Problem 417
Reciprocal cycles II
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.
倒数循环节 II
单位分数指分子为1的分数。分母为2至10的单位分数的十进制表示如下所示：
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
这里0.1(6)表示0.166666…，括号内表示有一位循环节。可以看出，1/7有六位循环节。
如果单位分数的分母的质因数只包含2和5，就不存在循环节。
我们记这些单位分数的循环节长度为0。
记L(n)为单位分数1/n的循环节长度。已知对于3 ≤ n ≤ 1 000 000，∑L(n)等于55535191115。
对于3 ≤ n ≤ 100 000 000，求∑L(n)。