53 lines
1.8 KiB
Markdown
53 lines
1.8 KiB
Markdown
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---
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id: 5900f4ab1000cf542c50ffbd
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title: 'Problem 318: 2011 nines'
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challengeType: 5
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forumTopicId: 301974
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dashedName: problem-318-2011-nines
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---
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# --description--
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Consider the real number $\sqrt{2} + \sqrt{3}$.
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When we calculate the even powers of $\sqrt{2} + \sqrt{3}$ we get:
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$$\begin{align} & {(\sqrt{2} + \sqrt{3})}^2 = 9.898979485566356\ldots \\\\ & {(\sqrt{2} + \sqrt{3})}^4 = 97.98979485566356\ldots \\\\ & {(\sqrt{2} + \sqrt{3})}^6 = 969.998969071069263\ldots \\\\ & {(\sqrt{2} + \sqrt{3})}^8 = 9601.99989585502907\ldots \\\\ & {(\sqrt{2} + \sqrt{3})}^{10} = 95049.999989479221\ldots \\\\ & {(\sqrt{2} + \sqrt{3})}^{12} = 940897.9999989371855\ldots \\\\ & {(\sqrt{2} + \sqrt{3})}^{14} = 9313929.99999989263\ldots \\\\ & {(\sqrt{2} + \sqrt{3})}^{16} = 92198401.99999998915\ldots \\\\ \end{align}$$
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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 $n$.
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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 $n$.
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Let $C(p,q,n)$ be the number of consecutive nines at the beginning of the fractional part of ${(\sqrt{p} + \sqrt{q})}^{2n}$.
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Let $N(p,q)$ be the minimal value of $n$ such that $C(p,q,n) ≥ 2011$.
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Find $\sum N(p,q)$ for $p + q ≤ 2011$.
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# --hints--
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`twoThousandElevenNines()` should return `709313889`.
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```js
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assert.strictEqual(twoThousandElevenNines(), 709313889);
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```
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# --seed--
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## --seed-contents--
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```js
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function twoThousandElevenNines() {
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return true;
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}
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twoThousandElevenNines();
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```
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# --solutions--
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```js
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// solution required
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```
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