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bcfca67faa
* [mypy] fix type annotations for problem003/sol1 and problem003/sol3 * [mypy] fix type annotations for project euler problem007/sol2 * [mypy] fix type annotations for project euler problem008/sol2 * [mypy] fix type annotations for project euler problem009/sol1 * [mypy] fix type annotations for project euler problem014/sol1 * [mypy] fix type annotations for project euler problem 025/sol2 * [mypy] fix type annotations for project euler problem026/sol1.py * [mypy] fix type annotations for project euler problem037/sol1 * [mypy] fix type annotations for project euler problem044/sol1 * [mypy] fix type annotations for project euler problem046/sol1 * [mypy] fix type annotations for project euler problem051/sol1 * [mypy] fix type annotations for project euler problem074/sol2 * [mypy] fix type annotations for project euler problem080/sol1 * [mypy] fix type annotations for project euler problem099/sol1 * [mypy] fix type annotations for project euler problem101/sol1 * [mypy] fix type annotations for project euler problem188/sol1 * [mypy] fix type annotations for project euler problem191/sol1 * [mypy] fix type annotations for project euler problem207/sol1 * [mypy] fix type annotations for project euler problem551/sol1
50 lines
1.5 KiB
Python
50 lines
1.5 KiB
Python
"""
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Problem 44: https://projecteuler.net/problem=44
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Pentagonal numbers are generated by the formula, Pn=n(3n−1)/2. The first ten
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pentagonal numbers are:
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1, 5, 12, 22, 35, 51, 70, 92, 117, 145, ...
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It can be seen that P4 + P7 = 22 + 70 = 92 = P8. However, their difference,
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70 − 22 = 48, is not pentagonal.
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Find the pair of pentagonal numbers, Pj and Pk, for which their sum and difference
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are pentagonal and D = |Pk − Pj| is minimised; what is the value of D?
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"""
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def is_pentagonal(n: int) -> bool:
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"""
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Returns True if n is pentagonal, False otherwise.
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>>> is_pentagonal(330)
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True
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>>> is_pentagonal(7683)
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False
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>>> is_pentagonal(2380)
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True
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"""
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root = (1 + 24 * n) ** 0.5
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return ((1 + root) / 6) % 1 == 0
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def solution(limit: int = 5000) -> int:
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"""
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Returns the minimum difference of two pentagonal numbers P1 and P2 such that
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P1 + P2 is pentagonal and P2 - P1 is pentagonal.
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>>> solution(5000)
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5482660
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"""
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pentagonal_nums = [(i * (3 * i - 1)) // 2 for i in range(1, limit)]
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for i, pentagonal_i in enumerate(pentagonal_nums):
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for j in range(i, len(pentagonal_nums)):
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pentagonal_j = pentagonal_nums[j]
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a = pentagonal_i + pentagonal_j
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b = pentagonal_j - pentagonal_i
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if is_pentagonal(a) and is_pentagonal(b):
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return b
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return -1
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if __name__ == "__main__":
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print(f"{solution() = }")
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