Math Puzzler #2

Take a power of two. Rearrange the digits. Can you ever have a second power of two? For simple problem, neglect leading zeros. For a harder one (which I haven't yet solved), don't.

Solutions after the jump.

The solution for the simple problem...

For proof by contradiction, assume there are two such powers:


Further assume amath n>m endmath.

Like the difference of any two rearrangements of the same digits, `n-m` must be a multiple of nine (A future puzzler will make clear why this must be so).

However, amath n-m=2^l-2^k endmath can be factored into a Mersenne number and a power of two ( amath 2^k(2^(l-k)-1) endmath ). If `n` and `m` are the same length, `l-k` must be less than `4` (`2^4>10`), so the Mersenne number can only be `1`, `3`, or `7`, none of which are multiples of nine.

Therefore such `n` and `m` cannot exist.

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