[Guest]

Determine all possible positive palindrome(s) N, such that the decimal representation of 2N² has no leading zeroes and contains each of the digits from 0 to 9 exactly once.

[Pickett]

Determine all possible positive palindrome(s) N, such that the decimal representation of 2N² has no leading zeroes and contains each of the digits from 0 to 9 exactly once.

Here are the only 2 INTEGER solutions to this problem...I'm sure there are some decimal answers as well...

46464: 2(46464)² = 4317806592

69696: 2(69696)² = 9715064832

Which then you can just add decimal places to, to get some REAL solutions:

4.6464

46.464

464.64

4646.4

6.9696

69.696

696.96

6969.6...So, I will guess that that is all of them, (since the result of .46464 and .69696 have leading zeros in the answer.

[Guest]

Here are the only 2 INTEGER solutions to this problem...I'm sure there are some decimal answers as well...

46464: 2(46464)² = 4317806592

69696: 2(69696)² = 9715064832

Which then you can just add decimal places to, to get some REAL solutions:

4.6464

46.464

464.64

4646.4

6.9696

69.696

696.96

6969.6...So, I will guess that that is all of them, (since the result of .46464 and .69696 have leading zeros in the answer.

Indeed, 46464, 69696 are the only possible positive decimal palindromes that satisfy the given conditions.

However, I do not know of an easy analytic method (and still looking for it) to substantiate this.

Notes:

1. It may be observed that a palindrome is always an integer according to this mathworld article , unless stated otherwise.

2. According to wikipedia, decimal denotes base ten.

In hindsight, I should have mentioned “base ten representation……” in the problem text rather than “decimal representation”.

Edited May 11, 2009 by K Sengupta

[Pickett]

Indeed, 46464, 69696 are the only possible positive decimal palindromes that satisfy the given conditions.

However, I do not know of an easy analytic method (and still looking for it) to substantiate this.

Notes:

1. It may be observed that a palindrome is always an integer according to this mathworld article , unless stated otherwise.

2. According to wikipedia, decimal denotes base ten.

In hindsight, I should have mentioned “base ten representation……” in the problem text rather than “decimal representation”.

I wrote a simple little program to solve this problem... And I knew the upper bound for the number was 70710, since any number greater than that would have more than 10 digits in the resulting answer (which invalidates the requirement that all numbers be present in answer)...same with the lower bound being 22360 (since the resulting number would have less than 10 digits). See the attached code:

List<String> palindromes = getPalindromes(22360, 70710);
for (int i = 0; i < palindromes.size(); i++) {
Long l = Long.valueOf(palindromes.get(i));
Long newVal = 2 * (l * l);
String newStr = newVal.toString();
boolean hasAll = true;
for (int j = 0; j < 10; j++) {
if (newStr.indexOf(Integer.toString(j)) < 0) {
hasAll = false;
break;
}
}
if (hasAll) {
System.out.println(l + ": (2 * " + l + " * " + l + ") = " + newStr);
}
}
}

private static List<String> getPalindromes(long lowerBound, long upperBound) {
List<String> palindromes = new ArrayList<String>();
for (long i = lowerBound; i <= upperBound; i++) {
String val = Long.toString(i);
boolean isPal = true;
for (int j = 0; j < val.length(); j++) {
if (val.charAt(j) != val.charAt(val.length() - (j + 1))) {
isPal = false;
break;
}
}
if (isPal) {
palindromes.add(val);
}
}
return palindromes;
}

    public static void main(final String[] args) {

Basically just get all numbers that are palindromes between those bounds, and then go through each of those palindromes, perform the operation, and check to make sure each resulting answer has every number. Fun problem, I enjoyed it.