1.8 KiB
id | title | challengeType | forumTopicId | dashedName |
---|---|---|---|---|
5900f3871000cf542c50fe9a | Problem 27: Quadratic primes | 5 | 301919 | problem-27-quadratic-primes |
--description--
Euler discovered the remarkable quadratic formula:
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:
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
.
--hints--
quadraticPrimes(200)
should return a number.
assert(typeof quadraticPrimes(200) === 'number');
quadraticPrimes(200)
should return -4925.
assert(quadraticPrimes(200) == -4925);
quadraticPrimes(500)
should return -18901.
assert(quadraticPrimes(500) == -18901);
quadraticPrimes(800)
should return -43835.
assert(quadraticPrimes(800) == -43835);
quadraticPrimes(1000)
should return -59231.
assert(quadraticPrimes(1000) == -59231);
--seed--
--seed-contents--
function quadraticPrimes(range) {
return range;
}
quadraticPrimes(1000);
--solutions--
// solution required