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63 lines
1.6 KiB
Python
63 lines
1.6 KiB
Python
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"""
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We shall say that an n-digit number is pandigital if it makes use of all the
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digits 1 to n exactly once; for example, the 5-digit number, 15234, is 1 through
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5 pandigital.
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The product 7254 is unusual, as the identity, 39 × 186 = 7254, containing
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multiplicand, multiplier, and product is 1 through 9 pandigital.
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Find the sum of all products whose multiplicand/multiplier/product identity can
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be written as a 1 through 9 pandigital.
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HINT: Some products can be obtained in more than one way so be sure to only
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include it once in your sum.
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"""
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import itertools
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def isCombinationValid(combination):
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"""
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Checks if a combination (a tuple of 9 digits)
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is a valid product equation.
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>>> isCombinationValid(('3', '9', '1', '8', '6', '7', '2', '5', '4'))
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True
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>>> isCombinationValid(('1', '2', '3', '4', '5', '6', '7', '8', '9'))
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False
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"""
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return (
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int(''.join(combination[0:2])) *
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int(''.join(combination[2:5])) ==
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int(''.join(combination[5:9]))
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) or (
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int(''.join(combination[0])) *
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int(''.join(combination[1:5])) ==
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int(''.join(combination[5:9]))
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)
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def solution():
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"""
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Finds the sum of all products whose multiplicand/multiplier/product identity
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can be written as a 1 through 9 pandigital
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>>> solution()
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45228
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"""
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return sum(
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set(
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[
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int(''.join(pandigital[5:9]))
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for pandigital
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in itertools.permutations('123456789')
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if isCombinationValid(pandigital)
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]
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)
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)
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if __name__ == "__main__":
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print(solution())
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