chore(i8n,learn): processed translations

curriculum/challenges/espanol/10-coding-interview-prep/project-euler/problem-21-amicable-numbers.md

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+---
+id: 5900f3811000cf542c50fe94
+title: 'Problem 21: Amicable numbers'
+challengeType: 5
+forumTopicId: 301851
+dashedName: problem-21-amicable-numbers
+---
+
+# --description--
+
+Let d(`n`) be defined as the sum of proper divisors of `n` (numbers less than `n` which divide evenly into `n`).
+
+If d(`a`) = `b` and d(`b`) = `a`, where `a` ≠ `b`, then `a` and `b` are an amicable pair and each of `a` and `b` are called amicable numbers.
+
+For example, the proper divisors of 220 are 1, 2, 4, 5, 10, 11, 20, 22, 44, 55 and 110; therefore d(220) = 284. The proper divisors of 284 are 1, 2, 4, 71 and 142; so d(284) = 220.
+
+Evaluate the sum of all the amicable numbers under `n`.
+
+# --hints--
+
+`sumAmicableNum(1000)` should return a number.
+
+```js
+assert(typeof sumAmicableNum(1000) === 'number');
+```
+
+`sumAmicableNum(1000)` should return 504.
+
+```js
+assert.strictEqual(sumAmicableNum(1000), 504);
+```
+
+`sumAmicableNum(2000)` should return 2898.
+
+```js
+assert.strictEqual(sumAmicableNum(2000), 2898);
+```
+
+`sumAmicableNum(5000)` should return 8442.
+
+```js
+assert.strictEqual(sumAmicableNum(5000), 8442);
+```
+
+`sumAmicableNum(10000)` should return 31626.
+
+```js
+assert.strictEqual(sumAmicableNum(10000), 31626);
+```
+
+# --seed--
+
+## --seed-contents--
+
+```js
+function sumAmicableNum(n) {
+
+  return n;
+}
+
+sumAmicableNum(10000);
+```
+
+# --solutions--
+
+```js
+const sumAmicableNum = (n) => {
+  const fsum = (n) => {
+    let sum = 1;
+    for (let i = 2; i <= Math.floor(Math.sqrt(n)); i++)
+      if (Math.floor(n % i) === 0)
+        sum += i + Math.floor(n / i);
+    return sum;
+  };
+  let d = [];
+  let amicableSum = 0;
+  for (let i=2; i<n; i++) d[i] = fsum(i);
+  for (let i=2; i<n; i++) {
+    let dsum = d[i];
+    if (d[dsum]===i && i!==dsum) amicableSum += i+dsum;
+  }
+  return amicableSum/2;
+};
+```