Decimal expansion - BEFORE repeats Given a fraction of two relatively prime integers of lengths m and n, what is the maximum number of decimal places (in the decimal expansion) before the expansion starts repeating?  For example I happened to compute the ratio of two four digit numbers and the answer to 15 places had no repeats.  I do not know when the repeats would start.
This question differs from a "duplicate" question.  I am asking for how many digits (maximum) BEFORE it starts repeating.  Referred question is length of repetition.
 A: Assuming you are just interested in the length of the initial block (before the start of the period),  write $n=2^a5^bN$ where $\gcd(10,N)=1$.  Then the length of the initial block is $\max(a,b)$.  
Example:  Consider $\frac 1{360}$.  We have $360=2^3\times 5^1\times 3^2$ so we expect the initial block to have length $\max(3,1)=3$.  Indeed, $$\frac 1{360}=.002\overline 7$$
To prove this, let $c=\max(a,b)$.  Then $10^c\times \frac mn$ is a rational number with denominator prime to $10$.  For those, the decimal repeats from the start, so we are done. 
If you want the total length before you see repetitions then you need to add the length of the period to this.  That's the order of $10\pmod N$.   Sticking with $\frac 1{360}$ we see that $N=9$ so the order of $10$ is $1$.  In general, if you just want an upper bound, take $\max (a,b)+ \varphi(N)$.
A: We have two integer, $p,q$ we divide $p$ by $q$ and get some remainder $r$
with $0\le r<q$
if $r = 0$ then the decimal is finite.
If we are only considering repeating decimals, $0<r<b$
And we divide $10$ by $q$ and we get a new remainder $r_2$ also with $1\le  r_2<q$ 
And we repeat the process until we get a $r_n = r_m$ where $r_m$ is one of the previous values of remainder we have already had.
There are $q-1$ integers such that $0<r<q$ so the maximum length of the cycle is $q-1$
There is a theorem that says that if $q$ is prime does not divide 10, the length of the cycle must divide $q-1$
