How can you find whether a number is odd or even in a different way(other than (x%2==0))?

Question

The main aim of this question is to find different unknown logics to find numbers are even or not!!

Answer 1

# Using the bitwise AND operator (&)
# The least significant bit (rightmost bit) of an even number is always 0.
# The least significant bit of an odd number is always 1.
# Performing a bitwise AND with 1 isolates this last bit.
if (value & 1) == 0:
print(f"'{value}' is an EVEN integer.")
else:
print(f"'{value}' is an ODD integer.")

Answer 2

Is this a question or a challenge?

The challenge has been posed before:
https://www.sololearn.com/post/1723667/?ref=app

Answer 3

if (value//2)*2 == value:
print("Even")
else:
print("Odd")

Answer 4

Shift bits rightward once to lose the rightmost bit. Then shift back leftward and compare the original value. If they are the same then the rightmost bit was 0, and therefore it is an even number.

if (value>>1<<1 == value):
print("Even")
else:
print("Odd")