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  1. #include <stdio.h>
  2.  
  3. int sum_of_series(int limit, int difference)
  4. {
  5. int n = 1 + limit / difference; // integer math gives number of terms including term 0
  6. return ((n * (n - 1)) / 2) * difference;
  7. }
  8.  
  9. int main(void){
  10.  
  11. int sum_of_series_3 = sum_of_series(999,3);
  12. int sum_of_series_5 = sum_of_series(999,5);
  13. int sum_of_series_15 = sum_of_series(999,15);
  14.  
  15. int answer = sum_of_series_3 + sum_of_series_5 - sum_of_series_15;
  16.  
  17. printf("sum of all the multiples of 3 or 5 below 1000\n");
  18. printf("answer = %d 3s=%d 5s=%d 15s=%d\n",answer, sum_of_series_3, sum_of_series_5, sum_of_series_15);
  19. return 0;
  20. }
Success #stdin #stdout 0s 2248KB
stdin
Standard input is empty
stdout
sum of all the multiples of 3 or 5 below 1000
answer = 233168  3s=166833 5s=99500 15s=33165