Question

An Emirp number is a number which is prime backwards and forwards.
Example: 13 and 31 are both prime numbers. Thus,13 is an emirp number.
Design a class Emirp to check if a given number is Emirp number or not. Some of the members of the class are given below:
Classname:Emirp
Data members / instance variables:
n:stores the number
rev:stores the reverse of the number
f:stores the divisor
Member functions:

Emirp (int nn):to assign n= nn, rev=0 and f=2
int isprime(int x):check if the number is prime using the recursive technique and return 1 if prime otherwise return 0
void isEmirp(): reverse the given number and check if both the original number and the reverse number are prime , by invoking the function isprime(int) and display the result with an appropriate message

Specify the class Emirp giving details of the constructor(int) , int isprime(int) and void isEmirp(). Define the main() function to create an object and call the methods to check for Emirp number

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Code

				
					import java.util.Scanner;
public class Emirp
{
    int n,rev,f;
    Emirp(int nn)
    {
        n=nn;
        rev=0;
        f=2;
    }
    
    int isprime(int x)
    {
        if(n==x)
        {
            return 1;
        }
        else if(n % x==0 || n==1)
        {
            return 0;
        }
        else
        {
            return isprime(x+1);
        }
    }
    void isEmirp()
    {
        int x=n;
        while(x!=0)
        {
            rev=(rev*10)+x%10;
            x=x/10;
        }
        int ansl=isprime(f);
        x=n;
        n=rev;
        f=2;
        int ans2=isprime(f);
        if(ansl==1 && ans2==1)
        {
            System.out.println(x+" is an Emirp number");
        }
        else
        {
            System.out.println(x+" is not an Emirp number");
        }
    }
    public static void main()
    {
        Scanner sc=new Scanner(System.in);
        System.out.println("Enter a number");
        int x=sc.nextInt();
        Emirp obj1=new Emirp(x);
        obj1.isEmirp();
    }
}