27.py

# Problem 27
# Quadratic Primes
# Euler discovered the remarkable quadratic formula:
# n^2 + n + 41
# It turns out that the formula will produce 40 primes for the consecutive integer values 0 ≤ n ≤ 39. However, when n = 40, 40^2 + 40 + 41 = 40(40 + 1) + 41 is divisible by 41, and certainly when n = 41, 41^2 + 41 + 41 is clearly divisible by 41.
# The incredible formula n^2 - 79n + 1601 was discovered, which produces 80 primes for the consecutive values 0 ≤ n ≤ 79. The product of the coefficients, -79 and 1601, is -126479.
# Considering quadratics of the form:
# n^2 + n + b, where |a| < 1000 and |b| ≤ 1000
# where |n| is the modulus/absolute value of n
# e.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

highest = 0

def checkPrime(number):
    if number <= 1:
        return False
    elif number == 2:
        return True
    maxDivisor = math.ceil(math.sqrt(number)) + 1
    if number % 2 == 0:
        return False
    for divisor in range (3, maxDivisor, 2):
        if number % divisor == 0:
            return False
    return True

def consecutivePrimes(a, b):
    n = 0
    while True:
        if checkPrime(n**2 + a * n + b):
            n += 1
        else:
            break
    return n

for a in range(-999, 1000):
    for b in range(0, 1001):
        consecutive = consecutivePrimes(a, b)
        if consecutive > highest:
            highest = consecutive
            answer = a * b

print(answer)

# -59231