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>
39 lines
814 B
Markdown
39 lines
814 B
Markdown
---
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id: 5900f3ef1000cf542c50ff01
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title: 问题129：重新划分可分性
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challengeType: 5
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videoUrl: ''
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dashedName: problem-129-repunit-divisibility
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---
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# --description--
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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。
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# --hints--
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`euler129()`应该返回1000023。
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```js
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assert.strictEqual(euler129(), 1000023);
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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 euler129() {
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return true;
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}
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euler129();
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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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