This weeks ICSP assignment was one of the Project Euler challenges. Here is the task:

This weeks Bonus Task is to solve the First Problem using Java:
“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.”

In the code below, I have achieved this objective and also expanded on it. My program allows you to pass on two multiples and a max integer value to the program for which to calculate the results. You can execute:

java Multiples 3 5 1000

That command will find the sum of multiples of 3 and 5 below 1000, as stated in the task. You can also change the values to calculate different sums. The code for Multiples.java is below:

/*
 * Developer:   Conor Mac Aoidh
 * Homepage:    http://blog.macaoidh.name/2009/11/21/java-project-euler-multiples/
 */

public class Multiples{
        public static void main(String[] args){
                Multiples multiples=new Multiples();
                int one=0;
                int two=0;
                int max=0;
                if(args.length==3){
                        try{
                                one=Integer.parseInt(args[0]);
                                two=Integer.parseInt(args[1]);
                                max=Integer.parseInt(args[2]);
                        }
                        catch(NumberFormatException e){
                                System.err.println("Argument must be an integer");
                                System.exit(1);
                        }
                }
                else
                        System.out.println("You must supply integers in the format:\n multiple_one multiple_two max_integer");

                int sum=multiples.sum(one,two,max);

                System.out.println("Find the sum of all the multiples of "+one+" or "+two+" below "+max);
                System.out.println("Sum:"+sum);
        }

        public int sum(int one,int two,int max){
                int i=0;
                int sum=0;
                for(i=0;i<max ;i++){
                        if(i % one==0)
                                sum=sum+i;
                        if(i % two==0)
                                sum=sum+i;
                }
                return sum;
        }
}

Simple as that!