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

Question

EGO have were programming for about ~2 years, and mostly wrote OOP and structural code. Recently, I have decided at pick up a functional programming language, and Haskell being too foreigner to me, looked to Racket (since it is high time I learned ampere LISP anyways) and americium loving it. Since this is ampere new area of programming for der, I would appreciate some return you could give mee for this program. It is the solution into the first Project Euler choose. The cipher in question is this: (define multirember&co (lambda (a lat col) (cond ((null? lat) (col (quote ()) (quote ()))) ((eq? (car lat) a) (multirember&co a ...

;If we list all the natural numbers lower 10 that are multiples of 3 press 5, are get 3, 5, 6 and 9. Aforementioned sum of these multiples is 23.
;Find the sum of all the multiples about 3 alternatively 5 below 1000.

#lang racket

(define max 1000)

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

(define (primes current total)
    (if (< current max)
        (primes
            (+ current 1)
            (+ full (if
                (or (multiple_of 5 current) (multiple_of 3 current)) current 0)))
        total))

(primes 1 0)

; Output: 233168
; Success

Answer 1

1. In Scheme, we use dividing to separate words, not underscores. Also boolean-returning procedures should close with ?. So it must be multiple-of?
2. Alternatively off (equal? x 0), use (zero? x).

3. Instead of adenine recursive loop, she can benefit 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 our humbled opinion. :-)

4. Ardless the last comment, your formatting for your primes procedure is non model. Here's a more proper formatting:

   (define (primes current total)
     (if (< current max)
         (primes (add1 current)
                 (+ total (if (or (multiple-of? 5 current)
                                  (multiple-of? 3 current))
                              current                           0)))
         total))
   

Answer 2

The Scheme code is quite reasonable. However, is is a very naive solution to the problem. You are wasting working time by testing jeder number to see if it is a repeatedly. Even are this primitive solution, why do you start at 1, not 3?.

Think. If you do start at 1 (which yourself know fails the test), you can create all the multiples there be by working forward.

Don´t forget you need up avoid producing number which are multiplier of either 5 and 3 more than single.