Question

Program to Check whether a given number is a Emirp number.
(An Emirp Number (prime spelt backwards) is a prime number that results in a different prime when its digits are reversed.)

				
					Example 1:
Enter a number:
113
113 is an Emirp number
Here 113 is a prime number 
Reverse of 113 is 311 which is also prime.
As both the given number and reverse of number is prime therefore,13 is an emirp number.

Example 2:
Enter a number:
19
19 is not an Emirp number
Here 19 is a prime number 
Reverse of 19 is 91 which is mot a prime as 91 is divisible by 13.
As only given number is prime but not reverse of number therefore,19 is not an emirp number.

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Code

				
					import java.util.Scanner;

public class emirpNumber
{
    //Function isPrime(int n) checks if number is Prime
    //If it is true then return true elese return false
    public static boolean isPrime(int n)
    {
        int i=0;
        if(n>1)
        {
            for(i=2;i< n;i++)
            {
                if(n%i==0)
                {
                    return false;

                }
            }
            return true;   
        }
        else
        {
            return false;   
        }

    }
    //Function reverse(int n) returns reverse of a number
    public static int reverse(int n)
    {
        int temp=0,rev=0,rem=0;
        temp=n;
        while(temp>0)
        {
            rem=temp%10;
            rev=rev*10+rem;
            temp=temp/10;
        }
        return rev;
    }
    public static void main()
    {
        int reverseNum=0,num=0;
        Scanner sc=new Scanner(System.in);
        System.out.println("Enter a number:");
        num=sc.nextInt();
        reverseNum=reverse(num);
        if(isPrime(num)==true && isPrime(reverseNum)==true)
        {
            System.out.println(num+" is an Emirp number");   
        }
        else
        {
            System.out.println(num+" is not an Emirp number");   

        }
    }
}