Solution to Project Euler #1 - multiples of 3 and 5

I have been programming for about ~2 years, and mostly wrote OOP and structural code. Recently, I have decided to pick up a functional programming language, and Haskell being too alien for me, looked to Racket (since it is high time I learned a LISP anyways) and am loving it. Since this is a new area of programming for me, I would appreciate any feedback you could give me on this program. It is the solution to the first Project Euler program.

`;If we list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and 9. The sum of these multiples is 23.
;Find the sum of all the multiples of 3 or 5 below 1000.

#lang racket

(define max 1000)

(define (multiple_of base test)
(equal? (remainder test base) 0))

(define (primes current total)
(if (

• In Scheme, we use hyphens to separate words, not underscores. Also boolean-returning procedures should end in ?. So it should be multiple-of?

• Instead of (equal? x 0), use (zero? x).

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Instead of a recursive loop, you can use for comprehensions:

(for/sum ((i (in-range 1000))
#:when (or (multiple-of? 5 i)
(multiple-of? 3 i)))
i)


Surely, that's much more readable. In my humble opinion. :-)

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Notwithstanding the last comment, your formatting for your primes procedure is not ideal. Here's a more proper formatting:

`(define (primes current total)
(if (

StackExchange Code Review Q#78401, answer score: 3