Question

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.

Answer

233168

---

C

#include <stdio.h>

int main(int argc, char **argv)
{
    int sum = 0;
    for (int i = 0; i < 1000; i++) {
        if (i % 3 == 0 || i % 5 == 0) {
            sum += i;
        }
    }
    printf("%d\n", sum);
    return 0;
}

---

Java

// This forces every solution class to implement a common interface,
// which is helpful for unit testing like in the EulerTest implementation.
public interface EulerSolution {
	
	public String run();
	
}

public final class p001 implements EulerSolution {
	
	public static void main(String[] args) {
		System.out.println(new p001().run());
	}
	
	
	/* 
	 * Computers are fast, so we can implement this solution directly without any clever math.
	 * A conservative upper bound for the sum is 1000 * 1000, which fits in a Java int type.
	 */
	public String run() {
		int sum = 0;
		for (int i = 0; i < 1000; i++) {
			if (i % 3 == 0 || i % 5 == 0)
				sum += i;
		}
		return Integer.toString(sum);
	}
	
}

---

Mathematica

Total[Select[Range[999], Function[x, Mod[x, 3] == 0 || Mod[x, 5] == 0]]]

---

Python

#!/usr/bin/env python
print(sum(i for i in range(1000) if (i % 3 == 0) or (i % 5 == 0)))

---

JavaScript

let s = 0
for (var i = 1; i < 1000; i++) {
  if (i % 3 === 0 || i % 5 === 0) {
    s += i
  }
}
console.log(s)

---

Ruby

#!/usr/bin/env ruby
sum = 0
1000.times do |i|
  if i % 3 == 0 or i % 5 == 0
    sum += i
  end
end
puts sum

---

Go

package main

import "fmt"

func main() {
	sum := 0
	for i := 0; i < 1000; i++ {
		if i%3 == 0 || i%5 == 0 {
			sum += i
		}
	}
	fmt.Println(sum)
}

---

Haskell

main ::  IO ()
main = print $ sum [n | n <- [1..999], or [(n `mod` 3 == 0), (n `mod` 5 == 0)]]

---

Clojure

#!/usr/bin/env clojure
(defn multiple? [n]
  (or (= (rem n 3) 0) (= (rem n 5) 0)))

(println (reduce + (filter multiple? (range 1000))))

---

Scheme

(display
  (reduce + 0
          (filter
            (lambda (n)
              (or (= (remainder n 3) 0) (= (remainder n 5) 0)))
            (iota 1000))))
(newline)