Solution to Project Euler Problem 37: Truncatable primes - 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.

• Java Code

package sikademy;
/**
 *
 * @author Archangel Macsika
 * Copyright (c) Sikademy. All rights reserved
 */

public class SikademyEulerSolution {
	
	public String run() {
		long sum = 0;
		for (int count = 0, n = 10; count < 11; n++) {
			if (isTruncatablePrime(n)) {
				sum += n;
				count++;
			}
		}
		return Long.toString(sum);
	}
	
	
	private static boolean isTruncatablePrime(int n) {
		// Test if left-truncatable
		for (long i = 10; i <= n; i *= 10) {
			if (!Library.isPrime(n % (int)i))
				return false;
		}
		
		// Test if right-truncatable
		for (; n != 0; n /= 10) {
			if (!isPrime(n))
				return false;
		}
		
		return true;
	}

        public static boolean isPrime(int x) {
		if (x < 0)
			throw new IllegalArgumentException("Negative number");
		if (x == 0 || x == 1)
			return false;
		else if (x == 2)
			return true;
		else {
			if (x % 2 == 0)
				return false;
			for (int i = 3, end = sqrt(x); i <= end; i += 2) {
				if (x % i == 0)
					return false;
			}
			return true;
		}
	}

        public static int sqrt(int x) {
		if (x < 0)
			throw new IllegalArgumentException("Square root of negative number");
		int y = 0;
		for (int i = 1 << 15; i != 0; i >>>= 1) {
			y |= i;
			if (y > 46340 || y * y > x)
				y ^= i;
		}
		return y;
	}

	public static void main(String[] args) {
                SikademyEulerSolution solution = new SikademyEulerSolution();
		System.out.println(solution.run());
	}	
}

• Python Code

# 
# @author Archangel Macsika
# Copyright (c) Sikademy. All rights reserved.
# 

import itertools

def compute():
    ans = sum(itertools.islice(filter(is_truncatable_prime, itertools.count(10)), 11))
    return str(ans)


def is_truncatable_prime(n):
    # Test if left-truncatable
    i = 10
    while i <= n:
        if not is_prime(n % i):
            return False
        i *= 10

    # Test if right-truncatable
    while n > 0:
        if not is_prime(n):
            return False
        n //= 10
    return True

def sqrt(x):
	assert x >= 0
	i = 1
	while i * i <= x:
		i *= 2
	y = 0
	while i > 0:
		if (y + i)**2 <= x:
			y += i
		i //= 2
	return y

# Tests whether the given integer is a prime number.
def is_prime(x):
	if x <= 1:
		return False
	elif x <= 3:
		return True
	elif x % 2 == 0:
		return False
	else:
		for i in range(3, sqrt(x) + 1, 2):
			if x % i == 0:
				return False
		return True

if __name__ == "__main__":
    print(compute())