# 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 2^{a} can be rearranged to give 2^{b}, where a < b.

Consider 2^{b} − 2^{a}.

Since the digits of 2^{b} are a rearrangement of the digits of 2^{a}, it follows that 9 is a factor of 2^{b} − 2^{a}.

(See Casting Out Nines - Proof for an explanation of why this is true.)

As 2^{3} < 10 < 2^{4}, there can be at most four powers of 2 with the same number of digits. (Example: 1, 2, 4, 8.)

Therefore we must have 2^{b} − 2^{a} = c × 2^{a}, where c = 1, 3, or 7.

That is, the prime factorization of 2^{b} − 2^{a} does not contain 3^{2} = 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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