Seeking proof that powers of 3 greater than 9 divided into powers of 10, decimal representation with period greater than 1

Question

Anybody have a good proof for this fact? 1/3 and 1/9 have decimal representations with period 1, but numbers of the form (10^a)/(3^b) where b > 2 have period 2 or higher from what I've been able to see. I've tried looking it up with no luck, was wondering if someone could point me in the right direction. I have a feeling modular arithmetic will play a role here.

Answer 1

The decimal with a non-repeating initial part $a$ that lasts $n$ digits and repeats the digit $d$ starting at place $n+1$ has a value $$\frac{a}{10^n} + \frac{d}{9 \times 10^n} = \frac{9a + d}{9 \times 10^n}$$ (for example, $0.27444\ldots = \frac{27}{100} + \frac{4}{900}$). The fractions that can be written in such a form are those whose lowest-terms denominators divide $9 \times 10^n$, and the highest power of $3$ that divides $9 \times 10^n$ is $9$.

Answer 2

The period of the decimal representation of fractions of the form $\ \frac{n}{3^b}\ $ with $\ n\ $ not divisible by $3$ is the order of $\ 10\ $ modulo $\ 3^b\ $—i.e. the smallest positive integer $\ p\ $ such that $\ 3^b\mid 10^p-1\ $. For $\ b\ge 2\ $, $\ 10^{3^{b-2}}=1+m_b3^b\ $ where $\ m_b\equiv 1\ (\hspace{-0.6em}\mod3)\ $, from which it follows that the order of $10$ modulo $\ 3^b\ $, and hence the period of the decimal representation of $\ \frac{n}{3^b}\ $, is $\ 3^{b-2}\ $.

The above assertion can be proved by induction. It's clearly true for $\ b=2\ $, so let $\ b\ge2\ $ be any integer for which the order of $10$ modulo $\ 3^b\ $ is $\ 3^{b-2}\ $ and $\ m_b=\frac{10^{3^{b-2}}-1}{3^b}\equiv 1\ (\hspace{-0.6em}\mod3)\ $. Then $\ 10^{3^{b-2}}=1 + 3^b m_b\ $, so \begin{align} 10^{3^{b-1}}&=\left(1 + 3^b m_b\right)^3\\ &=1+3^{b+1}m_b+3^{2b+1}m_b^2+3^{3b}m_b^3\\ &=1+3^{b+1}\left(m_b+3^bm_b^2+3^{2b-1}m_b^3\right)\\ &\equiv\ 1\ \left(\hspace{-0.7em}\mod3^{b+1}\right)\ ,\text{ but}\\ 10^{3^{b-2}}&\not\equiv 1\ \left(\hspace{-0.7em}\mod3^{b+1}\right)\ . \end{align} Therefore, the order of $10$ modulo $\ 3^{b+1}\ $ is $\ 3^{b-1}\ $, and \begin{align} \frac{10^{3^{b-1}}-1}{3^{b+1}}&=m_b+3^bm_b^2+3^{2b-1}m_b^3\\ &\equiv 1\ (\hspace{-0.9em}\mod3)\ , \end{align} which completes the induction.