--- id: 5900f3871000cf542c50fe9a challengeType: 5 title: 'Problem 27: Quadratic primes' forumTopicId: 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
```yml tests: - text: quadraticPrimes(200) should return a number. testString: assert(typeof quadraticPrimes(200) === 'number'); - text: quadraticPrimes(200) should return -4925. testString: assert(quadraticPrimes(200) == -4925); - text: quadraticPrimes(500) should return -18901. testString: assert(quadraticPrimes(500) == -18901); - text: quadraticPrimes(800) should return -43835. testString: assert(quadraticPrimes(800) == -43835); - text: quadraticPrimes(1000) should return -59231. testString: assert(quadraticPrimes(1000) == -59231); ```
## Challenge Seed
```js function quadraticPrimes(range) { return range; } quadraticPrimes(1000); ```
## Solution
```js // solution required ```