## Reciprocal cycles II

### Problem 417

Published on Saturday, 2nd March 2013, 04:00 pm; Solved by 323

A unit fraction contains 1 in the numerator. The decimal representation of the unit fractions with denominators 2 to 10 are given:

> 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

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.

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 1/n. You are given that L(n) for 3 n 1 000 000 equals 55535191115.

Find L(n) for 3 n 100 000 000