An emirp (prime spelled backwards) is a prime number that results in a different prime when its digits are reversed.[1] This definition excludes the related palindromic primes. Emirps are also called reversible primes.


this are the list of emirps
http://oeis.org/A006567/list

public class EmirpProgram {
	public static void  main(String args[]){
		int a = 13;
		int count = 0;
		while(count<100){
			if(isEmirp(a)){
				System.out.print(a + "    ");
				count++;
				if(count%10==0)
					System.out.println();
				
			}
			a++;
		}
		
	}
	public static boolean isPrime(int p){
		for(int i=2;i <p; i++){
			if( p%i==0){
				return false;
				
			}
		}
		return true;
			
	}
	public static boolean isEmirp(int x){
		return isPrime(x) && isPrime(reverse(x));
		
	}
	public static int reverse(int r){
		if(r<10) 
			return r;
		return flipper(r%10,r/10);
		
		
	}
	public static int flipper(int a, int b){
		if(b<1)
			return a;
		
		
		
		return flipper(a*10+b%10,b/10);
		
		
		
	}

}

here is my output

13    17    31    37    71    73    79    97    101    107    
113    131    149    151    157    167    179    181    191    199    
311    313    337    347    353    359    373    383    389    701    
709    727    733    739    743    751    757    761    769    787    
797    907    919    929    937    941    953    967    971    983    
991    1009    1021    1031    1033    1061    1069    1091    1097    1103    
1109    1151    1153    1181    1193    1201    1213    1217    1223    1229    
1231    1237    1249    1259    1279    1283    1301    1321    1381    1399    
1409    1429    1439    1453    1471    1487    1499    1511    1523    1559    
1583    1597    1601    1619    1657    1669    1723    1733    1741    1753

some numbers are right and some arent, no numbers are missing tho just extras
what did I do wrong

Are you asking whether your code follows your algorithm correctly?
Or are you asking if your algorithm is correct?

For the first question, post your algorithm so we can check your code against it.
For the second, ask Google.

this is my algorithm for flipping numbers

public static int reverse(int r){
        if(r<10) 
            return r;
        return flipper(r%10,r/10);

i mod it by 10 and divide it by 10

Edited 3 Years Ago by Dani: Formatting fixed

some numbers are right and some arent, no numbers are missing tho just extras

Which are the extra numbers?

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