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)))
        return String.valueOf(x);
    public class Permutations
        ArrayList<String> Perms = new ArrayList<String>();
        public ArrayList<String> makePerms (String s)
            return Perms;
        public void p1(String prefix, String s)
            int N = s.length();
            if (N == 0)
                if (Perms.contains(prefix) == false)
                for (int i = 0; i < N; i++)
                    p1(prefix + s.charAt(i), s.substring(0, i) + s.substring(i+1, N));

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