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* add default arguments and problem url * Update and rename problem_27_sol1.py to sol1.py Co-authored-by: Dhruv <dhruvmanila@gmail.com>
75 lines
2.0 KiB
Python
75 lines
2.0 KiB
Python
"""
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Project Euler Problem 27
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https://projecteuler.net/problem=27
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Problem Statement:
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Euler discovered the remarkable quadratic formula:
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n2 + n + 41
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It turns out that the formula will produce 40 primes for the consecutive values
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n = 0 to 39. However, when n = 40, 402 + 40 + 41 = 40(40 + 1) + 41 is divisible
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by 41, and certainly when n = 41, 412 + 41 + 41 is clearly divisible by 41.
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The incredible formula n2 − 79n + 1601 was discovered, which produces 80 primes
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for the consecutive values n = 0 to 79. The product of the coefficients, −79 and
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1601, is −126479.
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Considering quadratics of the form:
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n² + an + b, where |a| < 1000 and |b| < 1000
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where |n| is the modulus/absolute value of ne.g. |11| = 11 and |−4| = 4
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Find the product of the coefficients, a and b, for the quadratic expression that
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produces the maximum number of primes for consecutive values of n, starting with
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n = 0.
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"""
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import math
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def is_prime(k: int) -> bool:
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"""
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Determine if a number is prime
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>>> is_prime(10)
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False
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>>> is_prime(11)
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True
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"""
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if k < 2 or k % 2 == 0:
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return False
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elif k == 2:
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return True
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else:
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for x in range(3, int(math.sqrt(k) + 1), 2):
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if k % x == 0:
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return False
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return True
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def solution(a_limit: int = 1000, b_limit: int = 1000) -> int:
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"""
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>>> solution(1000, 1000)
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-59231
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>>> solution(200, 1000)
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-59231
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>>> solution(200, 200)
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-4925
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>>> solution(-1000, 1000)
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0
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>>> solution(-1000, -1000)
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0
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"""
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longest = [0, 0, 0] # length, a, b
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for a in range((a_limit * -1) + 1, a_limit):
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for b in range(2, b_limit):
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if is_prime(b):
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count = 0
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n = 0
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while is_prime((n ** 2) + (a * n) + b):
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count += 1
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n += 1
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if count > longest[0]:
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longest = [count, a, b]
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ans = longest[1] * longest[2]
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return ans
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if __name__ == "__main__":
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print(solution(1000, 1000))
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