title: Find Sum Odd Factors Number
signature: findSumOddFactorsNumber :: Integer -> Integer
  findSumOddFactorsNumber n = 
test_code: import Text.Printf
  import Control.Exception (catch, SomeException)
  import Data.Typeable
  import qualified Data.List as List
  import qualified Data.Map as Map
  import qualified Data.Set as Set
  
  -- SOLUTION CODE
  -- ============================================
  
  PLACEHOLDER_CODE_BODY
  
  -- TESTING CODE 
  -- ============================================
  validateSolution :: Integer -> Integer -> Bool
  validateSolution actual expected = do
      actual == expected
  
  
  driver :: Integer -> Integer -> String
  driver n expected = do
      if (validateSolution (PLACEHOLDER_FN_NAME n) expected)
          then "PASSED"
      else "FAILED"
  
  main = do
      printf "TEST-0...%s\n" (driver (20) (6) )
  
      printf "TEST-1...%s\n" (driver (6) (4) )
  
      printf "TEST-2...%s\n" (driver (39) (56) )
  
      printf "TEST-3...%s\n" (driver (80) (6) )
  
      printf "TEST-4...%s\n" (driver (88) (12) )
  
      printf "TEST-5...%s\n" (driver (7) (8) )
  
      printf "TEST-6...%s\n" (driver (16) (1) )
  
      printf "TEST-7...%s\n" (driver (27) (40) )
  
      printf "TEST-8...%s\n" (driver (83) (84) )
  
      printf "TEST-9...%s\n" (driver (6) (4) )
  
source_py: def find_sum_odd_factors_number(n):
      res = 1
      while ((n % 2) == 0):
          n = (n // 2)
      for i in range(3, int((math.sqrt(n) + 1))):
          count = 0
          curr_sum = 1
          curr_term = 1
          while ((n % i) == 0):
              count += 1
              n = (n // i)
              curr_term *= i
              curr_sum += curr_term
          res *= curr_sum
      if (n >= 2):
          res *= (1 + n)
      return res
source_cpp: int find_sum_odd_factors_number(int n) {
    int res = 1;
    while ( n % 2 == 0 ) n = n / 2;
    for ( int i = 3;
    i <= sqrt ( n );
    i ++ ) {
      int count = 0, curr_sum = 1;
      int curr_term = 1;
      while ( n % i == 0 ) {
        count ++;
        n = n / i;
        curr_term *= i;
        curr_sum += curr_term;
      }
      res *= curr_sum;
    }
    if ( n >= 2 ) res *= ( 1 + n );
    return res;
  }