Prime Palindrome ISC Computer Science 2012 Practical

A prime palindrome integer is a positive integer (without leading zeros) which is prime as well as a palindrome.

Given two integers m and n, where m < n, write a program to determine how many prime-palindrome integers are there in the range between m and n (both inclusive) and output them.

The input contains two positive integers m and n where m < 3000 and n < 3000. Display the number of prime-palindrome integers in the specified range along with their values in the format specified below:

Test your program with the sample data and some random data:

Example 1
INPUT:
m = 100
N = 1000
OUTPUT:
THE PRIME PALINDROME NUMBERS ARE:
101, 131, 151, 181, 191, 313, 353, 373, 383, 727, 757, 787, 797, 919, 929
FREQUENCY OF PRIME PALINDROME INTEGERS: 15

Example 2
INPUT:
m = 100
n = 5000
OUTPUT: OUT OF RANGE

import java.util.Scanner;
class PrimePalindrome{
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        System.out.print("m = ");
        int m = Integer.parseInt(in.nextLine());
        System.out.print("n = ");
        int n = Integer.parseInt(in.nextLine());
        if(m > n || m > 2999 || n > 2999){
            System.out.println("OUT OF RANGE");
            return;
        }
        int f = 0;
        System.out.println("THE PRIME PALINDROME NUMBERS ARE:");
        for(int i = m; i <= n; i++){
            if(isPrime(i) && isPalindrome(i)){
                if(f == 0)
                    System.out.print(i);
                else
                    System.out.print(", " + i);
                f++;
            }
        }
        if(f == 0)
            System.out.println("NIL");
        else
            System.out.println();
        System.out.println("FREQUENCY OF PRIME PALINDROME INTEGERS: " + f);
    }
    public static boolean isPrime(int n){
        for(int i = 2; i <= n / 2; i++){
            if(n % i == 0)
                return false;
        }
        return true;
    }
    public static boolean isPalindrome(int n){
        int rev = 0;
        for(int i = n; i != 0; i /= 10)
            rev = rev * 10 + i % 10;
        return n == rev;
    }
}