Java Program to Emirp Number

An Emirp number is a prime number that results in a different prime number when its digits are reversed. For example, 13 is a prime number and 31 is also a prime number. Hence, 13 is an Emirp number.

Here is the Java program to check if a number is an Emirp number or not :

Program :

import java.util.Scanner;

public class EmirpNumber {

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        System.out.print("Enter a number: ");
        int n = sc.nextInt();
        if (isEmirp(n)) {
            System.out.println(n + " is an Emirp number");
        } else {
            System.out.println(n + " is not an Emirp number");
        }
        sc.close();
    }

    public static boolean isEmirp(int n) {
        if (!isPrime(n)) {
            return false;
        }
        int reverse = reverseNumber(n);
        return n != reverse && isPrime(reverse);
    }

    public static boolean isPrime(int n) {
        if (n <= 1) {
            return false;
        }
        for (int i = 2; i <= Math.sqrt(n); i++) {
            if (n % i == 0) {
                return false;
            }
        }
        return true;
    }

    public static int reverseNumber(int n) {
        int reverse = 0;
        while (n != 0) {
            reverse = reverse * 10 + n % 10;
            n = n / 10;
        }
        return reverse;
    }
}

Output :

Enter a number: 13
13 is an Emirp number

Enter a number: 16
16 is not an Emirp number

In the above program, we first take an integer input from the user and check if it is an Emirp number or not using the isEmirp() method. The isEmirp() method first checks if the number is prime or not using the isPrime() method.

If it is not prime, then it is not an Emirp number. If it is prime, then we reverse the digits of the number using the reverseNumber() method and check if the reversed number is not equal to the original number and is also a prime number. If both the conditions are true, then the number is an Emirp number.