JavaScript Program to Print Even Numbers from 1 to n

Use a for loop from 1 to n and print numbers where number % 2 === 0, like: for(let i=1; i<=n; i++){ if(i % 2 === 0) console.log(i); }.

Examples

Input5

Output2 4

Input10

Output2 4 6 8 10

Input1

Output

How to Think About It

To print even numbers from 1 to n, start counting from 1 up to n. For each number, check if it divides evenly by 2 using the remainder operator %. If the remainder is zero, it means the number is even, so print it.

Algorithm

1

Get the input number n.

2

Start a loop from 1 to n.

3

For each number, check if it is divisible by 2 with no remainder.

4

If yes, print the number.

5

Continue until the loop reaches n.

Code

javascript

const n = 10;
for (let i = 1; i <= n; i++) {
  if (i % 2 === 0) {
    console.log(i);
  }
}

Output

2
4
6
8
10

Dry Run

Let's trace the program with n = 5 to see which numbers print.

1

Start loop

i = 1

2

Check if even

1 % 2 = 1 (not even, skip)

3

Next number

i = 2

4

Check if even

2 % 2 = 0 (even, print 2)

5

Next number

i = 3

6

Check if even

3 % 2 = 1 (not even, skip)

7

Next number

i = 4

8

Check if even

4 % 2 = 0 (even, print 4)

9

Next number

i = 5

10

Check if even

5 % 2 = 1 (not even, skip)

i	i % 2 === 0?	Printed
1	false
2	true	2
3	false
4	true	4
5	false

Why This Works

Step 1: Loop through numbers

The for loop goes from 1 to n, checking each number one by one.

Step 2: Check even condition

Using i % 2 === 0 tests if the number divides evenly by 2, meaning it is even.

Step 3: Print even numbers

If the condition is true, the number is printed using console.log.

Alternative Approaches

Increment by 2 starting from 2

javascript

const n = 10;
for (let i = 2; i <= n; i += 2) {
  console.log(i);
}

This skips odd numbers entirely, making the loop faster and simpler.

Using while loop

javascript

const n = 10;
let i = 2;
while (i <= n) {
  console.log(i);
  i += 2;
}

A while loop can also print even numbers by increasing by 2 each time.

Using Array and filter

javascript

const n = 10;
const evens = Array.from({length: n}, (_, i) => i + 1).filter(x => x % 2 === 0);
evens.forEach(x => console.log(x));

This creates an array and filters even numbers, but uses more memory.

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

Time Complexity

The loop runs from 1 to n, so it checks each number once, making it O(n).

Space Complexity

No extra space is used except a few variables, so space is O(1).

Which Approach is Fastest?

Incrementing by 2 is faster because it skips odd numbers, reducing iterations by half.

Approach	Time	Space	Best For
Check each number with % 2	O(n)	O(1)	Simple and clear
Increment by 2 starting at 2	O(n/2)	O(1)	Faster, skips odd numbers
Array filter method	O(n)	O(n)	When working with arrays