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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:

$$\begin{align} & \frac{1}{2} = 0.5 \\ & \frac{1}{3} = 0.(3) \\ & \frac{1}{4} = 0.25 \\ & \frac{1}{5} = 0.2 \\ & \frac{1}{6} = 0.1(6) \\ & \frac{1}{7} = 0.(142857) \\ & \frac{1}{8} = 0.125 \\ & \frac{1}{9} = 0.(1) \\ & \frac{1}{10} = 0.1 \\ \end{align}$$

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);

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--seed-contents--

function reciprocalCyclesTwo() {

  return true;
}

reciprocalCyclesTwo();

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// solution required