## **Note: this wiki is now retired and will no longer be updated!**

**The static final versions of the pages are left as a convenience for readers. Note that meta-pages such as "discussion," "history," etc., will not work.**

# SICP exercise 1.21

From Drewiki

## Problem

Use the `smallest-divisor` procedure from the text to find the smallest divisor of each of the following numbers: 199, 1999, 19999.

## Solution

This exercise is straightforward. Here's the `smallest-divisor` procedure (and its dependencies) from the text:

(define (square x) (* x x)) (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))

And here are the requested answers:

(smallest-divisor 199)

*Output:*

`
`

199

(smallest-divisor 1999)

*Output:*

`
`

1999

(smallest-divisor 19999)

*Output:*

`
`

7