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56 lines
1.5 KiB
Python
56 lines
1.5 KiB
Python
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"""
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Problem:
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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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def isDigitCancelling(num, den):
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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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def solve(digit_len: int) -> str:
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"""
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>>> solve(2)
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'16/64 , 19/95 , 26/65 , 49/98'
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>>> solve(3)
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'16/64 , 19/95 , 26/65 , 49/98'
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>>> solve(4)
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'16/64 , 19/95 , 26/65 , 49/98'
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>>> solve(0)
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''
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>>> solve(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 isDigitCancelling(num, den):
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solutions.append("{}/{}".format(num, den))
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den += 1
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num += 1
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den = 10
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solutions = " , ".join(solutions)
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return solutions
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if __name__ == "__main__":
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print(solve(2))
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