find the sum of the multiples of 3 and 5 below 1000

Question

Ok guys, so I\'m doing the Project Euler challenges and I can\'t believe I\'m stuck on the first challenge. I really can\'t see why I\'m getting the wrong answer despite my

Answer 1

public class Solution {
    public static void main(String[] args) {
        /* Enter your code here. Read input from STDIN. Print output to STDOUT. Your class should be named Solution. */
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();
        while (t>0){
            int sum = 0;
            int count =0;
            int n = sc.nextInt();
            n--; 
            System.out.println((n/3*(6+(n/3-1)*3))/2 + (n/5*(10+(n/5-1)*5))/2 - (n/15*(30+(n/15-1)*15))/2);
            t--;
    }
}
}

Answer 2

If number is 10 then multiple of 3 is 3,6,9 and multiple of 5 is 5,10 total sum is 33 and program gives same answer:

package com.parag;

/*
 * @author Parag Satav
 */
public class MultipleAddition {

    /**
     * @param args
     */
    public static void main( final String[] args ) {
    // TODO Auto-generated method stub

    ArrayList<Integer> x = new ArrayList<Integer>();
    ArrayList<Integer> y = new ArrayList<Integer>();
    int totalforthree = 0;
    int totalforfive = 0;
    int number = 8;

    int total = 0;

    for ( int temp = 1; temp <= number; temp++ ) {
        if ( temp % 3 == 0 ) {
            x.add( temp );
            totalforthree += temp;
        }

        else if ( temp % 5 == 0 ) {
            y.add( temp );
            totalforfive += temp;
        }
    }

    total = totalforfive + totalforthree;
    System.out.println( "multiples of 3 : " + x );
    System.out.println( "multiples of 5 : " + y );
    System.out.println( "The multiples of 3 or 5 up to " + number + " are: " + total );

}

}

Answer 3

How I solved this is that I took an integer value (initialized to zero) and kept on adding the incremented value of i, if its modulo with 3 or 5 gives me zero.

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

Answer 4

Logics given above are showing wrong answer, because multiples of 3 & 5 are taken for calculation. There is something being missed in above logic, i.e., 15, 30, 45, 60... are the multiple of 3 and also multiple of 5. then we need to ignore such while adding.

    public static void main(String[] args) {
    int Sum=0, i=0, j=0;
    for(i=0;i<=1000;i++)
        if (i%3==0 && i<=999)
            Sum=Sum+i;
    for(j=0;j<=1000;j++)
        if (j%5==0 && j<1000 && j*5%3!=0)
            Sum=Sum+j;
    System.out.println("The Sum is "+Sum);
}

Answer 5

Okay, so this isn't the best looking code, but it get's the job done.

public class Multiples {

public static void main(String[]args) {
    int firstNumber = 3;
    int secondNumber = 5;
    ArrayList<Integer> numberToCheck = new ArrayList<Integer>();
    ArrayList<Integer> multiples = new ArrayList<Integer>();
    int sumOfMultiples = 0;
    for (int i = 0; i < 1000; i++) {
       numberToCheck.add(i);

       if (numberToCheck.get(i) % firstNumber == 0 || numberToCheck.get(i) % secondNumber == 0) {
           multiples.add(numberToCheck.get(i));
       }

    }

    for (int i=0; i<multiples.size(); i++) {

     sumOfMultiples += multiples.get(i);

    }
    System.out.println(multiples);
    System.out.println("Sum Of Multiples: " + sumOfMultiples);

}

}

Answer 6

If you are using Java 8 you can do it in the following way:

Integer sum = IntStream.range(1, 1000) // create range
                  .filter(i -> i % 3 == 0 || i % 5 == 0) // filter out
                  .sum(); // output: 233168

To count the numbers which are divisible by both 3 and 5 twice you can either write the above line twice or .map() the 2 * i values:

Integer sum = IntStream.range(1, 1000)
                  .filter(i -> i % 3 == 0 || i % 5 == 0)
                  .map(i -> i % 3 == 0 && i % 5 == 0 ? 2 * i : i)
                  .sum(); // output: 266333