Problem #1 easy

Multiples of 3 and 5

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.

Additional Notes

This is one of the simplest Project Euler problems, often used as an introduction to programming challenges. The solution demonstrates basic loop iteration and conditional logic.

The mathematical approach can be optimized using the formula for the sum of an arithmetic series, but the brute force approach shown here is sufficient for small inputs like 1000.

cpp java javascript php ruby go rust

Implementations

O(n) time complexity

View Raw

#include <iostream>
int sum_natural_35(size_t upper)
{
  unsigned int sum(0);
  for( int i = upper ; --i; )
  {
    if( 0 == i % 3 )
    {
      sum += i;
    }
    else if ( 0 == i % 5 )
    {
      sum += i;
    }
  }
  return sum;
}

int main(int argc, char* argv[])
{
  std::cout << sum_natural_35(1000) << std::endl;
}

View Raw

public class Euler001 {
    public static void main(String[] args) {
        int sum = 0;
        for (int i = 1; i < 1000; i++) {
            if (i % 3 == 0 || i % 5 == 0) {
                sum += i;
            }
        }
        System.out.println(sum);
    }
}

View Raw

function euler001() {
    let sum = 0;
    for (let i = 1; i < 1000; i++) {
        if (i % 3 === 0 || i % 5 === 0) {
            sum += i;
        }
    }
    return sum;
}

console.log(euler001());

View Raw

<?php
$sum = 0;
for ($i = 1; $i < 1000; $i++) {
    if ($i % 3 == 0 || $i % 5 == 0) {
        $sum += $i;
    }
}
echo $sum;
?>

View Raw

#!/usr/bin/env ruby
def euler001_solution
  sum = 0
  (0..999).each do |num|
    if num % 3 == 0 || num % 5 == 0
      sum += num
    end
  end
  sum
end

puts euler001_solution

View Raw

package main

import "fmt"

func main() {
    sum := 0
    for i := 1; i < 1000; i++ {
        if i%3 == 0 || i%5 == 0 {
            sum += i
        }
    }
    fmt.Println(sum)
}

View Raw

fn main() {
    let mut sum = 0;
    for i in 1..1000 {
        if i % 3 == 0 || i % 5 == 0 {
            sum += i;
        }
    }
    println!("{}", sum);
}

All Solutions