freeCodeCamp/curriculum/challenges/english/08-coding-interview-prep/project-euler/problem-65-convergents-of-e.english.md

id	challengeType	title	forumTopicId
5900f3ad1000cf542c50fec0	5	Problem 65: Convergents of e	302177

Description

The square root of 2 can be written as an infinite continued fraction.

√2 = 1 + 1

2 + 1

2 + 1

2 + 1

2 + ...

The infinite continued fraction can be written, √2 = [1;(2)], (2) indicates that 2 repeats ad infinitum. In a similar way, √23 = [4;(1,3,1,8)]. It turns out that the sequence of partial values of continued fractions for square roots provide the best rational approximations. Let us consider the convergents for √2.

1 + 1 = 3/2

2

1 + 1 = 7/5

2 + 1

2

1 + 1 = 17/12

2 + 1

2 + 1

2

1 + 1 = 41/29

2 + 1

2 + 1

2 + 1

2

Hence the sequence of the first ten convergents for √2 are: 1, 3/2, 7/5, 17/12, 41/29, 99/70, 239/169, 577/408, 1393/985, 3363/2378, ... What is most surprising is that the important mathematical constant,e = [2; 1,2,1, 1,4,1, 1,6,1 , ... , 1,2k,1, ...]. The first ten terms in the sequence of convergents for e are: 2, 3, 8/3, 11/4, 19/7, 87/32, 106/39, 193/71, 1264/465, 1457/536, ... The sum of digits in the numerator of the 10th convergent is 1+4+5+7=17. Find the sum of digits in the numerator of the 100th convergent of the continued fraction for e.

Instructions

Tests

tests:
  - text: <code>euler65()</code> should return 272.
    testString: assert.strictEqual(euler65(), 272);

Challenge Seed

function euler65() {
  // Good luck!
  return true;
}

euler65();

Solution

// solution required