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 <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) {
System.out.println( "First 10 even numbers are:" );
first10Even();
}
static void first10Even() {
int cnt = 0 ;
int number = 0 ;
while (cnt < 10 ) {
if (number % 2 == 0 ) {
cnt++;
System.out.print(number + " " );
}
number++;
}
}
}
|
Python3
def first_10_even():
cnt = 0
number = 0
while cnt < 10 :
if number % 2 = = 0 :
cnt + = 1
print (number, end = " " )
number + = 1
print ( "First 10 even numbers are:" )
first_10_even()
|
C#
using System;
public class GFG {
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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