Problem 2.5

SICP Exercise 2.5

Great exercise! The procedure cons is defined as:

(define (cons a b)
  (* (exp 2 a) (exp 3 b)))

Using exp procedure from earlier. And for car and cdr:

(define (car x)
  (define (iter x count)
    (if (= (remainder x 2) 0)
        (iter (/ x 2) (+ count 1))
        count))
  (iter x 0))

(define (cdr x)
  (define (iter x count)
    (if (= (remainder x 3) 0)
        (iter (/ x 3) (+ count 1))
        count))
  (iter x 0))

Simple explanation:

car: everytime remainder of result and 2 is 0, increase count by 1
cdr: everytime remainder of result and 3 is 0, increase count by 1