By a pure repeating decimal (in base $10$), we mean a decimal $0.\overline{a_1\cdots a_k}$ which repeats in blocks of $k$ digits beginning at the decimal point. An example is $.243243243\cdots = \tfrac{9}{37}$. By a mixed repeating decimal we mean a decimal $0.b_1\cdots b_m\overline{a_1\cdots a_k}$ which eventually repeats, but which cannot be reduced to a pure repeating decimal. An example is $.011363636\cdots = \tfrac{1}{88}$. Prove that if a mixed repeating decimal is written as a fraction $\tfrac pq$ in lowest terms, then the denominator $q$ is divisible by $2$ or $5$ or both.
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Tags: AMC, USA(J)MO, USAMO, number theory, relatively prime, number theory unsolved
admin25
27.07.2011 05:12
Moderator note: the original wording of this post was as follows: Quote: The repeating decimal $ 0.ab \cdots k \overline {pq \cdots u} = \dfrac {m}{n} $, where $m$ and $n$ are relatively prime integers, and there is at least one decimal before the repeating part. Show that $n$ is divisible by $2$ or $5$ (or both). (For example, $ 0.011 \overline {36} = 0.01136363636 \cdots = \dfrac {1}{88} $, and $88$ is divisible by $2$.)
Let $ x=0.ab\cdots k\overline{pq\cdots u} $. Also, let there $ m $ non-repeating digits and $ n $ repeating digits. We have
$ (100\cdots)x=ab\cdots k.\overline{pq\cdots u} $
$ (100\cdots)x=ab\cdots k+\frac{pq\cdots u}{999\cdots} $
$ (100\cdots)x=\frac{(999\cdots)(ab\cdots k)+(pq\cdots u)}{999\cdots} $
$ x=\frac{(999\cdots)(ab\cdots k)+(pq\cdots u)}{(999\cdots)(100\cdots)} $
where there are $ m $ zeroes and $ n+1 $ nines. Assume for the sake of contradiction that the simplified denominator does not contain a factor of $ 2 $ or $ 5 $. This means that all of the factors of $ 10 $ must cancel out. Now take the unsimplified fractions and cancel out only the factors of $ 10 $. This means that the fraction can be expressed as $ \frac{vw\cdots z}{999\cdots} $, for some digits $ v, w, \cdots z $. However, in decimal form, this can be expressed as $ 0.\overline{vw\cdots z} $, contradicting the fact that there is at least one decimal digit before the repeating part. Therefore, there must be a factor of $ 2 $ or $ 5 $ in the simplified denominator.
$ \mathbb{Q.E.D.} $
Wave-Particle
16.04.2017 05:15
Suppose there are $x$ non-repeating digits and $y$ repeating digits. We have that \begin{align*} \frac{m}{n} &= \frac{\overline{ab\cdots k}}{10^x} + \frac{\overline{pq\cdots u}}{10^x \cdot (10^y - 1)}\\ &= \frac{(10^y-1)\cdot \overline{ab\cdots k} + \overline{pq\cdots u}}{10^x \cdot (10^y-1)}.\\ \end{align*}Consider the numerator modulo $10$. Note that $(10^y-1)\cdot \overline{ab\cdots k} + \overline{pq\cdots u}\equiv u - k\not\equiv 0\pmod{10}$. Thus, when simplified, the $10^x$ will still have either a $2$ or $5$ factor left.