# Check whether a number is Emirpimes or not

Given a number ‘n’, check whether it is an emirpimes or not.
An emirpimes(“semiprime” when spelled backwards) derives its definition from the way it is spelt. So, an emirpimes is a number that is a semiprime (product of two prime numbers) itself, and the reversal of its digits gives another new number, which too is a semi prime. Hence, by definition we can conclude that none of the palindrome numbers can be emirpimes, as the reversal of their digits does not give any new number, but forms the same number again.

Examples :

Input : 15
Output : Yes
Explanation : 15 is itself a semi prime number, since it is a product of two prime numbers 3 and 5. The reversal of its digits gives a new number 51, which too is a semi prime, it being the product of two prime numbers, viz., 3 and 17

Input : 49
Output : Yes
Explanation : 49 is itself a semi prime number, since it is a product of two prime numbers(not necessarily distinct) 7 and 7. The reversal of its digits gives a new number 94, which too is a semi prime, it being the product of two prime numbers, viz., 2 and 47

Input : 25
Output : No
Explanation : 25 is itself a semi prime number, since it is a product of two prime numbers(not necessarily distinct) 5 and 5. The reversal of its digits gives a new number 52, which is not a semi prime, it being the product of three and not two prime numbers, viz., 2, 2 and 13

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

Approach :

1. First check whether the entered number is semi prime, itself.
2. If yes, form a number by reversing its digits.
3. Now, compare this number with the initially entered number to ascertain if the number is palindrome or not.
4. If the number is not a palindrome, check whether this new number is also semi prime or not.
5. If yes, then the initially entered number is reported to be an emirpimes.

## C++

 `// CPP code to check whether ` `// a number is Emirpimes or not ` `#include ` `using` `namespace` `std; ` ` `  `// Checking whether a number ` `// is semi-prime or not ` `int` `checkSemiprime(``int` `num) ` `{ ` `    ``int` `cnt = 0; ` `    ``for` `(``int` `i = 2; cnt < 2 && ` `                    ``i * i <= num; ++i) ` `    ``{ ` `        ``while` `(num % i == 0) ` `        ``{ ` `            ``num /= i; ` ` `  `            ``// Increment count of ` `            ``// prime numbers ` `            ``++cnt; ` `        ``} ` `    ``} ` ` `  `    ``// If number is still greater than 1, after ` `    ``// exiting the for loop add it to the count ` `    ``// variable as it indicates the number is ` `    ``// a prime number ` `    ``if` `(num > 1) ` `        ``++cnt; ` ` `  `    ``// Return '1' if count is  ` `    ``// equal to '2' else return '0' ` `    ``return` `cnt == 2; ` `} ` ` `  `// Checking whether a number ` `// is emirpimes or not ` `bool` `isEmirpimes(``int` `n) ` `{ ` `    ``// Number itself is not semiprime. ` `    ``if` `(checkSemiprime(n) == ``false``) ` `        ``return` `false``; ` ` `  `    ``// Finding reverse of n. ` `    ``int` `r = 0; ` `    ``for` `(``int` `t=n; t!=0; t=t/n) ` `        ``r = r * 10 + t % 10; ` ` `  `    ``// The definition of emirpimes excludes ` `    ``// palindromes, hence we do not check ` `    ``// further, if the number entered is a ` `    ``// palindrome ` `    ``if` `(r == n) ` `        ``return` `false``; ` ` `  `    ``// Checking whether the reverse of the ` `    ``// semi prime number entered is also ` `    ``// a semi prime number or not ` `    ``return` `(checkSemiprime(r)); ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` `    ``int` `n = 15; ` `    ``if` `(isEmirpimes(n)) ` `    ``cout << ``"Yes"``; ` `    ``else` `        ``cout << ``"No"``; ` `    ``return` `0; ` `} `

