Does there exist any integer number which is a power of 2, and ...?

Question

... and by changing the place of its digits we come to another power of 2?

Thanks for your consideration

Answer 1

I believe I know why there are no such integers.

If you take any integer and change the place of its digits, then subtract it from the original number, you get an integral multiple of 9. (This is your lemma. I can't recall the proof; I think we remember that it is true, though, for all integers.)

Suppose there where two different powers of 2 that are "anagrams," i.e., that have the same digits but in a different sequence. Their difference is then 2 to the nth minus 2 to the mth power. (Sorry I don't know how to give a clearer notation using my keyboard.) By our lemma, this would have to be equal to 9x for some integer x, i.e., the difference of these two powers of 2 would have to be an integral multiple of 9.

The prime factorization of 9 is 3 x 3, and thus cannot divide 2 to the nth minus 2 to the mth power, which is necessarily a multiple of 2.

Let me know if this is correct. Thank you for an interesting question.

Answer 2

I think it's pretty unlikely. For each number of digits there are only around 3 possible powers of two, and it is very unlikely that they're numerical "anagrams".

Of course this isn't a proof and I can't think of a reason why somewhere in the infinite reaches of the integers such a situation couldn't exist.

Answer 3

Well, I am not sure that I can do a theoretical calculation and prove that it is possible or not. The only other alternative is to write down all the powers of 2 and check.