human
Write me a code to count of all N digit numbers such that num + Rev ( num ) = 10 ^ N and the other steps required given the following:
Count of all N digit numbers such that num + Rev ( num ) = 10 ^ N | Function to return the count of such numbers.
If n is odd.
Driver code"
Solution in Python:
assistant
#Here is code to count of all N digit numbers such that num + Rev ( num ) = 10 ^ N and the other steps required.
 def countNumbers ( n ) :
	if n % 2 == 1 :
		return 0
	return ( 9 * pow ( 10 , n // 2 - 1 ) )
if __name__ == " _ _ main _ _ " :
	n = 2
	print ( countNumbers ( n ) )