# Check if given number is Emirp Number or not

• Difficulty Level : Easy
• Last Updated : 08 Jun, 2022

An Emirp Number (prime spelled backwards) is a prime number that results in a different prime when its decimal digits are reversed. This definition excludes the related palindromic primes.
Examples :

Input : n = 13
Output : 13 is Emirp!
Explanation :
13 and 31 are both prime numbers.
Thus, 13 is an Emirp number.

Input : n = 27
Output : 27 is not Emirp.

Objective : Input a number and find whether the number is an emirp number or not.

Approach : Input a number and firstly check if its a prime number or not. If the number is a prime number, then we find the reverse of the original number and check the reversed number for being prime or not. If the reversed number is also prime, then the original number is an Emirp Number otherwise it is not.
Below is the implementation of above approach :.

## C++

 // C++ program to check if given// number is Emirp or not.#include using namespace std; // Returns true if n is prime.// Else false.bool isPrime(int n){    // Corner case    if (n <= 1)        return false;     // Check from 2 to n-1    for (int i = 2; i < n; i++)        if (n % i == 0)            return false;     return true;} // Function will check whether number// is Emirp or notbool isEmirp(int n){    // Check if n is prime    if (isPrime(n) == false)        return false;     // Find reverse of n    int rev = 0;    while (n != 0) {        int d = n % 10;        rev = rev * 10 + d;        n /= 10;    }     // If both Original and Reverse are Prime,    // then it is an Emirp number    return isPrime(rev);} // Driver codeint main(){    int n = 13; // Input number    if (isEmirp(n) == true)        cout << "Yes";    else        cout << "No";} // This code is contributed by Anant Agarwal.

## Java

 // Java program to check if given number is// Emirp or not.import java.io.*;class Emirp {    // Returns true if n is prime. Else    // false.    public static boolean isPrime(int n)    {        // Corner case        if (n <= 1)            return false;         // Check from 2 to n-1        for (int i = 2; i < n; i++)            if (n % i == 0)                return false;         return true;    }     // Function will check whether number    // is Emirp or not    public static boolean isEmirp(int n)    {        // Check if n is prime        if (isPrime(n) == false)            return false;         // Find reverse of n        int rev = 0;        while (n != 0) {            int d = n % 10;            rev = rev * 10 + d;            n /= 10;        }         // If both Original and Reverse are Prime,        // then it is an Emirp number        return isPrime(rev);    }     // Driver Function    public static void main(String args[]) throws IOException    {        int n = 13; // Input number        if (isEmirp(n) == true)            System.out.println("Yes");        else            System.out.println("No");    }}

## Python3

 # Python3 code to check if# given number is Emirp or not. # Returns true if n is prime.# Else false.def isPrime( n ):         # Corner case    if n <= 1:        return False         # Check from 2 to n-1    for i in range(2, n):        if n % i == 0:            return False         return True # Function will check whether# number is Emirp or notdef isEmirp( n):         # Check if n is prime    n = int(n)    if isPrime(n) == False:        return False                 # Find reverse of n    rev = 0    while n != 0:        d = n % 10        rev = rev * 10 + d        n = int(n / 10)                      # If both Original and Reverse    # are Prime, then it is an    # Emirp number    return isPrime(rev) # Driver Functionn = 13 # Input numberif isEmirp(n):    print("Yes")else:    print("No")     # This code is contributed by "Sharad_Bhardwaj".

## C#

 // C# program to check if given// number is Emirp or not.using System; class Emirp {    // Returns true if n is prime    // Else false.    public static bool isPrime(int n)    {        // Corner case        if (n <= 1)            return false;         // Check from 2 to n-1        for (int i = 2; i < n; i++)            if (n % i == 0)                return false;         return true;    }     // Function will check whether number    // is Emirp or not    public static bool isEmirp(int n)    {        // Check if n is prime        if (isPrime(n) == false)            return false;         // Find reverse of n        int rev = 0;        while (n != 0) {            int d = n % 10;            rev = rev * 10 + d;            n /= 10;        }         // If both Original and Reverse are Prime,        // then it is an Emirp number        return isPrime(rev);    }     // Driver Function    public static void Main()    {        int n = 13; // Input number        if (isEmirp(n) == true)            Console.WriteLine("Yes");        else            Console.WriteLine("No");    }} // This code is contributed by vt_m.



## Javascript



Output :

Yes

Time complexity: O(n)

Auxiliary Space:O(1)

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