Emirp Number in Java

In this section, we will learn what is an emirp number and also create Java programs to check if the given number is emirp or not. The emirp number Java program frequently asked in Java coding tests to check the logic of the programmer.

Emirp Number

A number is called an emirp number if we get another prime number on reversing the number itself. In other words, an emirp number is a number that is prime forwards or backward. It is also known as twisted prime numbers.

Note: Palindrome primes are excluded.

Emirp Number Example

Suppose, we want to check the number 79 is emirp or not.

We know that 79 is a prime number means that divisible by 1 and self only. On reversing the number, we get 97 which is another prime number. Therefore, 79 and 97 both are prime numbers. Hence, 79 is a prime number. Similarly, we can check other numbers also.

Some other emirp numbers are 13, 199, 107, 113, 1399, 1583, 1201, 3049, etc.

Steps to find Emirp Number

Read or initialize a number (n).
First, check the given number (n) is prime or not.
- If not, break the execution and exit.
- If prime, find the reverse (r) of the given number (n).
Check the reverse number (r) is prime or not.
- If not, print number (n) is not emirp.
- If prime, print the given number (n) as an emirp number.

Emirp Number Java Program

EmirpNumberExample1.java

import java.io.*; import java.util.*; public class EmirpNumberExample1 { //function that checks the given number is prime or not public static boolean isPrime(int n) { //base case if (n <= 1) return false; //loop executes from 2 to n-1 for (int i = 2; i < n; i++) if (n % i == 0) //returns false if the condition returns true return false; //returns true if the condition returns false return true; } //function that checks if the given number is emirp or not public static boolean isEmirp(int n) { //checks if the given number is prime or not if (isPrime(n) == false) return false; //variable that stores the reverse of the number int reverse = 0; //the while loop executes until the specified condition becomes false while (n != 0) { //finds the last digit of the number (n) int digit = n % 10; //finds the reverse of the given number reverse = reverse * 10 + digit; //removes the last digit n = n/10; } //calling the user-defined function that checks the reverse number is prime or not return isPrime(reverse); } //driver code public static void main(String args[]) { Scanner sc=new Scanner(System.in); System.out.print("Enter a number to check: "); //reading an integer from the user int n=sc.nextInt(); if (isEmirp(n) == true) System.out.println("Yes, the given number is an emirp number."); else System.out.println("No, the given number is not an emirp number."); } }

Output 1:

Enter a number to check: 1399
Yes, the given number is an emirp number.

Output 2:

Enter a number to check: 14
No, the given number is an emirp number.

Emirp Number in Java

Emirp Number

Note: Palindrome primes are excluded.

Emirp Number Example

Steps to find Emirp Number

All Answers

Emirp Number Java Program

Java Number Programs

This question belongs to these collections

Similar questions

need a help?

find thousands of online teachers now