56 lines
1.1 KiB
Markdown
56 lines
1.1 KiB
Markdown
---
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id: 5900f3ef1000cf542c50ff01
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challengeType: 5
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title: 'Problem 129: Repunit divisibility'
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videoUrl: ''
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localeTitle: 问题129：重新划分可分性
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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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testString: 'assert.strictEqual(euler129(), 1000023, "<code>euler129()</code> should return 1000023.");'
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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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</section>
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