15Jun/090
Exercise 1.21 of SICP
Exercise 1.21: Use the smallest-divisor procedure to find the smallest divisor of each of the following numbers: 199, 1999, 19999.
(define (smallest-divisor n) (find-divisor n 2)) (define (find-divisor n test-divisor) (cond ((> (square test-divisor) n) n) ((divides? test-divisor n) test-divisor) (else (find-divisor n (+ test-divisor 1))))) (define (divides? a b) (= (remainder b a) 0)))))
> (smallest-divisor 199)
199
> (smallest-divisor 1999)
1999
> (smallest-divisor 19999)
7