## Java

 `// Java code to check whether a ` `// number is Emirpimes or not ` `import` `java.io.*; ` ` `  `class` `GFG  ` `{ ` `     `  `    ``// Checking whether a number ` `    ``// is semi-prime or not ` `    ``static` `boolean` `checkSemiprime(``int` `num) ` `    ``{ ` `        ``int` `cnt = ``0``; ` `        ``for` `(``int` `i = ``2``; cnt < ``2` `&&  ` `                        ``i * i <= num; ++i) ` `        ``{ ` `            ``while` `(num % i == ``0``) ` `            ``{ ` `                ``num /= i; ` `     `  `                ``// Increment count of ` `                ``// prime numbers ` `                ``++cnt; ` `            ``} ` `        ``} ` `     `  `        ``// If number is still greater than 1, ` `        ``// after exiting the for loop add it  ` `        ``// to the count variable as it indicates ` `        ``// the number is a prime number ` `        ``if` `(num > ``1``) ` `            ``++cnt; ` `     `  `        ``// Return '1' if count is equal ` `        ``// to '2' else return '0' ` `        ``return` `cnt == ``2``; ` `    ``} ` `     `  `    ``// Checking whether a number ` `    ``// is emirpimes or not ` `    ``static` `boolean` `isEmirpimes(``int` `n) ` `    ``{ ` `        ``// Number itself is not semiprime. ` `        ``if` `(checkSemiprime(n) == ``false``) ` `            ``return` `false``; ` `     `  `        ``// Finding reverse of n. ` `        ``int` `r = ``0``; ` `        ``for` `(``int` `t = n; t != ``0``; t = t / n) ` `            ``r = r * ``10` `+ t % ``10``; ` `     `  `        ``// The definition of emirpimes excludes ` `        ``// palindromes, hence we do not check ` `        ``// further, if the number entered is a ` `        ``// palindrome ` `        ``if` `(r == n) ` `            ``return` `false``; ` `     `  `        ``// Checking whether the reverse of the ` `        ``// semi prime number entered is also ` `        ``// a semi prime number or not ` `        ``return` `(checkSemiprime(r)); ` `    ``} ` ` `  `    ``// Driver Code ` `    ``public` `static` `void` `main (String[] args) ` `    ``{ ` `        ``int` `n = ``15``; ` `        ``if` `(isEmirpimes(n)) ` `            ``System.out.println(``"Yes"``); ` `        ``else` `            ``System.out.println(``"No"``); ` `    ``} ` `} ` ` `  `// This code is contributed by Ajit. `

## Python3

 `# Python3 code to check whether  ` `# a number is Emirpimesor not  ` ` `  `# Checking whether a number  ` `# is semi-prime or not  ` `def` `checkSemiprime(num):  ` ` `  `    ``cnt ``=` `0``; ` `    ``i ``=` `2``; ` `    ``while` `(cnt < ``2` `and` `(i ``*` `i) <``=` `num):  ` `     `  `        ``while` `(num ``%` `i ``=``=` `0``): ` `            ``num ``/``=` `i;  ` ` `  `            ``# Increment count of  ` `            ``# prime numbers  ` `            ``cnt ``+``=` `1``; ` `        ``i ``+``=` `1``; ` ` `  `    ``# If number is still greater than 1,  ` `    ``# after exiting the add it to the  ` `    ``# count variable as it indicates  ` `    ``# the number is a prime number  ` `    ``if` `(num > ``1``):  ` `        ``cnt ``+``=` `1``;  ` ` `  `    ``# Return '1' if count is equal  ` `    ``# to '2' else return '0'  ` `    ``return` `cnt ``=``=` `2``;  ` ` `  `# Checking whether a number  ` `# is emirpimes or not  ` `def` `isEmirpimes(n):  ` `     `  `    ``# Number itself is not semiprime.  ` `    ``if` `(checkSemiprime(n) ``=``=` `False``):  ` `        ``return` `False``;  ` ` `  `    ``# Finding reverse of n.  ` `    ``r ``=` `0``; ` `    ``t ``=` `n; ` `    ``while` `(t !``=` `0``):  ` `        ``r ``=` `r ``*` `10` `+` `t ``%` `10``; ` `        ``t ``=` `t ``/` `n; ` ` `  `    ``# The definition of emirpimes excludes  ` `    ``# palindromes, hence we do not check  ` `    ``# further, if the number entered  ` `    ``# is a palindrome  ` `    ``if` `(r ``=``=` `n):  ` `        ``return` `false;  ` ` `  `    ``# Checking whether the reverse of the  ` `    ``# semi prime number entered is also  ` `    ``# a semi prime number or not  ` `    ``return` `(checkSemiprime(r));  ` ` `  `# Driver Code  ` `n ``=` `15``;  ` `if` `(isEmirpimes(n)):  ` `    ``print``(``"No"``);  ` `else``: ` `    ``print``(``"Yes"``);  ` ` `  `# This code is contributed by mits `

