4
$\begingroup$

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!

$\endgroup$
6
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.