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**Problem:**

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.

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**Solution:**

748317

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**Code:**

The solution may include methods that will be found here: Library.java .

The solution may include methods that will be found here: Library.java .

public interface EulerSolution{ public String run(); }

/* * Solution to Project Euler problem 37 * By Nayuki Minase * * http://nayuki.eigenstate.org/page/project-euler-solutions * https://github.com/nayuki/Project-Euler-solutions */ public final class p037 implements EulerSolution { public static void main(String[] args) { System.out.println(new p037().run()); } 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 (!Library.isPrime(n)) return false; } return true; } }

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