chore(i18n,learn): processed translations (#44851)

curriculum/challenges/japanese/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 $\sqrt{2} + \sqrt{3}$.
+
+When we calculate the even powers of $\sqrt{2} + \sqrt{3}$ we get:
+
+$$\begin{align} & {(\sqrt{2} + \sqrt{3})}^2 = 9.898979485566356\ldots \\\\ & {(\sqrt{2} + \sqrt{3})}^4 = 97.98979485566356\ldots \\\\ & {(\sqrt{2} + \sqrt{3})}^6 = 969.998969071069263\ldots \\\\ & {(\sqrt{2} + \sqrt{3})}^8 = 9601.99989585502907\ldots \\\\ & {(\sqrt{2} + \sqrt{3})}^{10} = 95049.999989479221\ldots \\\\ & {(\sqrt{2} + \sqrt{3})}^{12} = 940897.9999989371855\ldots \\\\ & {(\sqrt{2} + \sqrt{3})}^{14} = 9313929.99999989263\ldots \\\\ & {(\sqrt{2} + \sqrt{3})}^{16} = 92198401.99999998915\ldots \\\\ \end{align}$$
+
+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 ${(\sqrt{2} + \sqrt{3})}^{2n}$ approaches 1 for large $n$.
+
+Consider all real numbers of the form $\sqrt{p} + \sqrt{q}$ with $p$ and $q$ positive integers and $p < q$, such that the fractional part of ${(\sqrt{p} + \sqrt{q})}^{2n}$ approaches 1 for large $n$.
+
+Let $C(p,q,n)$ be the number of consecutive nines at the beginning of the fractional part of ${(\sqrt{p} + \sqrt{q})}^{2n}$.
+
+Let $N(p,q)$ be the minimal value of $n$ such that $C(p,q,n) ≥ 2011$.
+
+Find $\sum N(p,q)$ for $p + q ≤ 2011$.
+
+# --hints--
+
+`twoThousandElevenNines()` should return `709313889`.
+
+```js
+assert.strictEqual(twoThousandElevenNines(), 709313889);
+```
+
+# --seed--
+
+## --seed-contents--
+
+```js
+function twoThousandElevenNines() {
+
+  return true;
+}
+
+twoThousandElevenNines();
+```
+
+# --solutions--
+
+```js
+// solution required
+```