##
**Problem:**

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 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.

##
**Solution:**

-59231

##
**Code:**

The solution may include methods that will be found here: Library.java .

The solution may include methods that will be found here: Library.java .

public interface EulerSolution{ public String run(); }

/* * Solution to Project Euler problem 27 * By Nayuki Minase * * http://nayuki.eigenstate.org/page/project-euler-solutions * https://github.com/nayuki/Project-Euler-solutions */ public final class p027 implements EulerSolution { public static void main(String[] args) { System.out.println(new p027().run()); } public String run() { int bestNum = 0; int bestA = 0; int bestB = 0; for (int a = -1000; a <= 1000; a++) { for (int b = -1000; b <= 1000; b++) { int num = numberOfConsecutivePrimesGenerated(a, b); if (num > bestNum) { bestNum = num; bestA = a; bestB = b; } } } return Integer.toString(bestA * bestB); } private static int numberOfConsecutivePrimesGenerated(int a, int b) { for (int i = 0; ; i++) { int n = i * i + i * a + b; if (n < 0 || !Library.isPrime(n)) return i; } } }

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