How would I do the following in sicp/scheme/dr. racket?

(define (even? n) (= (% n 2) 0))

Currently it seems like that's not a primitive symbol: %: unbound identifier in: %.

This may be the stupidest way in the world to do it, but without a % or bitwise-&1 I am doing (without logs or anything else):

(define (even? n) (if (< (abs n) 2) (= n 0) (even? (- n 2))))
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2 Answers

mod is modulo in scheme:

(define (even? n) (= (modulo n 2) 0))
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I think it's a good practice to get comfortable writing your own procedures when it feels like they are "missing". You could implement your own mod as -

(define (mod a b) (if (< a b) a (mod (- a b) b))) 
(mod 0 3) ; 0 (mod 1 3) ; 1 (mod 2 3) ; 2 (mod 3 3) ; 0 (mod 4 3) ; 1 (mod 5 3) ; 2 (mod 6 3) ; 0 (mod 7 3) ; 1 (mod 8 3) ; 2

But maybe we make it more robust by supporting negative numbers and preventing caller from divi

(define (mod a b) (if (= b 0) (error 'mod "division by zero") (rem (+ b (rem a b)) b))) (define (rem a b) (cond ((= b 0) (error 'rem "division by zero")) ((< b 0) (rem a (neg b))) ((< a 0) (neg (rem (neg a) b))) ((< a b) a) (else (rem (- a b) b))))
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