title: Sum Of All Proper Divisors Of A Natural Number
signature: function sum_of_all_proper_divisors_of_a_natural_number(num)
test_code: using Printf
  
  # SOLUTION CODE
  # ============================================
  PLACEHOLDER_CODE_BODY
  
  # TESTING CODE 
  # ============================================
  function validate_solution(actual, expected)
      return abs(actual - expected) < 1e-06
  end
  
  function driver(num, expected)
      try
  
          if validate_solution(PLACEHOLDER_FN_NAME(num), expected)
              return "PASSED";
          end
          return "FAILED";
      catch exception_obj
          return string(Base.typename(typeof(exception_obj)).wrapper);
      end
  end
  
  function main()
      
      result = driver(2, 1.0);
      @printf("TEST-0...%s\n",result);
      
      result = driver(57, 23.0);
      @printf("TEST-1...%s\n",result);
      
      result = driver(28, 28.0);
      @printf("TEST-2...%s\n",result);
      
      result = driver(43, 1.0);
      @printf("TEST-3...%s\n",result);
      
      result = driver(38, 22.0);
      @printf("TEST-4...%s\n",result);
      
      result = driver(29, 1.0);
      @printf("TEST-5...%s\n",result);
      
      result = driver(45, 33.0);
      @printf("TEST-6...%s\n",result);
      
      result = driver(47, 1.0);
      @printf("TEST-7...%s\n",result);
      
      result = driver(44, 40.0);
      @printf("TEST-8...%s\n",result);
      
      result = driver(3, 1.0);
      @printf("TEST-9...%s\n",result);
      
  end
  
  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 );
  }