Check if a number is odd or even in python

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

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

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'

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.

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

Hope it helps,

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