ee1e8abd87
* feat(tools): add seed/solution restore script * chore(curriculum): remove empty sections' markers * chore(curriculum): add seed + solution to Chinese * chore: remove old formatter * fix: update getChallenges parse translated challenges separately, without reference to the source * chore(curriculum): add dashedName to English * chore(curriculum): add dashedName to Chinese * refactor: remove unused challenge property 'name' * fix: relax dashedName requirement * fix: stray tag Remove stray `pre` tag from challenge file. Signed-off-by: nhcarrigan <nhcarrigan@gmail.com> Co-authored-by: nhcarrigan <nhcarrigan@gmail.com>
69 lines
1.3 KiB
Markdown
69 lines
1.3 KiB
Markdown
---
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id: 5900f4ab1000cf542c50ffbd
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title: 问题318：2011个九
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challengeType: 5
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videoUrl: ''
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dashedName: problem-318-2011-nines
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---
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# --description--
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考虑实数√2+√3。
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当我们计算√2+√3的偶数幂时
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我们得到：
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（√2+√3）2 = 9.898979485566356 ...
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（√2+√3）4 = 97.98979485566356 ...
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（√2+√3）6 = 969.998969071069263 ...
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（√2+√3）8 = 9601.99989585502907 ...
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（√2+√3）10 = 95049.999989479221 ...
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（√2+√3）12 = 940897.9999989371855 ...
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（√2+√3）14 = 9313929.99999989263 ...
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（√2+√3）16 = 92198401.99999998915 ...
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这些幂的小数部分开头的连续九个数字似乎没有减少。 实际上，可以证明（√2+√3）2n的小数部分对于大n接近1。
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考虑形式为√p+√q的所有实数，其中p和q为正整数，且p <q，使得小数部分 （√p+√q）的2n对于大n接近1。
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令C（p，q，n）为（√p+√q）2n的小数部分开头的连续九个数字。
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令N（p，q）为n的最小值，以使C（p，q，n）≥2011。
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求p + q≤2011的∑N（p，q）。
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# --hints--
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`euler318()`应该返回709313889。
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```js
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assert.strictEqual(euler318(), 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 euler318() {
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return true;
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}
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euler318();
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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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