## How do you calculate/prove the length of n, the number of non-repeating digits preceeding a periodic sequence of a fractional repeating decimal

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!

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1/6 is 0.1666... ; 1/7 is 0.142857... – J. M. Oct 11 2010 at 1:53
Otherwise, this might be more suitable for math.stackexchange.com – J. M. Oct 11 2010 at 1:54
1/6 = 0.16666...thanks for the correction – unknown (google) Oct 11 2010 at 1:59
That 1/11 is really 1/12. – Gerry Myerson Oct 11 2010 at 2:28

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 $\max{\alpha, \beta }$.

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