Find If a Number is Even or Odd Using Bitwise Operator.

Understanding whether a number is even or odd is one of the most common tasks in programming. While most beginners use the arithmetic modulus operator (%), there’s a faster and more efficient alternative the bitwise AND operator (&).

In this post, we'll explore both approaches, compare them, and understand why bitwise is better.

Approach 1: Using Arithmetic Operator.

The most common and straightforward method is using the modulus operator (%). If a number divided by 2 leaves a remainder of 0, it's even. Otherwise, it's odd.

if (n % 2 == 0) → Even  
else → Odd

Example Code:

// C++ Example: Find Number is Odd or Even
#include <iostream>
using namespace std;

void checkEvenOrOdd_Arithmetic(int n) {
    if (n % 2 == 0)
        cout << n << " is Even" << endl;
    else
        cout << n << " is Odd" << endl;
}
        

# Python Example: Find Number is Even or Odd
def check_even_or_odd_arithmetic(n):
    if n % 2 == 0:
        print(f"{n} is Even")
    else:
        print(f"{n} is Odd")
        

// Java Example: Find Number is Even or Odd
public class CheckEvenOrOdd {
    public static void checkEvenOrOddArithmetic(int n) {
        if (n % 2 == 0) {
            System.out.println(n + " is Even");
        } else {
            System.out.println(n + " is Odd");
        }
    }

    public static void main(String[] args) {
        // Example usage
        int number = 10;
        checkEvenOrOddArithmetic(number);
    }
}
        

Output:

6 is Even

Approach 2: Using Bitwise AND Operator.

A faster and smarter way to check if a number is even or odd is using bitwise operators.

The & operator performs a bit-by-bit comparison between two numbers.

For each bit, it returns:

• 1 if both bits are 1
• 0 otherwise

When you perform n & 1, you're checking the Least Significant Bit (LSB) of the number.

• If LSB is 0 → number is even
• If LSB is 1 → number is odd

Because the binary of even number always end with 0 and binary of odd number always end with 1. Let's understand this with few examples:

Example 1: 

Number = 4
Binary of 4: 0100
Binary of 1: 0001

    0100
AND 0001
  = 0000 → 0 → Even

Example 2:

Number = 7
Binary of 7: 0111
Binary of 1: 0001

    0111
AND 0001
  = 0001 → 1 → Odd

Example Code:

        
//C++ Example: Find Even and Odd Using Bitwise AND
#include <iostream>
using namespace std;

void checkEvenOrOdd_Bitwise(int n) {
    if (n & 1)
        cout << n << " is Odd" << endl;
    else
        cout << n << " is Even" << endl;
}

int main() {
    checkEvenOrOdd_Bitwise(4);
    return 0;
}

        

        
#Python Example: Find Even and Odd Using def check_even_or_odd_bitwise(n):
    if n & 1:
        print(f"{n} is Odd")
    else:
        print(f"{n} is Even")

# Example usage
if __name__ == "__main__":
    check_even_or_odd_bitwise(4)


        

        
//Java Example: Find Even and Odd Using Bitwise AND
public class CheckEvenOrOddBitwise {
    public static void checkEvenOrOddBitwise(int n) {
        if ((n & 1) == 1) {
            System.out.println(n + " is Odd");
        } else {
            System.out.println(n + " is Even");
        }
    }

    public static void main(String[] args) {
        checkEvenOrOddBitwise(4);
    }
}

        

Output:

6 is Even

Using n & 1 is a fast and efficient way to check if a number is even or odd by examining its last binary bit. It's a foundational concept in bit manipulation that helps build strong binary thinking.