freeCodeCamp/curriculum/challenges/english/10-coding-interview-prep/project-euler/problem-417-reciprocal-cycles-ii.md

id, title, challengeType, forumTopicId, dashedName

id	title	challengeType	forumTopicId	dashedName
5900f50d1000cf542c51001f	Problem 417: Reciprocal cycles II	5	302086	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.

assert.strictEqual(reciprocalCyclesTwo(), 446572970925740);

--seed--

--seed-contents--

function reciprocalCyclesTwo() {

  return true;
}

reciprocalCyclesTwo();

--solutions--

// solution required