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Coding challenge | How to check if a number is odd or even using JS

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Determining whether a number is odd or even is a fundamental programming task. In JavaScript, there are several ways to accomplish this. This tutorial will walk you through two simple methods - using the modulus operator and leveraging the bitwise AND operator.






1. Using the modulus operator



The modulus operator (%) is the most common and straightforward way to determine if a number is odd or even. This operator returns the remainder of a division operation.



Key concept:




  • A number is even if it is divisible by 2 (remainder = 0).

  • A number is odd if it is not divisible by 2 (remainder ≠ 0).



Code example:




CODE
function isEven(number) {
return number % 2 === 0;
}

function isOdd(number) {
return number % 2 !== 0;
}

// Usage:
console.log(isEven(4)); // true
console.log(isOdd(4)); // false
console.log(isEven(7)); // false
console.log(isOdd(7)); // true






Output:




  • isEven(4) returns true, meaning 4 is even.

  • isOdd(7) returns true, meaning 7 is odd.






2. Using the bitwise AND operator



The bitwise AND operator (&) can also be used to determine odd or even numbers. This approach relies on binary representation:



Key concept:




  • Even numbers have a 0 as the least significant bit (e.g., 2 = 10 in binary).

  • Odd numbers have a 1 as the least significant bit (e.g., 3 = 11 in binary).



How it works:

Performing number & 1 checks the last bit of the number:




  • If number & 1 === 0, the number is even.

  • If number & 1 === 1, the number is odd.



Code example:




CODE
function isEven(number) {
return (number & 1) === 0;
}

function isOdd(number) {
return (number & 1) === 1;
}

// Usage:
console.log(isEven(4)); // true
console.log(isOdd(4)); // false
console.log(isEven(7)); // false
console.log(isOdd(7)); // true






Output:




  • isEven(4) returns true, meaning 4 is even.

  • isOdd(7) returns true, meaning 7 is odd.






Comparison of methods:




























Feature Modulus Operator % Bitwise AND Operator &
Readability Very easy to understand Less intuitive
Readability Slightly slower for large numbers Slightly faster for large numbers
Use Case General-purpose applications Optimised low-level operations





Which method should you use?




  • Use the Modulus operator if you want a solution that is simple and easy to understand. It’s the most common method and is suitable for most scenarios.

  • Use the bitwise AND operator if performance is critical (e.g., in high-performance code or embedded systems).






Final thoughts



Both methods are effective for checking if a number is odd or even in JavaScript. Choose the one that best fits your specific use case. Happy coding! 🎉

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