That's one big repeating binary expansion

If we multiply any number written in binary with $2^{S}$ , ( $S$ is an integer), then the decimal point of the number shifts right by $S$ digits.

If we multiply $2^S$ to $\frac{1}{3^{10}}$ , the decimal point will shift right by $S$ digits. Since the binary expansion of $\frac{1}{3^{10}}$ repeats after $S$ digits, it means fractional part of $\frac{2^{S}}{3^{10}} =$ fractional part of $\frac{1}{3^{10}}$

This means $2^{S} \equiv 1 \pmod{3^{10}}$ .

Since $2^{S}$ and $3^{10}$ are coprime integers, a solution to this congruence exists. By definition of multiplicative order , the minimum possible value of $S$ will be $\text{ord}_{3^{10}} (2)$ . We know that $2$ is a primitive root of $3$ as well as $3^2$ , so by lemma 14 in this document $2$ is a primitive root of $3^{10}$ also.
By definition of primitive root, $\text{ord}_{3^{10}} (2)= \phi(3^{10})$ .

$S = \phi (3^{10} )= 3^{10} \left( 1 - \frac{1}{3}\right) = 2 \times 3^9 = \boxed{39366}$ $_\square$

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Generalization: To find the smallest length $k$ , of the repetend of recurring decimal of a rational number $r$ when written in integral base $b \geq 2$ . Let the prime factorization of $b$ be $\prod _{i = 1} ^{n} p_{i} ^{a_{i}}$ .

We are given a rational number $r$ . Since we are studying the decimal part, it would easier if we talk about the fractional part of $r$ , i.e. $\{ r \}$ .

We can write $\{ r \} = \frac{m}{n}$ such that $m, n$ are positive integers such that $n \neq 0$ , and $0 \leq m < n$ . If $m$ is $0$ then the decimal won't repeat, $k$ is zero and $r$ is an integer.

Otherwise we will simplify $\frac{m}{n}$ into $\frac{m'}{n'}$ such that $m'$ and $n'$ are coprime integers. Then $k$ will be independent of $m'$ , it will only depend on $n'$ .

If $n'$ is divisible by any of the $p_{i}$ 's, then divide $n'$ by $p_{i}$ until it does not divide any of the $p_{i}$ 's. For example if $r$ was $\frac{5}{60}$ and base was $10 = 2 \times 5$ , then $60$ is divisible by $2$ and $5$ , so we divide it by $2$ and $5$ until it is not divisible by either, we will end up with $\frac{60}{2^{2} \times 5} = 3$ . Let the new number that we end up with be $q$ .

We can do this because multiplying a rational number by factors of the base will not change the value of $k$ , and by removing all the factors of $b$ from $n'$ , we are making the both of them coprime. This is important because if they are not coprime, then there will be no solution to the following congruence.

$b^{k} \equiv 1 \pmod {q}$

The smallest length of the repetend will be $k = \boxed{\text{ord}_{q} (b)}$ $_\square$

We want a similar summation with a power of two being raised to the nth power on the bottom to achieve a repeating decimal with a value of $\frac{1}{3^{10}}$ .In other words, let $\sum _{ n=1 }^{ \infty }{ \frac { a }{ { { ({ 2 }^{ k }) }^{ n } } } =\frac { 1 }{ { 3 }^{ 10 } } }$ We also can create an expression for the infinite sum in terms of $a$ and $k$ : $\sum _{ n=1 }^{ \infty }{ \frac { a }{ { { ({ 2 }^{ k }) }^{ n } } } =\frac { a }{ 1-\frac { 1 }{ { 2 }^{ k } } } -a } =\frac { a }{ { 2 }^{ k }-1 }$ hence we are looking for when ${3}^{10}$ divides $2^k-1$ , and we set $a=\frac{2^k-1}{{3}^{10}}$ . Suppose that $3^n$ divides $2^k-1$ , then ${ 2 }^{ 3a }-1=({ 2 }^{ a }-1)(1+{ 2 }^{ a }+{ 2 }^{ 2a })$ Since if we use this recurrence starting from $a=2$ every a is even, $1+{ 2 }^{ a }+{ 2 }^{ 2a }$ is 0 mod 3. Hence $3^{n+1}$ divides $2^{3k}-1$ . Applying this, the period of $\frac{1}{3^{10}}$ is $\frac{2}{3}3^{10}$ and by setting $a$ we may obtain the exact decimal sequence that is repeating.

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There exists no other solution $2^k$ for $k<3^{n}$ because 2 is a primitive root modulo $3^{n}$ . Simply put, this means that the sequence $\{2,2^2,2^3... 2^{3^{n}}\}$ contains all numbers relatively prime to $3^{n}$ when put in modulo $3^{n}$ , or in our case precisely $\frac{2}{3}3^{10}$ numbers. Suppose there exists another one in this sequence: then some sequence containing these $\frac{2}{3}3^{10}$ elements must repeat at least twice yet this is impossible because that would require more elements than there are in the sequence. Hence $\frac{2}{3}3^{10}$ is the minimum solution.