49 lines
1.4 KiB
Markdown
49 lines
1.4 KiB
Markdown
|
---
|
||
|
|
id: 5900f50d1000cf542c51001f
|
||
|
|
title: 'Problem 417: Reciprocal cycles II'
|
||
|
|
challengeType: 5
|
||
|
|
forumTopicId: 302086
|
||
|
|
dashedName: problem-417-reciprocal-cycles-ii
|
||
|
|
---
|
||
|
|
|
||
|
|
# --description--
|
||
|
|
|
||
|
|
A unit fraction contains 1 in the numerator. The decimal representation of the unit fractions with denominators 2 to 10 are given:
|
||
|
|
|
||
|
|
$$\begin{align} & \frac{1}{2} = 0.5 \\\\ & \frac{1}{3} = 0.(3) \\\\ & \frac{1}{4} = 0.25 \\\\ & \frac{1}{5} = 0.2 \\\\ & \frac{1}{6} = 0.1(6) \\\\ & \frac{1}{7} = 0.(142857) \\\\ & \frac{1}{8} = 0.125 \\\\ & \frac{1}{9} = 0.(1) \\\\ & \frac{1}{10} = 0.1 \\\\ \end{align}$$
|
||
|
|
|
||
|
|
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.
|
||
|
|
|
||
|
|
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.
|
||
|
|
|
||
|
|
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 $55\\,535\\,191\\,115$.
|
||
|
|
|
||
|
|
Find $\sum L(n)$ for $3 ≤ n ≤ 100\\,000\\,000$.
|
||
|
|
|
||
|
|
# --hints--
|
||
|
|
|
||
|
|
`reciprocalCyclesTwo()` should return `446572970925740`.
|
||
|
|
|
||
|
|
```js
|
||
|
|
assert.strictEqual(reciprocalCyclesTwo(), 446572970925740);
|
||
|
|
```
|
||
|
|
|
||
|
|
# --seed--
|
||
|
|
|
||
|
|
## --seed-contents--
|
||
|
|
|
||
|
|
```js
|
||
|
|
function reciprocalCyclesTwo() {
|
||
|
|
|
||
|
|
return true;
|
||
|
|
}
|
||
|
|
|
||
|
|
reciprocalCyclesTwo();
|
||
|
|
```
|
||
|
|
|
||
|
|
# --solutions--
|
||
|
|
|
||
|
|
```js
|
||
|
|
// solution required
|
||
|
|
```
|