# Euler Problem 49

The arithmetic sequence, 1487, 4817, 8147, in which each of the terms increases by 3330, is unusual in two ways: (i) each of the three terms are prime, and, (ii) each of the 4-digit numbers are permutations of one another.

There are no arithmetic sequences made up of three 1-, 2-, or 3-digit primes, exhibiting this property, but there is one other 4-digit increasing sequence.

What 12-digit number do you form by concatenating the three terms in this sequence?

﻿Nothing extrordinary here. I could have made the code tighter, but it still ran in 7 seconds.

import java.util.*;

public class Euler49
{
public static void main(String[] args)
{
Euler49 e = new Euler49();
System.out.format(“Problem: %s\n”, e.getClass().getName());
}
public String Problem ()
{
ArrayList<Integer> allP = new ArrayList<Integer>();
ArrayList<Integer> P4 = new ArrayList<Integer>();
allP = makePrimes (10000);
String S = “”;
for (int p: allP)
if (p > 1487 && Integer.toString(p).length() == 4)
A:for (int i: P4)
{
for (int n = 1000; n < (9999-i)/2; n++)
{
if (P4.contains(i+n))
if (isPerm(i,i+n))
if (P4.contains(i+n+n))
if (isPerm(i, i+n+n))
{
S += Integer.toString(i) + Integer.toString(i+n) + Integer.toString(i+n+n);
break A;
}
}
}
return String.valueOf(S);
}
public ArrayList<Integer> makePrimes (int lim)
{
ArrayList<Integer> P = new ArrayList<Integer>();
boolean [] isPrime = new boolean [lim+1];
for (int i = 2; i < lim; i++)
isPrime[i] = true;
for (int p = 2; p*p < lim; p++)
if (isPrime[p] == true)
for (int m = p*p; m < lim; m = m+p)
if (m % p == 0)
isPrime[m] = false;
for (int i = 2; i < lim; i++)
if (isPrime[i])
return P;
}
public boolean isPerm (int i1, int i2)
{
String S1 = Integer.toString(i1);
String S2 = Integer.toString(i2);
for (int n = 0; n <= 3; n++)
S2 = S2.replaceFirst(String.valueOf(S1.charAt(n)), “”);
if (S2.equals(“”))
return true;
else
return false;
}
}