title: Sum Of All Proper Divisors Of A Natural Number
signature: function sumOfAllProperDivisorsOfANaturalNumber($num) {
test_code: <?php
  
  
  // SOLUTION CODE 
  // ============================================
  PLACEHOLDER_CODE_BODY
  
  // TESTING CODE 
  // ============================================
  function validateSolution($actual, $expected){
      return abs($actual - $expected) < 1e-06;
  }
  
  function driver($num, $expected){
      try{
        if (validateSolution(PLACEHOLDER_FN_NAME($num), $expected)){
          return "PASSED";
        }
        return "FAILED";
      }
      catch (Exception $exception_obj) {
        return get_class($exception_obj);
      }
  }
  
  function main(){
      printf("TEST-0...%s\n",driver(
        2, 
        1.0));
      
      printf("TEST-1...%s\n",driver(
        57, 
        23.0));
      
      printf("TEST-2...%s\n",driver(
        28, 
        28.0));
      
      printf("TEST-3...%s\n",driver(
        43, 
        1.0));
      
      printf("TEST-4...%s\n",driver(
        38, 
        22.0));
      
      printf("TEST-5...%s\n",driver(
        29, 
        1.0));
      
      printf("TEST-6...%s\n",driver(
        45, 
        33.0));
      
      printf("TEST-7...%s\n",driver(
        47, 
        1.0));
      
      printf("TEST-8...%s\n",driver(
        44, 
        40.0));
      
      printf("TEST-9...%s\n",driver(
        3, 
        1.0));
      
  }
  
  main();
  ?>
source_py: def sum_of_all_proper_divisors_of_a_natural_number(num):
      result = 0
      i = 2
      while (i <= math.sqrt(num)):
          if ((num % i) == 0):
              if (i == (num / i)):
                  result = (result + i)
              else:
                  result = (result + (i + (num / i)))
          i = (i + 1)
      return (result + 1)
source_cpp: int sum_of_all_proper_divisors_of_a_natural_number(int num) {
    int result = 0;
    for ( int i = 2;
    i <= sqrt ( num );
    i ++ ) {
      if ( num % i == 0 ) {
        if ( i == ( num / i ) ) result += i;
        else result += ( i + num / i );
      }
    }
    return ( result + 1 );
  }