JavaScript Program to Check if Number is Even or Odd

Use the modulus operator % in JavaScript to check if a number is even or odd: if number % 2 === 0, it is even; otherwise, it is odd.

Examples

Input4

Output4 is even

Input7

Output7 is odd

Input0

Output0 is even

How to Think About It

To check if a number is even or odd, divide it by 2 and look at the remainder. If the remainder is 0, the number is even because it divides evenly by 2. If the remainder is 1, the number is odd because it leaves a leftover when divided by 2.

Algorithm

1

Get the input number.

2

Calculate the remainder when the number is divided by 2.

3

If the remainder is 0, the number is even.

4

Otherwise, the number is odd.

5

Print the result.

Code

javascript

const number = 7;
if (number % 2 === 0) {
  console.log(`${number} is even`);
} else {
  console.log(`${number} is odd`);
}

Output

7 is odd

Dry Run

Let's trace the number 7 through the code to see how it determines if it's even or odd.

1

Input number

number = 7

2

Calculate remainder

7 % 2 = 1

3

Check remainder

Since remainder is 1, number is odd

4

Print result

"7 is odd" is printed

Number	Number % 2	Even or Odd
7	1	Odd

Why This Works

Step 1: Using modulus operator

The % operator gives the remainder of division, which helps us find if a number divides evenly by 2.

Step 2: Checking remainder

If the remainder is 0, the number is even because it divides exactly by 2.

Step 3: Determining odd

If the remainder is not 0 (usually 1), the number is odd because it leaves a leftover when divided by 2.

Alternative Approaches

Using bitwise AND operator

javascript

const number = 7;
if ((number & 1) === 0) {
  console.log(`${number} is even`);
} else {
  console.log(`${number} is odd`);
}

This method uses bitwise operation to check the last binary digit; it's very fast but less readable for beginners.

Using ternary operator

javascript

const number = 7;
console.log(`${number} is ${number % 2 === 0 ? 'even' : 'odd'}`);

This is a concise way to write the check and print in one line, improving readability.

Complexity: O(1) time, O(1) space

Time Complexity

The operation uses a single modulus calculation which takes constant time regardless of input size.

Space Complexity

No extra memory is needed beyond the input and a few variables, so space is constant.

Which Approach is Fastest?

Both modulus and bitwise AND methods run in constant time; bitwise may be slightly faster but less clear.

Approach	Time	Space	Best For
Modulus operator (%)	O(1)	O(1)	Clarity and simplicity
Bitwise AND (&)	O(1)	O(1)	Performance in low-level code
Ternary operator	O(1)	O(1)	Concise code style