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curriculum/challenges/italian/10-coding-interview-prep/project-euler/problem-27-quadratic-primes.md

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