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Project Euler #37: Truncatable primes

Question
The number 3797 has an interesting property. Being prime itself, it is possible to continuously remove digits from left to right, and remain prime at each stage: 3797, 797, 97, and 7. Similarly we can work from right to left: 3797, 379, 37, and 3.
Find the sum of the only eleven primes that are both truncatable from left to right and right to left.
NOTE: 2, 3, 5, and 7 are not considered to be truncatable primes.

Answer : 748317

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Solution

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import java.util.*;

public class Solution {
    private static boolean[] prime = null;
    public static void main(String[] args) {
        /* Enter your code here. Read input from STDIN. Print output to STDOUT. Your class should be named Solution. */
        Scanner in = new Scanner(System.in);
        int n = in.nextInt();
        generatePrimes();
        long res = 0;

        for(int i=11;i<n;i++)
        {
            if(prime[i] && isRightTruncatable(i) && isLeftTruncatable(i))
                res+=i;
        }
        System.out.println(res);
    }
    
    public static boolean isLeftTruncatable(int n)
    {
        String s = String.valueOf(n);
        int i=0;
        while(i<s.length())
        {
            if(!prime[Integer.valueOf(s.substring(i, s.length()))]) return false;
            i+=1;
        }
        return true;
    }
    
    public static boolean isRightTruncatable(int n)
    {
        while(n!=0)
        {
            if(!prime[n]) return false;
            n/=10;
        }
        return true;
    }
    
    public static void generatePrimes()
    {
        int limit = 1000000;
        prime = new boolean[limit+1];
        prime[2] = true;
        prime[3] = true;
        int root = (int) Math.ceil(Math.sqrt(limit));

        //Sieve of Atkin for prime number generation
        for (int x = 1; x < root; x++)
        {
            for (int y = 1; y < root; y++)
            {
                int n = 4 * x * x + y * y;
                if (n <= limit && (n % 12 == 1 || n % 12 == 5))
                    prime[n] = !prime[n];

                n = 3 * x * x + y * y;
                if (n <= limit && n % 12 == 7)
                    prime[n] = !prime[n];

                n = 3 * x * x - y * y;
                if ((x > y) && (n <= limit) && (n % 12 == 11))
                    prime[n] = !prime[n];
            }
        }

        for (int i = 5; i <= root; i++)
        {
            if (prime[i])
            {
                for (int j = i * i; j < limit; j += i * i)
                {
                    prime[j] = false;
                }
            }
        }
    }
}
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