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 $c$, if and only if:
- $M \times n \equiv -1 \pmod {10^c}$
In other words, the last $c$ digits of $M$ coincide with that of $\sqbrk {mmm \dots m}$ if and only if 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:
| \(\ds M \times n\) | \(\equiv\) | \(\ds \sqbrk {mmm \dots m} \times n\) | \(\ds \pmod {10^c}\) | Congruence of Product | ||||||||||
| \(\ds \) | \(\equiv\) | \(\ds n \sum_{k \mathop = 0}^{K - 1} m 10^{k d}\) | \(\ds \pmod {10^c}\) | |||||||||||
| \(\ds \) | \(\equiv\) | \(\ds \paren {10^d - 1} \sum_{k \mathop = 0}^{K - 1} 10^{k d}\) | \(\ds \pmod {10^c}\) | Number times Recurring Part of Reciprocal gives 9-Repdigit | ||||||||||
| \(\ds \) | \(\equiv\) | \(\ds \paren {10^d - 1} \times \frac {10^{K d} - 1} {10^d - 1}\) | \(\ds \pmod {10^c}\) | Sum of Geometric Sequence | ||||||||||
| \(\ds \) | \(\equiv\) | \(\ds 10^{K d} - 1\) | \(\ds \pmod {10^c}\) | |||||||||||
| \(\ds \) | \(\equiv\) | \(\ds -1\) | \(\ds \pmod {10^c}\) | $10^c \divides 10^{K d}$ as $K > \dfrac c d$ | ||||||||||
| \(\ds \) | \(\equiv\) | \(\ds 10^c - 1\) | \(\ds \pmod {10^c}\) |
which is the $c$-digit repdigit number consisting of $9$s.
Therefore the last $c$ digits of $M \times n$ are all $9$'s.
$\Box$
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:
| \(\ds \gcd \set {m n, 10}\) | \(=\) | \(\ds \gcd \set {10^d - 1, 10}\) | Number times Recurring Part of Reciprocal gives 9-Repdigit | |||||||||||
| \(\ds \) | \(=\) | \(\ds \gcd \set {-1, 10}\) | GCD with Remainder | |||||||||||
| \(\ds \) | \(=\) | \(\ds 1\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds m n\) | \(\perp\) | \(\ds 10\) | Definition of Coprime Integers | ||||||||||
| \(\ds \leadsto \ \ \) | \(\ds m n\) | \(\perp\) | \(\ds 10^c\) | Powers of Coprime Numbers are Coprime | ||||||||||
| \(\ds \leadsto \ \ \) | \(\ds n\) | \(\perp\) | \(\ds 10\) | Divisor of One of Coprime Numbers is Coprime to Other |
Hence the result.
$\blacksquare$
Example
- $\dfrac 1 {27} = 0 \cdotp \dot 0 3 \dot 7$
The last $4$ digits of $17 \, 037$ coincides with that of $037 \, 037$.
We have:
- $17 \, 037 \times 27 = 459 \, 999$
It is observed that the last $4$ digits of $459 \, 999$ are all $9$'s.