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Emirp numbers | Function to find reverse of any number.
Sieve method used for generating emirp number ( use of sieve ofEratosthenes ).
Create a boolean array " prime [ 0 . . n ] " and initialize all entries it as true . A value in prime [ i ] will finally be false if i is Not a prime , else true ..
If prime [ p ] is not changed , then it is a prime.
Update all multiples of p.
Traverse all prime numbers.
Find reverse a number.
A number is emrip if it is not a palindrome number and its reverse is also prime ..
Mark reverse prime as false so that it 's  not printed again.
Driver Code"
Solution in Python:
assistant
#Here is code to emirp numbers and the other steps required.
 def reverse ( x ) :
	rev = 0 ;
	while ( x > 0 ) :
		rev = ( rev * 10 ) + x % 10 ;
		x = int ( x / 10 ) ;
	return rev ;
def printEmirp ( n ) :
	prime = [ 1 ] * ( n + 1 ) ;
	p = 2 ;
	while ( p * p <= n ) :
		if ( prime [ p ] == 1 ) :
			for i in range ( p * 2 , n + 1 , p ) :
				prime [ i ] = 0 ;
		p += 1 ;
	for p in range ( 2 , n + 1 ) :
		if ( prime [ p ] == 1 ) :
			rev = reverse ( p ) ;
			if ( p != rev and rev <= n and prime [ rev ] == 1 ) :
				print ( p , rev , end = " ▁ " ) ;
				prime [ rev ] = 0 ;
n = 100 ;
printEmirp ( n ) ;