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We write a power of 3 in bits in binary representation as follows. For example $3=(11)$, $3^2=(1001)$ which means that we let the $k$-th bit from the right be $1$ if the binary representation of this power of 3 contains $2^{k-1}$, and $0$ otherwise.

  1. Prove that the highest power of 3 that has a palindromic binary representation is $3^3 = (11011)$.

  2. Prove that $3 = (11)$ is the only power of 3 with a periodic binary representation (in the sense that it consists of a finite sequence of $1$s and $0$s repeated two or more times, like "$11$" consists of two repetitions of the bitstring "$1$").

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@minasteris' edit: $729_{10} = 1011011001_2$ is not palyndromic. And a quick verification with Mathematica shows that for $k \leq 10000$, the only number $3^k$ that are palyndromic are $3, 9, 27$, which suggests those are the only three. –  TMM Feb 29 '12 at 16:41
    
@TMM: thanx you are right i delete the edit –  minasteris Feb 29 '12 at 17:28
    
Any number with a periodic representation has a palindromic factor greater than the repeated bit, so 2 holds if 1 holds. –  Angela Richardson Mar 7 '12 at 11:08

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