## C#

 `// C# code to check whether a ` `// number is Emirpimes or not ` `using` `System; ` ` `  `class` `GFG ` `{ ` `     `  `    ``// Checking whether a number ` `    ``// is semi-prime or not ` `    ``static` `bool` `checkSemiprime(``int` `num) ` `    ``{ ` `        ``int` `cnt = 0; ` `        ``for` `(``int` `i = 2; cnt < 2 &&  ` `                        ``i * i <= num; ++i) ` `        ``{ ` `            ``while` `(num % i == 0) ` `            ``{ ` `                ``num /= i; ` `     `  `                ``// Increment count of ` `                ``// prime numbers ` `                ``++cnt; ` `            ``} ` `        ``} ` `     `  `        ``// If number is still greater than 1,  ` `        ``// after exiting the for loop add it  ` `        ``// to the count variable as it  ` `        ``// indicates the number is a prime number ` `        ``if` `(num > 1) ` `            ``++cnt; ` `     `  `        ``// Return '1' if count is equal ` `        ``// to '2' else return '0' ` `        ``return` `cnt == 2; ` `    ``} ` `     `  `    ``// Checking whether a number ` `    ``// is emirpimes or not ` `    ``static` `bool` `isEmirpimes(``int` `n) ` `    ``{ ` `        ``// Number itself is not semiprime. ` `        ``if` `(checkSemiprime(n) == ``false``) ` `            ``return` `false``; ` `     `  `        ``// Finding reverse of n. ` `        ``int` `r = 0; ` `        ``for` `(``int` `t = n; t != 0; t = t / n) ` `            ``r = r * 10 + t % 10; ` `     `  `        ``// The definition of emirpimes excludes ` `        ``// palindromes, hence we do not check ` `        ``// further, if the number entered is a ` `        ``// palindrome ` `        ``if` `(r == n) ` `            ``return` `false``; ` `     `  `        ``// Checking whether the reverse of the ` `        ``// semi prime number entered is also ` `        ``// a semi prime number or not ` `        ``return` `(checkSemiprime(r)); ` `    ``} ` ` `  `    ``// Driver Code ` `    ``public` `static` `void` `Main () ` `    ``{ ` `        ``int` `n = 15; ` `        ``if` `(isEmirpimes(n)) ` `            ``Console.WriteLine(``"Yes"``); ` `        ``else` `            ``Console.WriteLine(``"No"``); ` `    ``} ` `} ` ` `  `// This code is contributed by anuj_67. `

## PHP

 ` 1) ` `        ``++``\$cnt``; ` ` `  `    ``// Return '1' if count ` `    ``// is equal to '2' else ` `    ``// return '0' ` `    ``return` `\$cnt` `== 2; ` `} ` ` `  `// Checking whether a number ` `// is emirpimes or not ` `function` `isEmirpimes(``\$n``) ` `{ ` `     `  `    ``// Number itself is ` `    ``// not semiprime. ` `    ``if` `(checkSemiprime(``\$n``) == false) ` `        ``return` `false; ` ` `  `    ``// Finding reverse ` `    ``// of n. ` `    ``\$r` `= 0; ` `    ``for` `(``\$t` `= ``\$n``; ``\$t` `!= 0; ``\$t` `= ``\$t` `/ ``\$n``) ` `         ``\$r` `= ``\$r` `* 10 + ``\$t` `% 10; ` ` `  `    ``// The definition of emirpimes   ` `    ``// excludes palindromes,hence   ` `    ``// we do not check further,  ` `    ``// if the number entered  ` `    ``// is a palindrome ` `    ``if` `(``\$r` `== ``\$n``) ` `        ``return` `false; ` ` `  `    ``// Checking whether the  ` `    ``// reverse of the ` `    ``// semi prime number  ` `    ``// entered is also ` `    ``// a semi prime number  ` `    ``// or not ` `    ``return` `(checkSemiprime(``\$r``)); ` `} ` ` `  `    ``// Driver Code ` `    ``\$n` `= 15; ` `    ``if` `(isEmirpimes(``\$n``)) ` `     `  `    ``echo` `"No"``; ` `    ``else` `    ``echo` `"Yes"``; ` ` `  `// This code is contributed by Ajit. ` `?> `

Output :

```Yes
```

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