Project Euler 1 in Java

Question

I have following exercise:

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.

So I wrote below code which finds the sum of all multiples of 3 or 5 below 1000.

Main.java:

package pl.hubot.projecteuler.problem1;

public class Main {
    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.print(sum);
    }
}

How it code works?

I declared sum variable, iterated through i = 1 to i = 999 and if i is multiply of 3 or 5 then I add it to the sum. Finally, I printed out the sum variable.

Questions

• Is it a efficient method to solve this exercise?
• How can I simplify code?
• How can I increase performance of this code?

Answer 1

I think it is very concise, well done!

For this small problem this will be fast enough.

Code only

I would only add braces around the if body:

 for (int i = 1; i < 1000; i++) {
        if (i % 3 == 0 || i % 5 == 0) { 
            sum += i;
        }
 }

For performance, please note that modulo is kind of slow. You could implement two variables that hold multiples of 3 and 5.

    public static void main(String[] args) {
        int threes = 0;
        int fives  = 0;
        int sum    = 0;
        for (int i = 1; i < 1000; i++) {
            if (i>threes)
                threes+=3;
            if (i>fives)
                fives+=5;
            if (i == threes || i == fives )
                sum+=i;
        }
        System.out.print(sum);
    }

Math

Better will be to go purely mathemathical:

The formula for the sum of an arithmetic series is:

Sn = (n/2) * (a₁ + a_n)

Sn = the sum of the n terms in the sequence.

a₁ = the first term in the sequence.

a_n = the nth term in the sequence.

We need to determine the number of items 'n' in the sequence. For simplicity I name this occurrences.

This is calculated as dividing the limit - 1 by number. Because the integer division, it is rounded down.

The number a_n is the last multiple that fits below the limit, which is equal to occurrences * number

Calculate the sum om the series for threes, fives and fifteens. The solution is:

Sum(threes) + Sum(fives) - Sum(fifteens)

In code:

public static int sum( int limit, int number )
{
    int occurrences = (limit-1) / number;
    //Below is a rewrite ( thnx to Olivier Grégoire) of:
    // (int) ( ( occurrences / 2.0 ) * (  occurrences * number + number) );
    return number * occurrences * (occurrences + 1) / 2;
}
public static void main( String[] args )
{
    System.out.println(  sum(1000,3) + sum(1000,5) - sum(1000, 15));
}

Answer 2

Consider generating multiples of 3 or 5 that way you avoid needing to do % 3 or % 5

Other than that the code is neat and about as simple as you can get really.