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Name: Prime square remainders Let pn be the nth prime: 2, 3, 5, 7, 11, ..., and let r be the remainder when (pn−1)^n + (pn+1)^n is divided by pn^2. For example, when n = 3, p3 = 5, and 43 + 63 = 280 ≡ 5 mod 25. The least value of n for which the remainder first exceeds 10^9 is 7037. Find the least value of n for which the remainder first exceeds 10^10. Reference: https://projecteuler.net/problem=123 reference: #2695 Co-authored-by: Ravi Kandasamy Sundaram <rkandasamysundaram@luxoft.com>
100 lines
2.2 KiB
Python
100 lines
2.2 KiB
Python
"""
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Problem 123: https://projecteuler.net/problem=123
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Name: Prime square remainders
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Let pn be the nth prime: 2, 3, 5, 7, 11, ..., and
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let r be the remainder when (pn−1)^n + (pn+1)^n is divided by pn^2.
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For example, when n = 3, p3 = 5, and 43 + 63 = 280 ≡ 5 mod 25.
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The least value of n for which the remainder first exceeds 10^9 is 7037.
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Find the least value of n for which the remainder first exceeds 10^10.
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Solution:
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n=1: (p-1) + (p+1) = 2p
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n=2: (p-1)^2 + (p+1)^2
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= p^2 + 1 - 2p + p^2 + 1 + 2p (Using (p+b)^2 = (p^2 + b^2 + 2pb),
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(p-b)^2 = (p^2 + b^2 - 2pb) and b = 1)
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= 2p^2 + 2
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n=3: (p-1)^3 + (p+1)^3 (Similarly using (p+b)^3 & (p-b)^3 formula and so on)
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= 2p^3 + 6p
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n=4: 2p^4 + 12p^2 + 2
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n=5: 2p^5 + 20p^3 + 10p
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As you could see, when the expression is divided by p^2.
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Except for the last term, the rest will result in the remainder 0.
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n=1: 2p
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n=2: 2
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n=3: 6p
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n=4: 2
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n=5: 10p
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So it could be simplified as,
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r = 2pn when n is odd
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r = 2 when n is even.
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"""
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from typing import Dict, Generator
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def sieve() -> Generator[int, None, None]:
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"""
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Returns a prime number generator using sieve method.
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>>> type(sieve())
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<class 'generator'>
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>>> primes = sieve()
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>>> next(primes)
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2
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>>> next(primes)
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3
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>>> next(primes)
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5
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>>> next(primes)
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7
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>>> next(primes)
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11
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>>> next(primes)
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13
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"""
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factor_map: Dict[int, int] = {}
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prime = 2
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while True:
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factor = factor_map.pop(prime, None)
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if factor:
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x = factor + prime
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while x in factor_map:
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x += factor
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factor_map[x] = factor
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else:
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factor_map[prime * prime] = prime
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yield prime
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prime += 1
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def solution(limit: float = 1e10) -> int:
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"""
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Returns the least value of n for which the remainder first exceeds 10^10.
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>>> solution(1e8)
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2371
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>>> solution(1e9)
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7037
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"""
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primes = sieve()
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n = 1
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while True:
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prime = next(primes)
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if (2 * prime * n) > limit:
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return n
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# Ignore the next prime as the reminder will be 2.
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next(primes)
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n += 2
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if __name__ == "__main__":
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print(solution())
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