2018-10-10 18:03:03 -04:00
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id: 5900f4ab1000cf542c50ffbd
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2020-12-16 00:37:30 -07:00
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title: 问题318：2011个九
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2018-10-10 18:03:03 -04:00
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challengeType: 5
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videoUrl: ''
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2021-01-13 03:31:00 +01:00
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dashedName: problem-318-2011-nines
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2018-10-10 18:03:03 -04:00
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---
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2020-12-16 00:37:30 -07:00
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# --description--
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2020-02-18 01:40:55 +09:00
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考虑实数√2+√3。
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2020-12-16 00:37:30 -07:00
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2020-02-18 01:40:55 +09:00
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当我们计算√2+√3的偶数幂时
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2020-12-16 00:37:30 -07:00
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2020-02-18 01:40:55 +09:00
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我们得到：
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2020-12-16 00:37:30 -07:00
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2020-02-18 01:40:55 +09:00
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（√2+√3）2 = 9.898979485566356 ...
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（√2+√3）4 = 97.98979485566356 ...
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2020-02-18 01:40:55 +09:00
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（√2+√3）6 = 969.998969071069263 ...
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2020-02-18 01:40:55 +09:00
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（√2+√3）8 = 9601.99989585502907 ...
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（√2+√3）10 = 95049.999989479221 ...
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2020-02-18 01:40:55 +09:00
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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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2020-12-16 00:37:30 -07:00
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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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2020-02-18 01:40:55 +09:00
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2020-12-16 00:37:30 -07:00
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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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2020-02-18 01:40:55 +09:00
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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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2020-12-16 00:37:30 -07:00
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# --hints--
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2018-10-10 18:03:03 -04:00
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2020-12-16 00:37:30 -07:00
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`euler318()`应该返回709313889。
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2018-10-10 18:03:03 -04:00
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```js
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assert.strictEqual(euler318(), 709313889);
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2018-10-10 18:03:03 -04:00
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```
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2021-01-13 03:31:00 +01:00
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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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2020-12-16 00:37:30 -07:00
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# --solutions--
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2020-08-13 17:24:35 +02:00
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2021-01-13 03:31:00 +01:00
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```js
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// solution required
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```
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