## Project Euler & HackerRank Problem 27 Solution

##### Quadratic primes

by {BetaProjects} | APRIL 30, 2009 | Project Euler & HackerRank### Project Euler Problem 27 Statement

Euler published the remarkable quadratic formula:

`n`² + `n` + 41

It turns out that the formula will produce 40 primes for the consecutive values `n` = 0 to 39. However, when `n` = 40, 40^{2} + 40 + 41 = 40(40 + 1) + 41 is divisible by 41, and certainly when `n` = 41, 41² + 41 + 41 is clearly divisible by 41.

Using computers, the incredible formula `n`² − 79`n` + 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:

`n`² + `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.