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Compute the period of a decimal number a priori

How to find the period of a reciprocal $1$/$x$ ? in both cases when it is terminating i.e (the distinct prime factors of the denominator are only 2's and 5's ) and when it keeps repeating immediately as in $1$/$7$ or after some digits as in $1$/$6$

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marked as duplicate by Gerry Myerson, Sasha, Ross Millikan, Old John, t.b. Aug 29 '12 at 23:40

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

You find the period of the reciprocal by doing the division until it starts repeating. One can express this in terms of finding the smallest solution to $10^r\equiv10^s\pmod x$, but it's basically the same thing/ – Gerry Myerson Aug 29 '12 at 13:07
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Using this, we can figure out $\lambda(x)$, for any given $x$ such that $(x,10)=1$

So, $10^{\lambda(x)}≡1\pmod x\implies x\mid(10^{\lambda(x)}-1)$

Let $d$ the smallest positive integer such that $10^d≡1\pmod x$

Also let $\lambda(x)=de+g$ where $0≤g<d$

$a^g=a^{\lambda(x)-de}=a^{\lambda(x)}-(a^d)^e≡1\pmod x$, but $d$ is the smallest positive integer.

So, $g=0\implies d\mid \lambda(x)$, $d$ is called $ord_x{10}$.

Then we need to find out this $d$ .

Let $x\cdot t=10^d-1$ where t is a natural number so, $\frac{1}{x}=\frac{t}{10^d-1}$ and d will be period.

For the divisors of $10$, i.e, $2,5$ the minimum power of $10$ that is divisible by $2^m\cdot 5^n$ is $max(m,n)$ where m,n are non-negative integers.

So, the length of $\frac{1}{2^m\cdot 5^n}$ will be $max(m,n)$.

If $x=2^m\cdot 5^n \cdot Q$ where $(Q,10)=1$,

$\frac{1}{x}$ will have $max(m,n)$ non-recurring digits followed by $d$ recurring digits where d is $ord_x{10}$.

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