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