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409438d250
* Added solution for Project Euler problem 38. Fixes: #2695 * Update docstring and 0-padding in directory name. Reference: #3256 * Renamed is_9_palindromic to is_9_pandigital. * Changed just-in-case return value for solution() to None. * Moved exmplanation to module-level docstring and deleted unnecessary import
78 lines
2.3 KiB
Python
78 lines
2.3 KiB
Python
"""
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Project Euler Problem 38: https://projecteuler.net/problem=38
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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
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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,
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giving the pandigital, 918273645, which is the concatenated product of 9 and
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(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
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concatenated product of an integer with (1,2, ... , n) where n > 1?
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Solution:
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Since n>1, the largest candidate for the solution will be a concactenation of
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a 4-digit number and its double, a 5-digit number.
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Let a be the 4-digit number.
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a has 4 digits => 1000 <= a < 10000
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2a has 5 digits => 10000 <= 2a < 100000
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=> 5000 <= a < 10000
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The concatenation of a with 2a = a * 10^5 + 2a
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so our candidate for a given a is 100002 * a.
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We iterate through the search space 5000 <= a < 10000 in reverse order,
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calculating the candidates for each a and checking if they are 1-9 pandigital.
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In case there are no 4-digit numbers that satisfy this property, we check
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the 3-digit numbers with a similar formula (the example a=192 gives a lower
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bound on the length of a):
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a has 3 digits, etc...
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=> 100 <= a < 334, candidate = a * 10^6 + 2a * 10^3 + 3a
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= 1002003 * a
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"""
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from typing import Union
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def is_9_pandigital(n: int) -> bool:
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"""
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Checks whether n is a 9-digit 1 to 9 pandigital number.
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>>> is_9_pandigital(12345)
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False
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>>> is_9_pandigital(156284973)
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True
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>>> is_9_pandigital(1562849733)
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False
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"""
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s = str(n)
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return len(s) == 9 and set(s) == set("123456789")
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def solution() -> Union[int, None]:
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"""
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Return the largest 1 to 9 pandigital 9-digital number that can be formed as the
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concatenated product of an integer with (1,2,...,n) where n > 1.
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"""
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for base_num in range(9999, 4999, -1):
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candidate = 100002 * base_num
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if is_9_pandigital(candidate):
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return candidate
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for base_num in range(333, 99, -1):
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candidate = 1002003 * base_num
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if is_9_pandigital(candidate):
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return candidate
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return None
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if __name__ == "__main__":
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print(f"{solution() = }")
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