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Python Numbers: How to Check If a Number Is Odd

Determining whether an integer is even or odd is a fundamental logic operation in programming. An odd number is an integer that is not exactly divisible by 2; when divided by 2, it leaves a remainder of 1 (e.g., 1, 3, 5, -7).

This guide explains how to use the modulo operator (%) to check for odd numbers, how to handle user input, and how to validate data types to prevent errors.

Method 1: Using the Modulo Operator (Standard)​

The standard way to check for parity (even/odd) in Python is using the modulo operator (%). This operator returns the remainder of a division.

  • If number % 2 == 0, the number is Even.
  • If number % 2 != 0 (or == 1), the number is Odd.
number = 7

# ✅ Check if the remainder is NOT 0
if number % 2 != 0:
    print(f"{number} is an odd number.")
else:
    print(f"{number} is an even number.")

Output:

7 is an odd number.
note

In Python, the modulo operator handles negative numbers consistently. -3 % 2 returns 1, so checking if number % 2 == 1 is also a valid way to identify odd integers.

Method 2: Using Bitwise AND (Optimized)​

For high-performance scenarios or low-level algorithmic tasks, you can use the Bitwise AND operator (&).

In binary representation, all odd numbers have their least significant bit (the rightmost bit) set to 1. Even numbers have it set to 0.

  • 3 in binary is 011 (...1 -> Odd)
  • 4 in binary is 100 (...0 -> Even)
number = 9

# ✅ Check the last bit
if number & 1:
    print(f"{number} is Odd")
else:
    print(f"{number} is Even")

Output:

9 is Odd

Handling User Input and Errors​

When accepting input from users via input(), the data comes in as a string. You must convert it to an integer. If the user types text or a float, the program will crash unless handled.

Error Scenario​

user_input = "abc"

try:
    # ⛔️ Incorrect: Comparing string directly or converting invalid text
    if int(user_input) % 2 != 0:
        print("Odd")
except ValueError as e:
    print(f"Error: {e}")

Output:

Error: invalid literal for int() with base 10: 'abc'

Solution: Try-Except Block​

Wrap the conversion logic in a try-except block to catch ValueError.

user_input = input("Enter an integer: ")

try:
    # ✅ Convert to int first
    number = int(user_input)
    
    if number % 2 != 0:
        print(f"{number} is Odd.")
    else:
        print(f"{number} is Even.")
        
except ValueError:
    print("Invalid input. Please enter a whole number (e.g., 5, 10).")
tip

This logic only works for integers. If a user enters 5.5, int() will raise a ValueError. If you need to check floating point numbers (e.g., 5.0), you would need to convert to float first, then check is_integer().

Conclusion​

To check if a number is odd in Python:

  1. Use num % 2 != 0 for the standard, readable approach.
  2. Use num & 1 for a bitwise approach (often slightly faster).
  3. Validate Input: Always use try-except ValueError when converting user strings to integers.

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