59 lines
1.4 KiB
Markdown
59 lines
1.4 KiB
Markdown
---
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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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$$\begin{align}
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& \frac{1}{2} = 0.5 \\\\
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& \frac{1}{3} = 0.(3) \\\\
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& \frac{1}{4} = 0.25 \\\\
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& \frac{1}{5} = 0.2 \\\\
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& \frac{1}{6} = 0.1(6) \\\\
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& \frac{1}{7} = 0.(142857) \\\\
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& \frac{1}{8} = 0.125 \\\\
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& \frac{1}{9} = 0.(1) \\\\
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& \frac{1}{10} = 0.1 \\\\
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\end{align}$$
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Where $0.1(6)$ means $0.166666\ldots$, and has a 1-digit recurring cycle. It can be seen that $\frac{1}{7}$ 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 $\frac{1}{n}$. You are given that $\sum L(n)$ for $3 ≤ n ≤ 1\\,000\\,000$ equals $55\\,535\\,191\\,115$.
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Find $\sum L(n)$ for $3 ≤ n ≤ 100\\,000\\,000$.
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# --hints--
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`reciprocalCyclesTwo()` should return `446572970925740`.
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```js
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assert.strictEqual(reciprocalCyclesTwo(), 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 reciprocalCyclesTwo() {
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return true;
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}
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reciprocalCyclesTwo();
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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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