title: Find Sum Even Factors Number
signature: fun findSumEvenFactorsNumber(n: Int): Int {
test_code: package main
  
  import kotlin.text.*
  import kotlin.math.*
  
  
  // SOLUTION CODE
  // ===================================
  PLACEHOLDER_CODE_BODY
  
  // TESTING CODE
  // ===================================
  fun validateSolution(actual:Int, expected:Int):Boolean{
    return actual.equals(expected)
  }
  fun driver(n: Int, expected:Int):String {
      try{
        if (validateSolution(PLACEHOLDER_FN_NAME(n), expected)) {
                return "PASSED"
        }
        return "FAILED"
      } catch (e: Exception){
        return e::class.simpleName as String
      }
  
  }
  
  fun main() {
      var result: String = ""
      result = driver(71, 0)
      println("TEST-0...${result}")
      
      result = driver(78, 112)
      println("TEST-1...${result}")
      
      result = driver(39, 0)
      println("TEST-2...${result}")
      
      result = driver(36, 78)
      println("TEST-3...${result}")
      
      result = driver(49, 0)
      println("TEST-4...${result}")
      
      result = driver(17, 0)
      println("TEST-5...${result}")
      
      result = driver(53, 0)
      println("TEST-6...${result}")
      
      result = driver(66, 96)
      println("TEST-7...${result}")
      
      result = driver(92, 144)
      println("TEST-8...${result}")
      
      result = driver(71, 0)
      println("TEST-9...${result}")
      
          
  }
  
  main()
source_py: def find_sum_even_factors_number(n):
      if ((n % 2) != 0):
          return 0
      res = 1
      for i in range(2, (int(math.sqrt(n)) + 1)):
          count = 0
          curr_sum = 1
          curr_term = 1
          while ((n % i) == 0):
              count = (count + 1)
              n = (n // i)
              if ((i == 2) and (count == 1)):
                  curr_sum = 0
              curr_term = (curr_term * i)
              curr_sum = (curr_sum + curr_term)
          res = (res * curr_sum)
      if (n >= 2):
          res = (res * (1 + n))
      return res
source_cpp: int find_sum_even_factors_number(int n) {
    if ( n % 2 != 0 ) return 0;
    int res = 1;
    for ( int i = 2;
    i <= sqrt ( n );
    i ++ ) {
      int count = 0, curr_sum = 1, curr_term = 1;
      while ( n % i == 0 ) {
        count ++;
        n = n / i;
        if ( i == 2 && count == 1 ) curr_sum = 0;
        curr_term *= i;
        curr_sum += curr_term;
      }
      res *= curr_sum;
    }
    if ( n >= 2 ) res *= ( 1 + n );
    return res;
  }