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

Answer 1

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]

Answer 2

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

Answer 3

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.

Answer 4

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.

Answer 5

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 6

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

Hope it helps,