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id	challengeType	title	forumTopicId
5900f3871000cf542c50fe9a	5	Problem 27: Quadratic primes	301919

Description

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 \le n \le 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 \le n \le 79. The product of the coefficients, −79 and 1601, is −126479.

Considering quadratics of the form:

$n^2 + an + b$, where $|a| < range$ and $|b| \le range$
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.

Instructions

Tests

tests:
  - text: <code>quadraticPrimes(200)</code> should return a number.
    testString: assert(typeof quadraticPrimes(200) === 'number');
  - text: <code>quadraticPrimes(200)</code> should return -4925.
    testString: assert(quadraticPrimes(200) == -4925);
  - text: <code>quadraticPrimes(500)</code> should return -18901.
    testString: assert(quadraticPrimes(500) == -18901);
  - text: <code>quadraticPrimes(800)</code> should return -43835.
    testString: assert(quadraticPrimes(800) == -43835);
  - text: <code>quadraticPrimes(1000)</code> should return -59231.
    testString: assert(quadraticPrimes(1000) == -59231);

Challenge Seed

function quadraticPrimes(range) {

  return range;
}

quadraticPrimes(1000);

Solution

// solution required

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