1.1 KiB
1.1 KiB
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:
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
--hints--
euler417()
should return 446572970925740.
assert.strictEqual(euler417(), 446572970925740);
--seed--
--seed-contents--
function euler417() {
return true;
}
euler417();
--solutions--
// solution required