##
**Problem:**

The number, 1406357289, is a 0 to 9 pandigital number because it is made up of each of the digits 0 to 9 in some order, but it also has a rather interesting sub-string divisibility property.

Let d1 be the 1st digit, d2 be the 2nd digit, and so on. In this way, we note the following:

* d2d3d4=406 is divisible by 2

* d3d4d5=063 is divisible by 3

* d4d5d6=635 is divisible by 5

* d5d6d7=357 is divisible by 7

* d6d7d8=572 is divisible by 11

* d7d8d9=728 is divisible by 13

* d8d9d10=289 is divisible by 17

Find the sum of all 0 to 9 pandigital numbers with this property.

##
**Solution:**

16695334890

##
**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 43 * By Nayuki Minase * * http://nayuki.eigenstate.org/page/project-euler-solutions * https://github.com/nayuki/Project-Euler-solutions */ public final class p043 implements EulerSolution { public static void main(String[] args) { System.out.println(new p043().run()); } private static int[] DIVISIBILITY_TESTS = {2, 3, 5, 7, 11, 13, 17}; // First 7 primes public String run() { long sum = 0; int[] digits = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}; outer: do { for (int i = 0; i < DIVISIBILITY_TESTS.length; i++) { if (toInteger(digits, i + 1, 3) % DIVISIBILITY_TESTS[i] != 0) continue outer; } sum += toInteger(digits, 0, digits.length); } while (Library.nextPermutation(digits)); return Long.toString(sum); } private static long toInteger(int[] digits, int off, int len) { long result = 0; for (int i = off; i < off + len; i++) result = result * 10 + digits[i]; return result; } }

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