Problem 37—Truncatable primes

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题目描述如下：

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.

public class Problem37{    public static void main(String[] args)    {        int sum = 0;        for (int i = 10; i < 1000000; i++)        {            if (!contain24680(i) && isTruncatable(i))            {                System.out.print(i + "\t");                sum += i;            }        }        System.err.println("\nsum = " + sum);    }    /**     * 保证传入的为质数     *      * @param num     * @return     */    public static boolean isTruncatable(int num)    {        int temp = num;        while (temp > 0)        {            if (!MathUtils.isPrime(temp))            {                return false;            }            int length = MathUtils.getLength(temp);            temp = temp % (int) Math.pow(10, length - 1);        }        temp = num;        while (temp > 0)        {            if (!MathUtils.isPrime(temp))            {                return false;            }            temp = temp / 10;        }        return true;    }    public static boolean contain24680(int num)    {        String temp = String.valueOf(num);        if (temp.contains("0") || temp.contains("4")                || temp.contains("6") || temp.contains("8"))        {            return true;        }        return false;    }}

result:

23 37 53 73 313 317 373 797 3137 3797 739397
sum = 748317

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