# Program to print first 10 even numbers

Last Updated : 29 Jan, 2024

Program to print first 10 even numbers. A number is even if it is divisible by 2 for example 4, 100, 24 etc.

Output Format:

0 2 4 6 8 10 12 14 16 18

Approach: Checking Parity using Modulo operator(%)

Using the modulo % operator we can find the remainder of any number when divided by 2, giving us the parity according to two cases:

• remainder = 0: Even number
• remainder = 1: Odd number

While we do not get first 10 even numbers, we can use the above method to check the parity and print the even numbers.

Step-by-step algorithm:

• Create a function first10Even() which prints the first 10 even numbers.
• Keep count of total even numbers printed using a variable cnt initialized to 0.
• Until cnt reaches 10, iterate on whole numbers:
• if a whole number is even print that whole number and increment cnt by 1.

## C++

 #include using namespace std;   void first10Even() {     int cnt = 0;     int number = 0;     while (cnt < 10) {         if (number % 2 == 0) {             cnt++;             cout << number << " ";         }         number++;     } }   int main() {     cout << "First 10 even numbers are:\n";     first10Even(); }

## Java

 public class First10Even {     public static void main(String[] args) {         // Display a message indicating the purpose of the program         System.out.println("First 10 even numbers are:");           // Call the method to print the first 10 even numbers         first10Even();     }       // Method to print the first 10 even numbers     static void first10Even() {         // Initialize a counter for the number of even numbers and a variable to store the current number         int cnt = 0;         int number = 0;           // Continue the loop until 10 even numbers are printed         while (cnt < 10) {             // Check if the current number is even             if (number % 2 == 0) {                 // Increment the counter and print the even number                 cnt++;                 System.out.print(number + " ");             }               // Increment the number for the next iteration             number++;         }     } }

## Python3

 def first_10_even():     # Initialize a counter for the number of even numbers and a variable to store the current number     cnt = 0     number = 0       # Continue the loop until 10 even numbers are printed     while cnt < 10:         # Check if the current number is even         if number % 2 == 0:             # Increment the counter and print the even number             cnt += 1             print(number, end=" ")           # Increment the number for the next iteration         number += 1   # Display a message indicating the purpose of the program print("First 10 even numbers are:")   # Call the function to print the first 10 even numbers first_10_even()

## C#

 using System;   public class GFG {     // Function to print the first 10 even numbers     static void First10Even()     {         int cnt = 0;         int number = 0;         while (cnt < 10) {             if (number % 2 == 0) {                 cnt++;                 Console.Write(number + " ");             }             number++;         }     }       public static void Main()     {         Console.WriteLine("First 10 even numbers are:");         First10Even();     } }

## Javascript

 function first10Even() {     let cnt = 0;     let number = 0;     while (cnt < 10) {         if (number % 2 === 0) {             cnt++;             process.stdout.write(number + " ");         }         number++;     } }   console.log("First 10 even numbers are:"); first10Even();

Output

First 10 even numbers are:
0 2 4 6 8 10 12 14 16 18

Time Complexity: O(1)
Auxiliary Space: O(1)

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