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

Solution 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 

Solution 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.