##
**Problem:**

Take the number 192 and multiply it by each of 1, 2, and 3:

192 [×] 1 = 192

192 [×] 2 = 384

192 [×] 3 = 576

By concatenating each product we get the 1 to 9 pandigital, 192384576. We will call 192384576 the concatenated product of 192 and (1,2,3)

The same can be achieved by starting with 9 and multiplying by 1, 2, 3, 4, and 5, giving the pandigital, 918273645, which is the concatenated product of 9 and (1,2,3,4,5).

What is the largest 1 to 9 pandigital 9-digit number that can be formed as the concatenated product of an integer with (1,2, ... , n) where n [>] 1?

##
**Solution:**

932718654

##
**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 38 * By Nayuki Minase * * http://nayuki.eigenstate.org/page/project-euler-solutions * https://github.com/nayuki/Project-Euler-solutions */ import java.util.Arrays; public final class p038 implements EulerSolution { public static void main(String[] args) { System.out.println(new p038().run()); } public String run() { int max = -1; for (int n = 2; n <= 9; n++) { for (int i = 1; i < Library.pow(10, 9 / n); i++) { String concat = ""; for (int j = 1; j <= n; j++) concat += i * j; if (isPandigital(concat)) max = Math.max(Integer.parseInt(concat), max); } } return Integer.toString(max); } private static boolean isPandigital(String s) { if (s.length() != 9) return false; char[] temp = s.toCharArray(); Arrays.sort(temp); return new String(temp).equals("123456789"); } }

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