## 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?

932718654

## Code: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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