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.
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;
}
$ gcc -march=native -Ofast -std=c11 -o natural natural.c
$ time ./natural
real 0m0.001s
user 0m0.000s
sys 0m0.001s
Clojure
#!/usr/bin/env clojure
(defn multiple? [n]
(or (= (rem n 3) 0) (= (rem n 5) 0)))
(println (reduce + (filter multiple? (range 1000))))
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
JavaScript
let s = 0
for (var i = 1; i < 1000; i++) {
if (i % 3 === 0 || i % 5 === 0) {
s += i
}
}
console.log(s)
Python
Ruby
Rust
fn main() {
let result: u64 = (1..1000).filter(|x| x % 3 == 0 || x % 5 == 0).sum();
println!("{}", result);
}
$ rustc -C target-cpu=native -C opt-level=3 -o natural natural.rs
$ time ./natural
real 0m0.001s
user 0m0.000s
sys 0m0.001s