Find the sum of all the multiples of 3 or 5 below 1000

Question

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. I have the following code but the answer do

Answer 1

It should be sum = sum + i instead of 1.

Answer 2

Interesting fact of the difference in time between this 2 methods... The second method just calculate only with needed numbers and dont iterate through a loop. Example: 3 * (1+2+3+4...333) (Because 3 * 333 = 999 and 999 < 1000.

   static void Main(string[] args)
    {
        Stopwatch sw = new Stopwatch();

        Calc calculator = new Calc();
        long target = 999999999;
        sw.Start();
        Console.WriteLine("Result: " +  (calculator.calcMultipleSumFast(3,target) + calculator.calcMultipleSumFast(5, target) - calculator.calcMultipleSumFast(15, target)));
        sw.Stop();
        Console.WriteLine("Runtime: " + sw.Elapsed);
        Console.WriteLine();
        sw.Start();
        Console.WriteLine("Result: " + (calculator.calcMultiplesSum(3, 5,1000000000)));
        sw.Stop();
        Console.WriteLine("Runtime: " + sw.Elapsed);

        Console.ReadKey();
    }

    public class Calc
{
    public long calcMultiplesSum(long n1, long n2, long rangeMax)
    {
        long sum = 0;
        for (long i = 0; i < rangeMax; i++)
        {
            if ((i % n1 == 0) || (i % n2 == 0))
            {
                sum = sum + i;
            }
        }
        return sum;
    }

    public long calcMultipleSumFast(long n, long rangeMax)
    {
        long p = rangeMax / n;

        return (n * (p * (p + 1))) / 2;
    }
}

Answer 3

Python implementation of the problem. Short, precise and fat.

sum=0
for i in range(1000):
    if (i%3==0) or (i%5==0):
        sum=sum+i
print sum

Answer 4

int Sum(int N) {
long long c = 0;
    N--; //because you want it less than 1000 if less than or equal delete this line
    int n = N/3,b = N/5,u = N/15;
    c+= (n*(n+1))/2 * 3;
    c+= (b*(b+1))/2 * 5;
    c-= (u*(u+1))/2 * 15;
    return c;
}

Answer 5

Two things:

• you're including 1000 in the loop, and
• you're adding one to the sum each time, rather than the value itself.

Change the loop to

for(i=0;i<1000;i++)

And the sum line to

sum=sum+i;

Answer 6

package com.venkat.test;

public class CodeChallenge {

    public static void main(String[] args) {

        int j, sum=0;

        for ( j = 0; j <=1000; j++) {               
            if((j%5==0)||(j%3==0))
            {
                sum=sum+j;
            }               
        }           
        System.out.println(sum);            
    }    
}