Does there exist a fraction whose repeating decimal has a period of some highest number or can the number of this period take infinitely many values? It is well known that every decimal of a fraction becomes periodic at one point or the other, e.g: 1/3 starts repeating at period 1, 27/91 at period 6 and 3923/6173 at period 3086. 
Now, as there are an infinite number of non-zero values that we may place in the numerator or denominator of a fraction, the following question arises: does there exist a fraction whose repeating decimal has a period of some highest number or can the number of this period take infinitely many values? 
In other words, does there exist some fraction whose repeating decimal has a period of 577291485 or even 5*10^444? And furthermore, is it true that there are an infinite number of fractions whose decimal number has the period 5?
It should be kept in mind however that the numerator/denominator of the fractions in question may contain numbers not repeating thrice/twice, e.g: 586192/247591, or 1/43817 and 1/39820, so it's not just 1/9 or 1/999.
 A: Consider $\cfrac 19, \cfrac 1{99}, \cfrac 1{999} \dots$
For the second part, what happens to the period of the fraction when you multiply the denominator by $10$?
In general a fraction which repeats will have the form $$f=\frac p{10^r}+\frac q{10^{r}}\left(\frac 1{10^n}+\frac 1{10^{2n}}+\frac 1{10^{3n}}+\dots\right)$$ Where $p$ is an integer with up to $r$ digits ($r$ and $p$ may be zero) which represents the non-recurring part of the fraction, and $q$ is the repeating part, with up to $n$ digits. Now we see that $$10^rf-p=qs$$ where $s$ is the sum of the series. Multiply by $10^n$ and obtain $$10^{r+n}f-10^np=10^nqs=q+qs=q+10^rf-p$$ and $$f=\frac p{10^r}+\frac q{10^r(10^n-1)}$$
If we take $p=r=0$ for the moment so that $f=\cfrac q{10^n-1}$ is the form of a fraction which recurs without a non-recurring part, we see that a fraction which has a denominator which is a factor of $10^n-1$ will recur after $n$ places. It may recur after fewer places than that, but not after more.
Every odd prime $k$ is a factor of $10^n-1$ for some minimum value of $n$, and $\frac 1k$ will then recur with period $n$.
There are more answers on this site which go into more detail than this, and there is a full treatment in Number Theory book by Hardy and Wright. If you explore it a bit yourself you may find some interesting things. 
The reason denominators $9, 99, 999$ etc come into it (they correspond to $q=1$) is that they are the values of $10^r-1$.
A: $\frac{1}{10^p - 1}$ will have period $p$ for any positive integer $p$.
