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()

Answers:

Thank you for visiting the Q&A section on Magenaut. Please note that all the answers may not help you solve the issue immediately. So please treat them as advisements. If you found the post helpful (or not), leave a comment & I’ll get back to you as soon as possible.

Method 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

Method 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.

Method 3

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

Method 4

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

Hope it helps,

Method 5

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]

Method 6

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.

All methods was sourced from stackoverflow.com or stackexchange.com, is licensed under cc by-sa 2.5, cc by-sa 3.0 and cc by-sa 4.0