I am trying to solve the following problem:

Consider the decimal representations of the numbers $x_n=3^{-n}, \; n=1,2,3,\cdots$. Find the length of the repeating part of the decimal expansion of $x_n$. And prove that the repeating part of the decimal expansion of $x_{20}$ contains the sequence $20182019$.

The first part of the question is quite straightforward, without going too much in the details, we find that the length of the repeating decimal expansion of $x_n$ can be defined as:

$l_{n} = \begin{cases} 1 \; \text{for} \; n \leq 2 \\ 3^{n-2} \; \text{for} \; n >2. \end{cases}$

Which gives $l_{20}=3^{18}$, i.e., the length of the repeating part of the decimal expansion of $x_{20}$ is $3^{18}$.

The second part of the question is the one where I am having trouble figuring what to do. If I understand the question properly, we have this number $x_{20}$ which has a repeating decimal expansion of length $3^{18}$, and we are trying to prove that a certain sequence of numbers appears in its repeating decimal expansion.

This number would look something like:

$x_{20}= 0.\overline{a_1a_2\cdots 20182019 \cdots a_{3^{18}}}$, with $a_i$ being the $i$-th digit of the decimal expansion.

My first thought was to multiply $x_{20}$ by $10^{3^{18}}$ in order to have the whole repeating part of the decimal expansion shifted to the integer part of $x_{20}$ but I quickly realised this didn't help much.

I imagine there is no general way to find if a certain "sequence" of number appears in another number (or is there?), so my guess is that the solution would be a trick working solely with this example.

Thank you for reading, any help or advice would be greatly appreciated!


Consider the base $10^8$ decimals. This can be read off from the base $10$ decimals by packing $8$ base $10$ digits into one base $10^8$ digit.

Then by dividing $0.\overline{11111111}$ by $3^{18}$, by your part one, it still has period $3^{18}$, since the period is relatively prime to $8$. When you calculate the base $10^8$ digits, you need to do divisions of the form $10^8\times n$ by $3^{18} = 3,8742,0489$, and since the period is still the largest possible, each possible remainder should occur (from $0$ to $3^{18}-1$) without repetition, and each quotient should occur (from $00000000$ to $99999999$), since $10^8 < 3^{18}$.

In particular, $20182019$ should occur.


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