""" Project Euler Problem 27 https://projecteuler.net/problem=27 Problem Statement: Euler discovered the remarkable quadratic formula: n2 + n + 41 It turns out that the formula will produce 40 primes for the consecutive values n = 0 to 39. However, when n = 40, 402 + 40 + 41 = 40(40 + 1) + 41 is divisible by 41, and certainly when n = 41, 412 + 41 + 41 is clearly divisible by 41. The incredible formula n2 − 79n + 1601 was discovered, which produces 80 primes for the consecutive values n = 0 to 79. The product of the coefficients, −79 and 1601, is −126479. Considering quadratics of the form: n² + an + b, where |a| < 1000 and |b| < 1000 where |n| is the modulus/absolute value of ne.g. |11| = 11 and |−4| = 4 Find the product of the coefficients, a and b, for the quadratic expression that produces the maximum number of primes for consecutive values of n, starting with n = 0. """ import math def is_prime(number: int) -> bool: """Checks to see if a number is a prime in O(sqrt(n)). A number is prime if it has exactly two factors: 1 and itself. Returns boolean representing primality of given number num (i.e., if the result is true, then the number is indeed prime else it is not). >>> is_prime(2) True >>> is_prime(3) True >>> is_prime(27) False >>> is_prime(2999) True >>> is_prime(0) False >>> is_prime(1) False >>> is_prime(-10) False """ if 1 < number < 4: # 2 and 3 are primes return True elif number < 2 or number % 2 == 0 or number % 3 == 0: # Negatives, 0, 1, all even numbers, all multiples of 3 are not primes return False # All primes number are in format of 6k +/- 1 for i in range(5, int(math.sqrt(number) + 1), 6): if number % i == 0 or number % (i + 2) == 0: return False return True def solution(a_limit: int = 1000, b_limit: int = 1000) -> int: """ >>> solution(1000, 1000) -59231 >>> solution(200, 1000) -59231 >>> solution(200, 200) -4925 >>> solution(-1000, 1000) 0 >>> solution(-1000, -1000) 0 """ longest = [0, 0, 0] # length, a, b for a in range((a_limit * -1) + 1, a_limit): for b in range(2, b_limit): if is_prime(b): count = 0 n = 0 while is_prime((n**2) + (a * n) + b): count += 1 n += 1 if count > longest[0]: longest = [count, a, b] ans = longest[1] * longest[2] return ans if __name__ == "__main__": print(solution(1000, 1000))