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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'
title: '問題 318: 2011 個の 9'
challengeType: 5
forumTopicId: 301974
dashedName: problem-318-2011-nines
@ -8,25 +8,25 @@ dashedName: problem-318-2011-nines
# --description--
Consider the real number $\sqrt{2} + \sqrt{3}$.
実数 $\sqrt{2} + \sqrt{3}$ について考えます。
When we calculate the even powers of $\sqrt{2} + \sqrt{3}$ we get:
$\sqrt{2} + \sqrt{3}$ の偶数乗を計算すると、次のようになります。
$$\begin{align} & {(\sqrt{2} + \sqrt{3})}^2 = 9.898979485566356\ldots \\\\ & {(\sqrt{2} + \sqrt{3})}^4 = 97.98979485566356\ldots \\\\ & {(\sqrt{2} + \sqrt{3})}^6 = 969.998969071069263\ldots \\\\ & {(\sqrt{2} + \sqrt{3})}^8 = 9601.99989585502907\ldots \\\\ & {(\sqrt{2} + \sqrt{3})}^{10} = 95049.999989479221\ldots \\\\ & {(\sqrt{2} + \sqrt{3})}^{12} = 940897.9999989371855\ldots \\\\ & {(\sqrt{2} + \sqrt{3})}^{14} = 9313929.99999989263\ldots \\\\ & {(\sqrt{2} + \sqrt{3})}^{16} = 92198401.99999998915\ldots \\\\ \end{align}$$
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 $n$.
これらの累乗の分数部を見ると、先頭で連続している 9 の個数が非減少であるように見えます。 実際に、$n$ が大きいと ${(\sqrt{2} + \sqrt{3})}^{2n}$ の小数部が 1 に近付くということを証明できます。
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 $n$.
正の整数 $p$ と $q$ ($p < q$) があるとき、$n$ が大きいと ${(\sqrt{p} + \sqrt{q})}^{2n}$ の小数部 が 1 に近付くようなすべての実数 $\sqrt{p} + \sqrt{q}$ について考えます。
Let $C(p,q,n)$ be the number of consecutive nines at the beginning of the fractional part of ${(\sqrt{p} + \sqrt{q})}^{2n}$.
${(\sqrt{p} + \sqrt{q})}^{2n}$ の小数部の先頭で連続している 9 の個数を $C(p,q,n)$ とします。
Let $N(p,q)$ be the minimal value of $n$ such that $C(p,q,n) ≥ 2011$.
$C(p,q,n) ≥ 2011$ を満たす $n$ の最小値を $N(p,q)$ とします。
Find $\sum N(p,q)$ for $p + q ≤ 2011$.
$p + q ≤ 2011$ のとき、$\sum N(p,q)$ を求めなさい。
# --hints--
`twoThousandElevenNines()` should return `709313889`.
`twoThousandElevenNines()` は `709313889` を返す必要があります。
```js
assert.strictEqual(twoThousandElevenNines(), 709313889);