57 lines
1.3 KiB
Python
57 lines
1.3 KiB
Python
'''
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Take the number 192 and multiply it by each of 1, 2, and 3:
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192 × 1 = 192
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192 × 2 = 384
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192 × 3 = 576
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By concatenating each product we get the 1 to 9 pandigital, 192384576. We will call 192384576 the concatenated product of 192 and (1,2,3)
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The same can be achieved by starting with 9 and multiplying by 1, 2, 3, 4, and 5, giving the pandigital, 918273645, which is the concatenated product of 9 and (1,2,3,4,5).
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What is the largest 1 to 9 pandigital 9-digit number that can be formed as the concatenated product of an integer with (1,2, ... , n) where n > 1?
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'''
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def conLength(n):
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i = 1
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p = len(str(n))
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while p < 9:
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i += 1
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p += len(str(i * n))
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if p == 9:
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return True
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return False
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def isPandigital(n):
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digits = {1, 2, 3, 4, 5, 6, 7, 8, 9}
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if set(map(int, str(n))) == digits:
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return True
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return False
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def concatProduct(n):
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i = 1
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d = n
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while len(str(d)) < 9:
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i += 1
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d = int(str(d) + str(n * i))
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return d
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def main():
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print("Hello this is Patrick")
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m = 0
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for i in range(1, 54321):
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if conLength(i):
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d = concatProduct(i)
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if isPandigital(d) and d > m:
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m = d
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print(m)
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if __name__ == "__main__":
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main() |