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if $2^{s}$ is found by rearranging the digits of $2^r$ prove that $r=s$. I suspect that this question requires congruence but i need help.the 2 numbers have same digits, and the base 2 must have something to with this.

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Expanding on Thomas Andrew's answer, $2^s-2^r=2^r(2^{s-r}-1)$, but 9 does not divide $2^k-1$ until $k=6$, contradicting his observation that $s-r \le 4$.

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Hint: The numbers are congruent modulo nine, and the difference between $r$ and $s$ has to be less than $4$.

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