Decimal representation of $\frac{1}{p^k}$

Here I talk about the decimal representations of numbers of the form $\frac{1}{p^k}$ where $p$ is a prime.

If $p= 2 $ or $5$ the decimal expansion is a finite one otherwise it is recurring. The aim is to study the length of recurring cycle of the decimal expansion of $\frac{1}{p^k}$ $(p \ne 2,5)$. 

The expansion consists of some initial number say $M$ of $m$ digits after which some number say $N$ (which has $n$ digits) starts repeating as shown:

$ \frac{1}{p^k} = 0.MNNNN... =\frac{M}{10^m}+ \frac{N}{10^{m+n}}+ \frac{N}{10^{m+2n}}+..... = \frac{M}{10^m} + \frac{N}{10^{m}(10^{n}-1)}$

$\Rightarrow $ $Np^k + M (10^{n}-1)p^k = 10^m(10^n-1) $

Thus

$10^n \equiv 1 (mod \; p^k) $ ($p \neq 2,5$) where $n$ is the recurring cycle of the decimal expansion and is thus the least non zero solution of the above congruence i.e. the order of $10$ modulo $p^k$.

CLAIM: If $r $ is the order of $10$ modulo $p$ then the order of $10$ modulo $p^k$ is of the form $rp^{j}$ where $j$ is an integer. 

PROOF: 

Since $ 10^r \equiv 1(mod \; p) $ we can write $10^r = pu+1$ where $u$ is an integer.

$\Rightarrow$ $10^{rp^{k-1}}\equiv {(pu+1)}^{p^{k-1}} \equiv 1+p^ku+..... \equiv 1 (mod \; p^k)

Thus if $t$ is the order of $10$ modulo $p^k$ then $ t \mid rp^{k-1}$. Also $10^t \equiv 1 ( mod \; p) $ implies $ r \mid  t$.

Thus $t$ must be of the form $ rp^j$ where $0\leq j\leq k-1 $.

We enhance the result further.....

Let $s_1$ be the minimum value of $i$ for which $r$ is the order of $10$ modulo $p^i$. Then what is the order of $10$ modulo $p^{s_1 +1}$ ?....... It is $rp$ .

Call the order modulo $p^i$ as $r_i$. ( $r_1 = r_2 = ... r_{s_1} = r $). Let $10^r = p^{s_1}u_1 +1$

We know $r_{s_1+1}$ is of the form $rp^j$, $j \ne 0$. We show that $j=1$ works.

$10^{rp} \equiv {(p^{s_1}u_1 +1)}^p \equiv 1+p^{s_1+1}u_1 +... \equiv 1 (mod \; p^{s_1 +1})$

Thus rp is the order.

Now if $s_2$ is the minimum value of $i$ for which $rp$ is the order of $10$ modulo $p^i$, then using similar arguments as above we can show that $rp^2$ is the order modulo $p^{s_2 +1}$ and so on.......

We thus obtain that:

Order of $10$ modulo $p,p^2,...,p^{s_1}$ is $r$

Order of $10$ modulo $p^{s_1+1},...,p^{s_2}$ is $rp$

Order of $10$ modulo $p^{s_2 +1},..,p^{s_3}$ is $rp^2$

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and the length of recurrence cycle is order itself.

NOTE:

1. By Fermats theorem $r$ is clearly a divisor of $p-1$.
2. There is nothing special about 10 in the whole discussion except of course that $(10,p)=1$.So the discussion  can be extended to representation in any general base $b$ and p chosen such that $(b,p)=1$