（Problem 1）Multiples of 3 and 5

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.

#include <stdio.h>#include <string.h>#include <ctype.h>   void solve(){  int sum,i;  sum=0;  for(i=3; i<1000; i++)  {     if(i%3==0 || i%5==0)     {        sum+=i;     }  }  printf("%d\n",sum);     }   int main(){  solve();  return 0;}

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