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: 

Class name: 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.

Answer 1

import java.util. Scanner; 

public class Emirp 

{ 

int n,rev,f; 

Emirpfint nn) 

{ 

n=nn; 

rev=0; 

f=2; 

} 

intisprime(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 ans1=isprime(f); 

n=rev; f=2; 

int ans2=isprime(f); 

if(ans 1 ==1 && ans2==1) 

System. out.println(n+" is anEmirp number"); 

else 

System.out.println(n+" is not an Emirp number"); 

} 

public static void main() 

{ 

Scanner sc=new Scanner(System.in); 

System.out.println("\n Enter a number"); 

int x=sc.nextInt(); 

Emirp obj = new Emirp(x); 

obj.isEmirp(); 

} 

}