Euler Problem 1: Multiples of 3 and 5 : C programming solution

No Comments

Problem Statement :

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.

Solution Approach:

Being the first problem of the series, its quite a simple one. there are many ways to approach this problem.

Most simplest one is  to iterate through all the numbers from 3 to 1000 and check if its divisible by 3 or 5. If they are, just add them to the sum.

Solution :

#include "stdio.h"

#define LIMIT 1000

int main ()
{

   int iter = 3;
   int sum = 0;

   while ( iter < LIMIT )
   {
      if( !(iter % 3) || !(iter % 5))
      {
         sum+=iter;
      }
   iter++;
   }
   
   printf(" Sum is %d \n",sum);

   return 0;
}

Answer:  233168

Let me know your approach you happen to find a better way to solve this using C programming.

Categories: Programming Tags: Tags: , , ,

Leave a Reply

Your email address will not be published. Required fields are marked *

*

code