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>>> fraction_list(4)
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['16/64', '19/95', '26/65', '49/98']
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>>> fraction_list(0)
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[]
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>>> fraction_list(5)
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['16/64', '19/95', '26/65', '49/98']
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"""
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solutions = []
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den = 11
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last_digit = int("1" + "0" * digit_len)
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for num in range(den, last_digit):
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while den <= 99:
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if (num != den) and (num % 10 == den // 10) and (den % 10 != 0):
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if is_digit_cancelling(num, den):
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solutions.append(f"{num}/{den}")
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den += 1
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num += 1
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den = 10
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return solutions
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def solution(n: int = 2) -> int:
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"""
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Return the solution to the problem
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"""
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result = 1.0
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for fraction in fraction_list(n):
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frac = Fraction(fraction)
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result *= frac.denominator / frac.numerator
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return int(result)
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if __name__ == "__main__":
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print(solution())
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