project_euler.problem_065.sol1

Project Euler Problem 65: https://projecteuler.net/problem=65

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

sqrt(2) = 1 + 1 / (2 + 1 / (2 + 1 / (2 + 1 / (2 + …))))

The infinite continued fraction can be written, sqrt(2) = [1;(2)], (2) indicates that 2 repeats ad infinitum. In a similar way, sqrt(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 sqrt(2).

1 + 1 / 2 = 3/2 1 + 1 / (2 + 1 / 2) = 7/5 1 + 1 / (2 + 1 / (2 + 1 / 2)) = 17/12 1 + 1 / (2 + 1 / (2 + 1 / (2 + 1 / 2))) = 41/29

Hence the sequence of the first ten convergents for sqrt(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 the digits in the numerator of the 100th convergent of the continued fraction for e.

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The solution mostly comes down to finding an equation that will generate the numerator of the continued fraction. For the i-th numerator, the pattern is:

n_i = m_i * n_(i-1) + n_(i-2)

for m_i = the i-th index of the continued fraction representation of e, n_0 = 1, and n_1 = 2 as the first 2 numbers of the representation.

For example: n_9 = 6 * 193 + 106 = 1264 1 + 2 + 6 + 4 = 13

n_10 = 1 * 193 + 1264 = 1457 1 + 4 + 5 + 7 = 17

Functions

solution(→ int)	Returns the sum of the digits in the numerator of the max-th convergent of
sum_digits(→ int)	Returns the sum of every digit in num.

Module Contents

project_euler.problem_065.sol1.solution(max_n: int = 100) → int

Returns the sum of the digits in the numerator of the max-th convergent of the continued fraction for e.

>>> solution(9)
13
>>> solution(10)
17
>>> solution(50)
91

project_euler.problem_065.sol1.sum_digits(num: int) → int

Returns the sum of every digit in num.

>>> sum_digits(1)
1
>>> sum_digits(12345)
15
>>> sum_digits(999001)
28