79 lines
1.8 KiB
Markdown
79 lines
1.8 KiB
Markdown
---
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id: 5900f3871000cf542c50fe9a
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title: 'Problem 27: Quadratic primes'
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challengeType: 5
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forumTopicId: 301919
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dashedName: problem-27-quadratic-primes
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---
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# --description--
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Euler discovered the remarkable quadratic formula:
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<div style='margin-left: 4em;'>$n^2 + n + 41$</div>
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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.
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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.
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Considering quadratics of the form:
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<div style='margin-left: 4em;'>
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$n^2 + an + b$, where $|a| < range$ and $|b| \le range$<br>
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where $|n|$ is the modulus/absolute value of $n$<br>
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e.g. $|11| = 11$ and $|-4| = 4$<br>
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</div>
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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 $n = 0$.
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# --hints--
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`quadraticPrimes(200)` should return a number.
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```js
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assert(typeof quadraticPrimes(200) === 'number');
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```
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`quadraticPrimes(200)` should return -4925.
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```js
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assert(quadraticPrimes(200) == -4925);
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```
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`quadraticPrimes(500)` should return -18901.
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```js
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assert(quadraticPrimes(500) == -18901);
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```
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`quadraticPrimes(800)` should return -43835.
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```js
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assert(quadraticPrimes(800) == -43835);
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```
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`quadraticPrimes(1000)` should return -59231.
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```js
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assert(quadraticPrimes(1000) == -59231);
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```
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# --seed--
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## --seed-contents--
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```js
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function quadraticPrimes(range) {
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return range;
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}
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quadraticPrimes(1000);
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```
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# --solutions--
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```js
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// solution required
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```
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