Emirp Number

import java.io.*;
class Emirp
    {
static BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
int n, rev, f;

Emirp(int nn) //parameterised constructor
{
n=nn;
rev=0;
f=2;
}

int isprime(int x) //recursive function for checking whether 'x' is Prime or not
{
if(f<=x)
{
     if(x%f!=0) // f is increasing everytime we are not getting any factor
     {
         f++;
         isprime(x);
     }
}
/* Since, 'f' was increasing everytime we did not get any factor,
* Hence, if 'f' becomes equal to the number, then that means that the number
* is not divisible by any other number except 1 and itself
* Hence it is Prime */
if(f==x)
     return 1;
else
     return 0;
}

void isEmirp()
{
int copy=n, d;
while(copy>0) // code for reversing a number
{
d=copy%10;
rev=rev*10+d;
copy=copy/10;
}
int a=isprime(n); //checking whether the Original number is Prime or not
f=2; //resetting the value of f for checking whether the reverse number is Prime or not
int b=isprime(rev); //checking whether the Reverse number is Prime or not
if(a==1 && b==1) //If both Original and Reverse are Prime, then it is an Emirp number
System.out.println(n+" is an Emirp Number");
else
System.out.println(n+" is Not an Emirp Number");
}

public static void main()throws IOException
{
System.out.print("Enter any number : "); //inputting the original number
int n=Integer.parseInt(br.readLine());
Emirp ob=new Emirp(n);
ob.isEmirp();
} 
    }