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 <iostream> 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)
Article Tags :
Recommended Articles