feat: enable new langs (#42491)

Enable italian and portuguese

curriculum/challenges/italian/10-coding-interview-prep/project-euler/problem-78-coin-partitions.md

@@ -0,0 +1,103 @@
+---
+id: 5900f3ba1000cf542c50fecd
+title: 'Problem 78: Coin partitions'
+challengeType: 5
+forumTopicId: 302191
+dashedName: problem-78-coin-partitions
+---
+
+# --description--
+
+Let ${p}(n)$ represent the number of different ways in which `n` coins can be separated into piles. For example, five coins can be separated into piles in exactly seven different ways, so ${p}(5) = 7$.
+
+<div style='text-align: center;'>
+
+| Coin piles        |
+| ----------------- |
+| OOOOO             |
+| OOOO   O          |
+| OOO   OO          |
+| OOO   O   O       |
+| OO   OO   O       |
+| OO   O   O   O    |
+| O   O   O   O   O |
+
+</div><br>
+
+Find the least value of `n` for which ${p}(n)$ is divisible by `divisor`.
+
+# --hints--
+
+`coinPartitions(7)` should return a number.
+
+```js
+assert(typeof coinPartitions(7) === 'number');
+```
+
+`coinPartitions(7)` should return `5`.
+
+```js
+assert.strictEqual(coinPartitions(7), 5);
+```
+
+`coinPartitions(10000)` should return `599`.
+
+```js
+assert.strictEqual(coinPartitions(10000), 599);
+```
+
+`coinPartitions(100000)` should return `11224`.
+
+```js
+assert.strictEqual(coinPartitions(100000), 11224);
+```
+
+`coinPartitions(1000000)` should return `55374`.
+
+```js
+assert.strictEqual(coinPartitions(1000000), 55374);
+```
+
+# --seed--
+
+## --seed-contents--
+
+```js
+function coinPartitions(divisor) {
+
+  return true;
+}
+
+coinPartitions(7);
+```
+
+# --solutions--
+
+```js
+function coinPartitions(divisor) {
+  const partitions = [1];
+
+  let n = 0;
+  while (partitions[n] !== 0) {
+    n++;
+    partitions.push(0);
+
+    let i = 0;
+    let pentagonal = 1;
+    while (pentagonal <= n) {
+      const sign = i % 4 > 1 ? -1 : 1;
+      partitions[n] += sign * partitions[n - pentagonal];
+      partitions[n] = partitions[n] % divisor;
+
+      i++;
+
+      let k = Math.floor(i / 2) + 1;
+      if (i % 2 !== 0) {
+        k *= -1;
+      }
+      pentagonal = Math.floor((k * (3 * k - 1)) / 2);
+    }
+  }
+  return n;
+}
+```