Check if a number is odd or even in python [duplicate]

Question

I'm trying to make a program which checks if a word is a palindrome and I've gotten so far and it works with words that have an even amount of numbers. I know how to make it do something if the amount of letters is odd but I just don't know how to find out if a number is odd. Is there any simple way to find if a number is odd or even?

Just for reference, this is my code:

a = 0  while a == 0:     print("\n \n" * 100)     print("Please enter a word to check if it is a palindrome: ")     word = input("?: ")      wordLength = int(len(word))     finalWordLength = int(wordLength / 2)     firstHalf = word[:finalWordLength]     secondHalf = word[finalWordLength + 1:]     secondHalf = secondHalf[::-1]     print(firstHalf)     print(secondHalf)      if firstHalf == secondHalf:         print("This is a palindrom")     else:         print("This is not a palindrom")       print("Press enter to restart")     input() 

Thanks

Answer 1

if num % 2 == 0:     pass # Even  else:     pass # Odd 

The % sign is like division only it checks for the remainder, so if the number divided by 2 has a remainder of 0 it's even otherwise odd.

Or reverse them for a little speed improvement, since any number above 0 is also considered "True" you can skip needing to do any equality check:

if num % 2:     pass # Odd else:     pass # Even  

Answer 2

Similarly to other languages, the fastest "modulo 2" (odd/even) operation is done using the bitwise and operator:

if x & 1:     return 'odd' else:     return 'even' 

Using Bitwise AND operator

• The idea is to check whether the last bit of the number is set or not. If last bit is set then the number is odd, otherwise even.
• If a number is odd & (bitwise AND) of the Number by 1 will be 1, because the last bit would already be set. Otherwise it will give 0 as output.