chore(i8n,learn): processed translations

curriculum/challenges/espanol/10-coding-interview-prep/project-euler/problem-318-2011-nines.md

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+---
+id: 5900f4ab1000cf542c50ffbd
+title: 'Problem 318: 2011 nines'
+challengeType: 5
+forumTopicId: 301974
+dashedName: problem-318-2011-nines
+---
+
+# --description--
+
+Consider the real number √2+√3.
+
+When we calculate the even powers of √2+√3
+
+we get:
+
+(√2+√3)2 = 9.898979485566356...
+
+(√2+√3)4 = 97.98979485566356...
+
+(√2+√3)6 = 969.998969071069263...
+
+(√2+√3)8 = 9601.99989585502907...
+
+(√2+√3)10 = 95049.999989479221...
+
+(√2+√3)12 = 940897.9999989371855...
+
+(√2+√3)14 = 9313929.99999989263...
+
+(√2+√3)16 = 92198401.99999998915...
+
+It looks like that the number of consecutive nines at the beginning of the fractional part of these powers is non-decreasing. In fact it can be proven that the fractional part of (√2+√3)2n approaches 1 for large n.
+
+Consider all real numbers of the form √p+√q with p and q positive integers and p<q, such that the fractional part of (√p+√q)2n approaches 1 for large n.
+
+Let C(p,q,n) be the number of consecutive nines at the beginning of the fractional part of (√p+√q)2n.
+
+Let N(p,q) be the minimal value of n such that C(p,q,n) ≥ 2011.
+
+Find ∑N(p,q) for p+q ≤ 2011.
+
+# --hints--
+
+`euler318()` should return 709313889.
+
+```js
+assert.strictEqual(euler318(), 709313889);
+```
+
+# --seed--
+
+## --seed-contents--
+
+```js
+function euler318() {
+
+  return true;
+}
+
+euler318();
+```
+
+# --solutions--
+
+```js
+// solution required
+```