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Subject: "power of 2"     Previous Topic | Next Topic
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Member since Oct-20-02
Oct-19-02, 00:21 AM (EST)
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"power of 2"
   Hi. I'm trying to settle the following problem but having difficult making further progress. Any hints or suggestions would be really appreciated. Thank you~

Does there exist a positive integer which is a power of 2, such that we can obtain another power of 2 by rearranging it's digits?

So far I only have obvious approach, that is the sum of digits of 2^a and 2^b must match if they're equal by rearranging digits of one of them. In this approach we have 2^a = 2^b (mod 9) so tha a = b (mod 6) since 2^(|a-b|) = 1 (mod 9) and 6 is the order of 2 modulo 9. This takes care of the case when the number of digits of 2^a and 2^b are equal and furthermore reveals that one of 2^a and 2^b must contain a digit 0.

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Nov-05-02, 11:15 AM (EST)
1. "RE: power of 2"
In response to message #0
   Well... i can give you an example of one.
12^2=144 and 21^2=441(if this is what you ment).

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Nov-10-02, 07:07 PM (EST)
2. "RE: power of 2"
In response to message #1
   what i meant by power of 2 was that the number has a form 2^n where n is an integer. Neither 12^2 nor 21^2 is a power of two since one can't find an integer n so that 2^n = 12^2 or 2^n = 21^2.

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