# Project Euler

### A Taste of Number Theory

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.

#### The Catch

How to speed up the search from |a| < 1000 and |b| < 1000. That is looping from -999 to 999 for a and b, totaling 1999 * 1999 = 3,996,001 iterations!

#### The Light

By doing a little mathematical analysis, it is obvious that for n = 0, n2 + an + b = 02 + 0 + b = b. Since this function must return a prime number, b must be prime. Thus, we can skip all non-prime values for b and keep track of the streak. It can be shown that this simple tweak lowers the number of iterations to 355,822. Use the prime checking method discussed in Problem 3 to check for prime number.

#### The Code

```public class Problem27
{
public static void main(String[] args)
{
int maxA = 0, maxB = 0 , maxN = 0;
int count = 0;

for(int a = -999; a < 1000; a++)
{
for(int b = -999; b < 1000; b++)
{
if(!isPrime(b))
continue;

int n = 0;
while(isPrime( n*n + a*n + b ))
{
n++;
}

if(n > maxN)
{
maxN = n;
maxA = a;
maxB = b;
}
}
}
System.out.println("a * b = " + (maxA * maxB));
}

public static boolean isPrime(int n)
{
if( n < 2 )
return false;

if(n % 2 == 0 && n != 2)
return false;

for(int i = 3; i * i < n; i++)
{
if(n % i == 0)
return false;
}
return true;
}
}
```