81 lines
1.9 KiB
Python
81 lines
1.9 KiB
Python
#!/usr/bin/env python3
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# The number, 1406357289, is a 0 to 9 pandigital number because it is made up of each of the digits 0 to 9 in some order, but it also has a
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# rather interesting sub-string divisibility property.
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#
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# Let d1 be the 1st digit, d2 be the 2nd digit, and so on. In this way, we note the following:
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#
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# d2d3d4=406 is divisible by 2
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# d3d4d5=063 is divisible by 3
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# d4d5d6=635 is divisible by 5
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# d5d6d7=357 is divisible by 7
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# d6d7d8=572 is divisible by 11
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# d7d8d9=728 is divisible by 13
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# d8d9d10=289 is divisible by 17
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#
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# Find the sum of all 0 to 9 pandigital numbers with this property.
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from typing import Tuple
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from itertools import permutations
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from projecteuler import timing
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# Function to check if the value has the desired property.
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def has_property(n: Tuple[str, ...]) -> bool:
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value = int(n[1]) * 100 + int(n[2]) * 10 + int(n[3])
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if value % 2 != 0:
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return False
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value = int(n[2]) * 100 + int(n[3]) * 10 + int(n[4])
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if value % 3 != 0:
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return False
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value = int(n[3]) * 100 + int(n[4]) * 10 + int(n[5])
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if value % 5 != 0:
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return False
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value = int(n[4]) * 100 + int(n[5]) * 10 + int(n[6])
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if value % 7 != 0:
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return False
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value = int(n[5]) * 100 + int(n[6]) * 10 + int(n[7])
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if value % 11 != 0:
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return False
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value = int(n[6]) * 100 + int(n[7]) * 10 + int(n[8])
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if value % 13 != 0:
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return False
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value = int(n[7]) * 100 + int(n[8]) * 10 + int(n[9])
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if value % 17 != 0:
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return False
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return True
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@timing
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def p043():
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# Find all the permutations
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perm = list(permutations('0123456789'))
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_sum = 0
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# For each permutation, check if it has the required property
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for i in perm:
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if has_property(i):
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_sum = _sum + int(''.join(map(str, i)))
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print('Project Euler, Problem 43')
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print(f'Answer: {_sum}')
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if __name__ == '__main__':
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p043()
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