# Euler Problem 52

It can be seen that the number, 125874, and its double, 251748, contain exactly the same digits, but in a different order.

Find the smallest positive integer, x, such that 2x, 3x, 4x, 5x, and 6x, contain the same digits.

This one seemed pretty straight forward … and it was. The only thing I had to straighten out first was my Permutations routine. I created a custom class for it (given that its recursive) and then instantiated it in the Problem method. Nice and tidy.

import java.util.*;

public class Euler52
{
public static void main(String[] args)
{
Euler52 e = new Euler52();
System.out.format(“Problem: %s\n”, e.getClass().getName());
System.out.format(“Answer = %s\n”, e.Problem());
}
public String Problem ()
{
ArrayList<String> P = new ArrayList<String>();
Permutations p = new Permutations();
int x = 0;
for (int i = 100000; i <= 1000000; i++)
{
P = p.makePerms(Integer.toString(i));
if (i % 5000 == 0)
System.out.format(“%d\n”, i);
if (P.contains(Integer.toString(2*i)))
if (P.contains(Integer.toString(3*i)))
if (P.contains(Integer.toString(4*i)))
if (P.contains(Integer.toString(5*i)))
if (P.contains(Integer.toString(6*i)))
{
x=i;
break;
}
}
return String.valueOf(x);
}
public class Permutations
{
ArrayList<String> Perms = new ArrayList<String>();
public ArrayList<String> makePerms (String s)
{
Perms.clear();
p1(“”,s);
return Perms;
}
public void p1(String prefix, String s)
{
int N = s.length();
if (N == 0)
{
if (Perms.contains(prefix) == false)
Perms.add(prefix);
}
else
for (int i = 0; i < N; i++)
p1(prefix + s.charAt(i), s.substring(0, i) + s.substring(i+1, N));
}
}
}