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* Enable ruff RUF002 rule * Fix --------- Co-authored-by: Christian Clauss <cclauss@me.com>
33 lines
824 B
Python
33 lines
824 B
Python
"""
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Problem 120 Square remainders: https://projecteuler.net/problem=120
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Description:
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Let r be the remainder when (a-1)^n + (a+1)^n is divided by a^2.
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For example, if a = 7 and n = 3, then r = 42: 6^3 + 8^3 = 728 ≡ 42 mod 49.
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And as n varies, so too will r, but for a = 7 it turns out that r_max = 42.
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For 3 ≤ a ≤ 1000, find ∑ r_max.
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Solution:
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On expanding the terms, we get 2 if n is even and 2an if n is odd.
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For maximizing the value, 2an < a*a => n <= (a - 1)/2 (integer division)
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"""
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def solution(n: int = 1000) -> int:
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"""
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Returns ∑ r_max for 3 <= a <= n as explained above
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>>> solution(10)
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300
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>>> solution(100)
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330750
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>>> solution(1000)
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333082500
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"""
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return sum(2 * a * ((a - 1) // 2) for a in range(3, n + 1))
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if __name__ == "__main__":
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print(solution())
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