Consider, for example, $\frac{1}{N} = \frac{1}{195} = \frac{1}{3\cdot5\cdot13} = 0.0051282051282051282... = 0.0\overline{051282}$,
compared to, say,
$\frac{1}{N} = \frac{1}{77} = \frac{1}{7\cdot11} = 0.012987012987... = 0.\overline{012987}$.
QUESTION: Under what conditions will the repetend of $1/N$ include all of the zeros following the decimal point, such as in the example $\frac{1}{77}$? Is there a particular class of integers $N$ which this must always happen? I have noticed that this seems to be the case when $N$ is a semiprime, but it is not the case, as we see above, when $N$ has three prime factors.
Let's say that $1/N=0.00\dots 0\overline{a_1a_2\dots a_k}$, where there are $m$ zeroes, and $A$ is the number represented by $a_1a_2\dots a_k$ in base $10$. Then $$0.\overline{a_1a_2\dots a_k}=\frac{A}{10^k-1}$$ and so $$\frac 1N=\frac{A}{10^m(10^k-1)}.$$ As a result, $N$ should divide $10^m(10^k-1)$. So, $m$ can't be $0$ if $2$ or $5$ divide $n$; see if you can convince yourself that $m$ can be $0$ in all other cases.