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

you should use the same for loop for both to aviod double counting numbers that are multiple of both. such as 15,30...

   for(int temp =0; temp < 1000 ; temp++){
        if(temp % 3 == 0){
            x.add(temp);
            totalforthree += temp;
        }else if(temp % 5 == 0){
            y.add(temp);
            totalforfive += temp;
        }
    }

Answer 2

    int count = 0;
for (int i = 1; i <= 1000 / 3; i++)
{
count = count + (i * 3);
if (i < 1000 / 5 && !(i % 3 == 0))
{
count = count + (i * 5);
}
}

Answer 3

Don't you all think instead of using loops to compute the sum of multiples, we can easily compute sum of n terms using a simple formula of Arithmetic Progression to compute sum of n terms.

I evaluated results on both using loop and formula. Loops works simply valuable to short data ranges. But when the data ranges grows more than 10¹⁰ program takes more than hours to process the result with loops. But the same evaluates the result in milliseconds when using a simple formula of Arithmetic Progression.

What we really need to do is:
Algorithm :

1. Compute the sum of multiples of 3 and add to sum.
2. Compute the sum of multiples of 5 and add to sum.
3. Compute the sum of multiples of 3*5 = 15 and subtract from sum.

Here is code snippet in java from my blog post CodeForWin - Project Euler 1: Multiples of 3 and 5

n--; //Since we need to compute the sum less than n.
//Check if n is more than or equal to 3 then compute sum of all divisible by
//3 and add to sum.  
if(n>=3) {  
    totalElements = n/3;  
    sum += (totalElements * ( 3 + totalElements*3)) / 2;  
}  

//Check if n is more than or equal to 5 then compute sum of all elements   
//divisible by 5 and add to sum.  
if(n >= 5) {  
    totalElements = n/5;  
    sum += (totalElements * (5 + totalElements * 5)) / 2;  
}  

//Check if n is more than or equal to 15 then compute sum of all elements  
//divisible by 15 and subtract from sum.  
if(n >= 15) {  
    totalElements = n/15;  
    sum -= (totalElements * (15 + totalElements * 15)) / 2;  
}  

System.out.println(sum); 

Answer 4

In a mathematical perspective,
You did not consider about common factors between 3 and 5.
Because there is double counting.

ex; number 15 , 30 , 45 , 60 , 75 , 90 , 105 , 120 , 135 , 150 , 165 , 180 , 195 , 210 , 225 , 240 , 255 , 270 , 285 , 300 , 315 , 330 , 345 , 360 , 375 , 390 , 405 , 420 , 435 , 450 , 465 , 480 , 495 , 510 , 525 , 540 , 555 , 570 , 585 , 600 , 615 , 630 , 645 , 660 , 675 , 690 , 705 , 720 , 735 , 750 , 765 , 780 , 795 , 810 , 825 , 840 , 855 , 870 , 885 , 900 , 915 , 930 , 945 , 960 , 975 , 990 , are common factors.

total of common factors = 33165.
Your answer is 266333
Correct answer is 233168.
Your Answer - Total of common factors
266333-33165=233168.

(this is a code for getting common factors and Total of common factors )

public static void main(String[] args) {

System.out.println("The sum of the Common Factors : " + getCommonFactorSum());

}

private static int getCommonFactorSum() {
int sum = 0;
for (int i = 1; i < 1000; i++) {
    if (i % 3 == 0 && i % 5 == 0) {
        sum += i;
        System.out.println(i);
    }
}

Answer 5

I did this several ways and several times. The fastest, cleanest and simplest way to complete the required code for Java is this:

public class MultiplesOf3And5 {

public static void main(String[] args){

System.out.println("The sum of the multiples of 3 and 5 is: " + getSum());

}

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;
}

If anyone has a suggestion to get it down to fewer lines of code, please let me know your solution. I'm new to programming.

Answer 6

You are counting some numbers twice. What you have to do is add inside one for loop, and use an if-else statement where if you find multiples of 3, you do not count them in 5 as well.

 if(temp % 3 == 0){
     x.add(temp);
     totalforthree += temp;
 } else if(temp % 5 == 0){
     y.add(temp);
     totalforfive += temp;
 }