# Check if given number is Emirp Number or not

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.

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 not ` `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 code ` `int` `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 not ` `def` `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 Function ` `n ``=` `13` `# Input number ` `if` `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. `

## PHP

 ` `

Output :

`Yes`

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.

My Personal Notes arrow_drop_up Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.

Improved By : jit_t

Article Tags :
Practice Tags :

Be the First to upvote.

Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.