Is there a way to calculate the number of non-repeating digits that precede the periodic repeating portion of a decimal expansion? For example:


1/6 = 0.1666.... (there is 1 non repeating digit) \*\*(Correction)
1/12 = 0.08333... (there are 2 non repeating digits)
7/12 = 0.58333....(there are 2 non repeating digits)
1/96 = 0.01041666..(there are 5 non repeating digits)


Do any forumulas exist for predicting the maximum length n, of the number of non repeating digits preceding the repeating portion? 


I know that if the denominator of a fraction is n, the maximum length of the repeating periodic portion is n-1. Must also the length of the preceding portion before the cycle be n-1?


Thank you! 



How do you calculate/prove the length of n, the number of non-repeating digits preceeding a periodic sequence of a fractional repeating decimal
When one writes an irreducible fraction $m/n$ as a periodic digit number all one does is to write


$m/n=\frac{a}{999...9000.00}$


So the number of digits before the period is the maximum of the power of $2$ and $5$ in $n$, 
i.e. wirting $n=2^\alpha 5^\beta k$ with $k$ relatively prime to $10$, the number of digits before the
period is $\max\{\alpha, \beta \}$.


I think that this will follow for free from the following lemma, whose proof is trivial:


Lemma: if gcd$(k,10) =1$ then $k$ has a multiple of the form $999...9$.


