Project Euler Problem 65

http://projecteuler.net/problem=65

Problem 65

12 March 2004

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 10^th convergent is 1+4+5+7=17.

Find the sum of digits in the numerator of the 100^th convergent of the continued fraction for e.

Answer:

272

* Racket

#lang racket
; e
; (2;1,2,1,1,4,1,1,6,1)
; (+ 2 (/ (+ 1 (/ (+ 2 (/ 1))))))

(define (e digit)
  (let ([m (modulo digit 3)])
    (if (= m 1)
        (* 2 (add1 (quotient digit 3)))
        1)))

(define digits (map e (range 0 99)))

(define (calc digits)
  (if (null? digits)
      0
      (/ (+ (first digits) (calc (rest digits))))))

(define result (+ 2 (calc digits)))
result
(define numerator* (numerator result))
numerator*
(let loop ([n numerator*]
           [sum 0])
  (if (> n 0)
      (loop (quotient n 10) (+ sum (modulo n 10)))
      sum))