2018-10-10 22:03:03 +00:00
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---
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id: 5900f3ef1000cf542c50ff01
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challengeType: 5
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videoUrl: ''
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2020-10-01 15:54:21 +00:00
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title: 问题129:重新划分可分性
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2018-10-10 22:03:03 +00:00
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---
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## Description
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<section id="description">完全由1组成的数字称为repunit。我们将R(k)定义为长度k的重新定位;例如,R(6)= 111111.假设n是正整数且GCD(n,10)= 1,则可以证明总是存在一个值k,其中R(k)可被n整除让A(n)成为k的最小值;例如,A(7)= 6且A(41)= 5.A(n)首先超过10的n的最小值是17.求出A(n)首先超过1的n的最小值 - 百万。 </section>
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## Instructions
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<section id="instructions">
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</section>
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## Tests
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<section id='tests'>
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```yml
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tests:
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- text: <code>euler129()</code>应该返回1000023。
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2020-02-17 16:40:55 +00:00
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testString: assert.strictEqual(euler129(), 1000023);
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2018-10-10 22:03:03 +00:00
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```
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</section>
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## Challenge Seed
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<section id='challengeSeed'>
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<div id='js-seed'>
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```js
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function euler129() {
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// Good luck!
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return true;
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}
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euler129();
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```
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</div>
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</section>
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## Solution
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<section id='solution'>
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```js
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// solution required
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```
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2020-08-13 15:24:35 +00:00
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/section>
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