JAVASCRIPT

odd even condition

let number = 15;
if(number % 2 ==0){
	console.log( number + 'is  Even Number');
}
else{
	console.log(number " is Odd Number");
}

odd or even checker

a = int(input("Enter the number: "))
if (a % 2) == 0:
    print("{0} is even".format(a))
else:
    print(" {0} is odd ".format(a))

odd or even

public class EvenOdd{
    public static void evenOddValue(int value) {
    if(value % 2 == 0){
        System.out.println("The given value is even: " + value);
    }
    else{
        System.out.println("The given value is odd: " + value);
    }
}
    public static void main(String [] args) {
        int num = 25;
        EvenOdd obj = new EvenOdd();
        obj.evenOddValue(num);
     }
}

odd and even

Numbers -- odd & even
Write a method which can identifies given number is even or odd 
EX: 
identify(5) ->  "Odd"   
identify(6) ->  "Even"
 
Solution :
public  static  String  identifyOddEven( int  n ) {
return  n%2==0 ? "Even" : "odd" ; 
}