Euler Problem 1

This one should be easy.

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.

A simple loop and a remainder check.

public class Euler1
{
    public static void main(String[] args)
    {
        Euler1 e = new Euler1();
        System.out.print(“Problem 1:\nSum = ” + e.Problem1()+ “\n”);
    }

    public String Problem1 ()
    {
        int sum = 0;
        for (int i = 1; i < 1000; i++)
        {
            if (i%3 == 0 || i%5 == 0)
            {
                sum += i;
            }
        }
        return String.valueOf(sum);
    }
}

 

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s