Computer Science, asked by Wulonglei2, 1 year ago

Java program of emirp number

Answers

Answered by mangharam
51
Emirp Number:
An emirp number is a number which is prime backwards and forwards.
Example: 13 and 31 are both prime numbers. Thus, 13 is an emirp number. Design a class Emirp to check if a given number is Emirp number or not.

Code: 
import java.util.*;
class Emirp
{
  public static void main(String args[])
  {
    Scanner sc=new Scanner(System.in);
    char ch;
    do{
int i,c=0,r,n2=0,c1=0;
    System.out.println("Enter The Number: ");
    int n=sc.nextInt();
    
    for(i=1;i<=n;i++)
    {
      if(n % i == 0)
        c++;
    }
    if(c == 2)
    {
      r = n % 10;
      n2 = (n2 * 10) + r;
      n = n / 10;
    }
    for(i=1;i<=n2;i++)
    {
      if(n2 % i == 0)
        c1++;
    }
    if ( c ==2 && c1 ==2)
    {
          System.out.println("***Emirp Number***\n");
    }
    else
    {
          System.out.println("***Not Emirp Number***\n");
    }
    System.out.println("Do You Want To Check More Numbers ? Press y / n: ");
    ch=sc.next().charAt(0);
    }
    while(ch!='n');
    System.exit(0);
    }
}
Output:
Sample Output
Input the number
13
31
Number is Emirp

Input the number
17
71
Number is Emirp

Input the number
14
41
Number is not Emirp
Answered by jithinayalloor
28

Answer:

import java.util.*;

class emirp

{    

       public static void main()

       

       {  

           Scanner sc= new Scanner(System.in);

           System.out.println("enter the integer");

           int n = sc.nextInt(),f=0,cpy=n,rev=0,c=0;

           

           for (int i=1;i<=n;i++)

           {if(n%i==0)

               c++;

           

           }

           if(c==2)

           {

               while(n>0)

           {int ld=n%10;

               rev=rev*10+ld;

               n=n/10;

               

           }

           System.out.println("rev="+rev);

       }

           for (int j=1;j<=rev;j++)

           {if(rev%j==0)

               f++;

           }

           

               if(f==2)            

          System.out.println("it is an emirp number");

           else

           System.out.println("it is not an emirp number");

       }

   }

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