49 lines
1.1 KiB
Markdown
49 lines
1.1 KiB
Markdown
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---
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id: 5900f50d1000cf542c51001f
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title: 'Problem 417: Reciprocal cycles II'
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challengeType: 5
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forumTopicId: 302086
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dashedName: problem-417-reciprocal-cycles-ii
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---
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# --description--
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A unit fraction contains 1 in the numerator. The decimal representation of the unit fractions with denominators 2 to 10 are given:
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1/2= 0.5 1/3= 0.(3) 1/4= 0.25 1/5= 0.2 1/6= 0.1(6) 1/7= 0.(142857) 1/8= 0.125 1/9= 0.(1) 1/10= 0.1
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Where 0.1(6) means 0.166666..., and has a 1-digit recurring cycle. It can be seen that 1/7 has a 6-digit recurring cycle.
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Unit fractions whose denominator has no other prime factors than 2 and/or 5 are not considered to have a recurring cycle. We define the length of the recurring cycle of those unit fractions as 0.
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Let L(n) denote the length of the recurring cycle of 1/n. You are given that ∑L(n) for 3 ≤ n ≤ 1 000 000 equals 55535191115.
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Find ∑L(n) for 3 ≤ n ≤ 100 000 000
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# --hints--
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`euler417()` should return 446572970925740.
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```js
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assert.strictEqual(euler417(), 446572970925740);
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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 euler417() {
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return true;
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}
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euler417();
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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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