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@@ -0,0 +1,36 @@
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+# Set to check if a number is pandigital
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+pan = set("123456789")
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+
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+# Not used in this problem but may come in handy
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+# def setmaker(n):
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+# return set(x for x in str(n))
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+
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+def concatProduct(n, pan):
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+ """
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+ Input is a fixed number and a list, function will return concatenated
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+ product of the fixed number with individual elements of the list
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+ """
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+ product = ""
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+ for i in pan:
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+ product += str(n*i)
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+ return product
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+
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+def isPandigital(n):
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+ """Returns a boolean indicating whether a number is pandigital"""
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+ nset = set(str(n))
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+ if nset == pan and len(n) == 9:
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+ return True
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+ else:
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+ return False
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+
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+maxProduct = 918273645
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+
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+# Main loop, solution set contained to 4 digit numbers that are greater than the
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+# result given in the prompt by 9 and (1,2,3,4,5)
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+for fixed in range(9000,10000):
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+ concatenatedProduct = concatProduct(fixed, (1,2))
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+ if isPandigital(concatenatedProduct):
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+ if concatenatedProduct > maxProduct:
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+ maxProduct = concatenatedProduct
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+
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+print maxProduct
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