JavaScript: How to check if a number is even without using the modulo operator?

For a given integer number, write a program in JavaScript to check whether a number is odd or even and return the same to the user. It's pretty easy to check whether a number is even by using the modulo operator. But, in this article, we will be checking whether a number is even or not without using the modulo operator.

Using the For Loop

In this approach, we are going to use the for loop to check whether a number is even or not. The idea is to take a Boolean flag variable as true and toggle it n times. After n iterations, if the flag returns to its original state (true), the number is even.

How It Works

We start with isEven = true. Each iteration flips the boolean value. After an even number of flips, we return to true; after an odd number, we get false.

// Returns true if n is even
function isEven(n) {
   let flag = true;
   for (let i = 1; i <= n; i++)
      flag = !flag;
   
   if (flag)
      console.log(n + " is an Even number");
   else
      console.log(n + " is Odd");
}

// function calls
isEven(101);
isEven(158);

101 is Odd
158 is an Even number

Using Multiplication and Division

Here, we divide a number by 2, convert the result to an integer, then multiply by 2. If the final result equals the original number, it's even.

How It Works

For even numbers: 6 ÷ 2 = 3, 3 × 2 = 6 (matches original)
For odd numbers: 7 ÷ 2 = 3.5, parseInt(3.5) = 3, 3 × 2 = 6 (doesn't match 7)

// Returns true if n is even
function isEven(n) {
   // Check if n/2 multiplied by 2 equals original number
   if (parseInt(n / 2, 10) * 2 == n) {
      console.log(n + " is an Even number");
   } else {
      console.log(n + " is Odd");
   }
}

// function calls
isEven(101);
isEven(158);

101 is Odd
158 is an Even number

Using the Bitwise AND Operator

In this approach, we use the bitwise AND operator (&) to check if a number is even. We perform n & 1 - if the result is 0, the number is even; if 1, it's odd.

How It Works

In binary, even numbers end with 0, odd numbers end with 1. The bitwise AND with 1 isolates the last bit:

• Even: 6 & 1 = 110 & 001 = 000 = 0
• Odd: 7 & 1 = 111 & 001 = 001 = 1

// Returns true if n is even
function isEven(n) {
   // n&1 is 0 for even, 1 for odd
   if (!(n & 1)) {
      console.log(n + " is an Even number");
   } else {
      console.log(n + " is Odd");
   }
}

// function calls
isEven(101);
isEven(158);

101 is Odd
158 is an Even number

Comparison of Methods

Conclusion

The bitwise AND method is the most efficient approach for checking even numbers without modulo. For readability, the multiplication/division method is preferred, while the for loop approach is mainly educational.