# Proof for the uniqueness of all the combination of the digits of $2^k$, where k>3

A long time ago I found a question on the internet that went a little like this:

Suppose that we have $n=2^k$ where $k\gt 3$. If $m$ is another number that is a combination of the digits of $2^k$, prove that $m$ cannot be a power of $2$.

I gave up on it a long time ago, but have now become interested in number theory and hope that someone could shed some light on this problem.

Edit: This is only for base 10.

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When you say "a combination of the digits", you mean it is obtained by "shuffling them around", but using all of them? The usual mathematical term of that is "permutation" rather than "combination". – Arturo Magidin May 2 '11 at 2:56
@Arturo Yes, you are correct. I meant to say that m is a permutation of the digits of n. – Hautdesert May 3 '11 at 2:09

Assuming you mean shuffle the digits around (in base $10$)

For the case where you don't bring zeroes to the front:

We look at the remainder when you divide by $9$.

If $2^m = 2^n$ modulo $9$, then $m-n$ is divisible by $6$. In which case, they have different number of digits.

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I got stuck trying to follow this. The order of 2 mod 9 is 6, and there are no more than 4 consecutive powers of 2 with the same number of digits, so these all differ mod 9 and therefore no one of them is a digit-permutation of another; but how to handle the case when the digit 0 is permuted into the most-significant position? Maybe I am not interpreting the question correctly because I am allowing for e.g. "1024" to be permuted to "0124" which is a valid representation of 124. – Dan Brumleve May 2 '11 at 3:27
@Dan: Good point. That case is not considered in this answer and I have edited the answer to reflect that. – Aryabhata May 2 '11 at 3:29
I have heard a version of this problem and I don't recall zeroes being an issue, so I think the correct statement assumes that they aren't at the front. – Qiaochu Yuan May 2 '11 at 3:52
A variation which makes it false is including more than one base: 2^20 is a digit-permutation of 7^8, and 2^10 is a digit-permutation of 7^4. – Dan Brumleve May 2 '11 at 5:27
@Dan, no one knows. Guy, Unsolved Problems In Number Theory, F24, lists a bunch of numbers $n$ such that $2^n$ has no zero digit - the biggest in the list is $n=86$ - asks whether that's it, and notes that Dan Hoey has checked there are no others with $n\le2500000000$. – Gerry Myerson May 3 '11 at 6:44