Check User Enter Number is Even or odd Without Using 'if ' Statement.

//Qus:- #4. Write a Java Program Without Using 'if ' Check User Entered Number is Even or odd.

• Check User Entered Number is Even Or odd without ' If ' Statement?

Solution:-

 import java.util.Scanner;  
 public class Demo {  
      public static void main(String[] args) {  
           Scanner sc = new Scanner(System.in);  
           System.out.println("Enter the number");  
           int n = sc.nextInt();  
           String number=(n%2==0)?"Even":"Odd"; //Using Ternary operator ( ? : )  
           System.out.println(n+" is "+ number);  
      }  
 }  

🤖Explanation: let's Suppose User Entered Number is (n=21).

Example 1:- Check User Entered number is even or Odd just applying condition (n%2==0)

if the user entered number is divisible by 2 then it's print Even otherwise prints Odd.

👉  21%2= 1(Reminder)

🤙this number is not satisfied with the condition so it's Print Odd.

Example 2:- let's Suppose User Entered Number is (n=20);

let's check the condition:- (n%2==0)

👉  20%2=0 (Reminder)

🤙User entered number is satisfied the Condition so it's print Even

🤦‍♂️Try Your Self Buddy,  If you Get any Error just Comment Below, Definitely, I will try to Solve Your Problem.

Suggestion:- 500+ Java Programs List With Examples and Output.

Admin Words:- I am always tried to solve this program in a simple way. Which helps you to understand easily?  Hopefully, now you can understand How to check Even Or odd without ' If ' Statement. If you find any error in this program please remind us to comment below of this post. And stay away to learn programs J with Us. and also Read our Next Program.

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