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Solution to puzzle 90: Powers of 2: rearranged digits

Does there exist an integral power of 2 such that it is possible to rearrange the digits giving another power of 2?  Numbers are written in standard decimal notation, with no leading zeroes.


We argue by reductio ad absurdum.

Suppose the digits of 2a can be rearranged to give 2b, where a < b.
Consider 2b − 2a.
Since the digits of 2b are a rearrangement of the digits of 2a, it follows that 9 is a factor of 2b − 2a.
(See Casting Out Nines - Proof for an explanation of why this is true.)

As 23 < 10 < 24, there can be at most four powers of 2 with the same number of digits.  (Example: 1, 2, 4, 8.)
Therefore we must have 2b − 2a = c × 2a, where c = 1, 3, or 7.

That is, the prime factorization of 2b − 2a does not contain 32 = 9.
Therefore our original supposition, that there exists an integral power of 2 such that it is possible to rearrange the digits giving another power of 2, is false.


Generalization

Suppose we allow leading zeroes.  Can we then find an integral power of 2 such that it is possible to rearrange the digits giving another power of 2?

Source: Power of 2 on The CTK Exchange Forums

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