Emirp number in java, program to check emirp number

Emirp is simply the reverse prononciation of word "prime". An emirp number is the number which is prime and whose reverse is also a prime number.

EXAMPLE-1

17 

Reverse of the number = 71

Both the numbers are prime so 17 is an emirp number.

EXAMPLE-2

37

Reverse of the number = 73

Both the numbers are prime so 37 is an emirp number.

EXAMPLE-3

13

Reverse of the number = 31

Both the numbers are prime so 13 is an emirp number.

Emrip number can be easily made in java with the help of methods/functions.

Firstly, take an input of number, then check it is prime or not, then reverse the number and again check it is prime or not, if both are true, print the suitable output.

ALGORITHM-

STEP 1 : start STEP 2 : print "enter a number" STEP 3 : take input of number in e STEP 4 : int n=num, int rev =0 STEP 5 : repeat STEP to STEP until n>0 STEP 6 : int d=n%10 STEP 7 : rev=(rev*10)+d STEP 8 : n=n/10 STEP 9 :  int p= input value from STEP 13 STEP 10 : int d=0, x=num STEP 11 : int i=2, repeat STEP 12 until i<=x/2, i++ STEP 12 : if (x/i==0) then d=1; break STEP 13 : return d STEP 14 : input value in q from STEP 18 STEP 15 :int d=0, x=rev STEP 16 : int i=2, repeat STEP 12 until i<=x/2, i++ STEP 17 : if (x/i==0) then d=1; break STEP 18 : return d STEP 19 : if (p==1&&q==1) then goto STEP 20 else goto STEP 21 STEP 20 : print num+" is an emirp number" STEP 21 : print num+" is not an emirp number" STEP 22 : end

JAVA CODE-

import java.util.*;

public class emrip

{
    int num;
    int rev;
    emrip()
    {
    }
    emrip(int nn)
    {
        num=nn;
    }
    int isPrime(int x)
    {
        int d=0;
        for(int i=2;i<=x/2;i++)
        {
            if(x%i==0)
            {
            d=1;
            break;
           }
        }
        return d;
    }
    void isEmrip()
    {
        int n=num;
        rev=0;
        while(n>0)
         }
            int d=n%10;
            rev=(rev*10)+d;
            n=n/10;
         }
        emrip obj=new emrip();
        int p=obj.isPrime(num);
        int q=obj.isPrime(rev);
        if(p==1&&q==1)
        System.out.println(num+" is an emirp number");
        else
        System.out.println(num+" is not an emirp number");
    }
    public static void main(String[] arg)
    {
        Scanner sc=new Scanner(System.in);
        System.out.println("enter a number");
        int e=sc.nextInt();
        emrip obj=new emrip(e);
        obj.isEmrip();
    }   
}

output-

Enter a number

37

37 is an emirp number

SIMILAR JAVA CODES-

1-Disarium number

2-Magic number

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