I have a modified version of the first Project Euler problem  where I find the sum of all the numbers below the value 2100000000 that are multiples of 3 or 5 using a multi-threaded program.The concept of the partially-multi-threaded version: we have a value sum, where we add together all the multiples of 3 (that are not multiples of 3 and 5). Then, we add all the multiples of 5 (that were multiples of 3 and 5) to sum.On my computer, this code executes at (an average)  4050 milliseconds.  How can the execution time be reduced?MultipleOf3.java//Calculates the sum of multiples of 3 from 3 to 2100000000.public class MultipleOf3  implements Runnable {    private volatile int sum = 0;    public void run() {        int counter = 1;        for(int i = 3; i < 2100000000; i += 3) {          //If the multiple of 3 ends with a 5 or 0, then skip it because           // it is a multiple of 5.          if(counter == 5) {                counter = 1;                continue;           }           sum += i;           counter++;        }    }   public int getSum() {       return sum;   }}MultipleOf5.java// Calculates the sum of multiples of 5 from 5 to 2100000000.    public class MultipleOf5  implements Runnable {         private volatile int sum = 0;         public void run() {              //This time we added all multiples of 5 - including the              //values that were skipped that were multiples of 3 and 5             for(int i = 5; i < 2100000000; i += 5) {                sum += i;             }         }         public int getSum() {             return sum;         }    }Main.javapublic class Main {    private static int parallelSolution() throws InterruptedException{        int totalSum = 0;        MultipleOf3 multipleOf3 = new MultipleOf3();        Thread newThread = new Thread(multipleOf3);        MultipleOf5 multipleOf5 = new MultipleOf5();        Thread secondThread = new Thread(multipleOf5);        newThread.start();        secondThread.start();        newThread.join();        secondThread.join();        totalSum += multipleOf3.getSum();        totalSum += multipleOf5.getSum();         return totalSum;   }    public static void main(String[] args) throws InterruptedException  {        long startTime = System.currentTimeMillis();            int sum = parallelSolution();           long stopTime = System.currentTimeMillis();        long elapsedTime = stopTime - startTime;        System.out.println(The total sum =  + sum);        System.out.println(Elapsed time is  + elapsedTime);    }}
Counting multiples of 3 or 5 using threads
java;performance;multithreading
BugBoth sums are much too large to fit into a 32-bit int.  So this code returns an incorrect answer.  Gaussian solutionIf you first calculate how many multiples of whatever there are, you can use the Gaussian formula to get the sum directly.  public class MultipleOf implements Runnable {    private final int interval;    private final long ceiling;    private long sum;    public MultipleOf(int interval, long ceiling) {        this.interval = interval;        this.ceiling = ceiling;    }    @Override    public void run() {        // n is the number of multiples of interval that are less than ceiling        long n = (ceiling - 1) / interval;        sum = interval * (n * (n+1) / 2);    }    public long getSum() {        return sum;    }}This also requires only one multiples class and removes the magic numbers from the class.  You can run the code like     private static long parallelSolution(long ceiling) throws InterruptedException {        long totalSum = 0;        MultipleOf multipleOf3 = new MultipleOf(3, ceiling);        Thread thread3 = new Thread(multipleOf3);        MultipleOf multipleOf5 = new MultipleOf(5, ceiling);        Thread thread5 = new Thread(multipleOf5);        MultipleOf multipleOf15 = new MultipleOf(15, ceiling);        Thread thread15 = new Thread(multipleOf15);        thread3.start();        thread5.start();        thread15.start();        thread3.join();        thread5.join();        thread15.join();        totalSum += multipleOf3.getSum();        totalSum -= multipleOf15.getSum();        totalSum += multipleOf5.getSum();         return totalSum;    }When I ran this with almost your original main, I got an answer in 2 milliseconds.  Your original code ran in 6000 milliseconds in my test.  Of course, running without threads takes 0 milliseconds.  I also changed the thread names to be more descriptive.  OverheadI'd try running your solution without threads and see if it is faster or slower.  