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

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

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.

Use the modulo operator:

if wordLength % 2 == 0:
    print "wordLength is even"
else:
    print "wordLength is odd"

For your problem, the simplest is to check if the word is equal to its reversed brother. You can do that with word[::-1], which create the list from word by taking every character from the end to the start:

def is_palindrome(word):
    return word == word[::-1]

It shouldn't matter if the word has an even or odd amount fo letters:

def is_palindrome(word):
    if word == word[::-1]:
        return True
    else:
        return False

One of the simplest ways is to use de modulus operator %. If n % 2 == 0, then your number is even.

Hope it helps,

The middle letter of an odd-length word is irrelevant in determining whether the word is a palindrome. Just ignore it.

Hint: all you need is a slight tweak to the following line to make this work for all word lengths:

secondHalf = word[finalWordLength + 1:]

P.S. If you insist on handling the two cases separately, if len(word) % 2: ... would tell you that the word has an odd number of characters.