Number times Recurring Part of Reciprocal gives 9-Repdigit/Generalization

Theorem
Let a (strictly) positive integer $n$ be such that the decimal expansion of its reciprocal has a recurring part of period $d$ and no non-recurring part.

Let $m$ be the integer formed from the $d$ digits of the recurring part. Let $M$ be an arbitrary integer.

Then:
 * $M \equiv \sqbrk {mmm \dots m} \pmod {10^c}$

for some positive integer $c$, :
 * $M \times n \equiv -1 \pmod {10^c}$

In other words, the last $c$ digits of $M$ coincide with that of $\sqbrk {mmm \dots m}$ the last $c$ digits of $M \times n$ are all $9$s.

Proof
$\sqbrk {mmm \dots m}$ can be expressed as:
 * $\ds \sum_{k \mathop = 0}^{K - 1} m 10^{k d}$

for some sufficiently large $K > \dfrac c d$.

Sufficient Condition
Suppose:
 * $M \equiv \sqbrk {mmm \dots m} \pmod {10^c}$

We have:

which is the $c$-digit repdigit number consisting of $9$s.

Therefore the last $c$ digits of $M \times n$ are all $9$'s.

Necessary Condition
It is observed that all implications but the first one:
 * $M \equiv \sqbrk {mmm \dots m} \pmod {10^c} \implies M \times n \equiv \sqbrk {mmm \dots m} \times n \pmod {10^c}$

trivially reverses.

By Common Factor Cancelling in Congruence/Corollary 2, the converse of the above is true if $n$ and $10^c$ are coprime.

We have:

Hence the result.

Example

 * $\dfrac 1 {27} = 0 \cdotp \dot 0 3 \dot 7$

The last $4$ digits of $17 \, 037$ coincides with that of $027 \, 027$.

We have:
 * $17 \, 037 \times 27 = 459 \, 999$

It is observed that the last $4$ digits of $459 \, 999$ are all $9$'s.