# Euler Problem 38

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?

This one was easier than it sounded. Just created a list of all the concatenated products. Then filtered the list for pandigital numbers. The highest one was the winner.

import java.util.*;

public class Euler38
{
public static void main(String[] args)
{
System.out.format(“Problem 38:\n”);
Euler38 e = new Euler38();
}
public String Problem ()
{
ArrayList<String> list = new ArrayList<String>();
ArrayList<String> list2 = new ArrayList<String>();
String[] digits = {“1″,”2″,”3″,”4″,”5″,”6″,”7″,”8″,”9”};
for (int i = 9; i <= 1000000; i++)
{
String S = “”;
for (int n = 1; n <= 100; n++)
{
if (S.length() < 9)
S += i*n;
if (S.contains(“0”))
break;
if (S.length() == 9)
if (S.length() > 9)
break;
}
}
for (String n: list)
{
boolean flag = true;
for (String d: digits)
{
if (n.contains(d))
if (n.replace(d,””).length() != 8)
flag = false;
}
if (flag)