Number times Recurring Part of Reciprocal gives 9-Repdigit

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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.

Then $m \times n$ is a $d$-digit repdigit number consisting of $9$s.


Generalization

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

Let $x = \dfrac 1 n = \sqbrk {0. mmmm \dots}$.

Then:

$10^d x = \sqbrk {m.mmmm \dots}$

Therefore:

\(\ds 10^d x - x\) \(=\) \(\ds \sqbrk {m.mmmm \dots} - \sqbrk {0. mmmm \dots}\)
\(\ds \leadsto \ \ \) \(\ds \frac 1 n \paren {10^d - 1}\) \(=\) \(\ds m\)
\(\ds \leadsto \ \ \) \(\ds m n\) \(=\) \(\ds 10^d - 1\)

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

$\blacksquare$


Examples

Example: $37$

\(\ds \dfrac 1 {37}\) \(=\) \(\ds 0 \cdotp \dot 0 2 \dot 7\)
\(\ds 27 \times 37\) \(=\) \(\ds 10^3 - 1\)


Example: $41$

\(\ds \dfrac 1 {41}\) \(=\) \(\ds 0 \cdotp \dot 0 243 \dot 9\)
\(\ds 2439 \times 41\) \(=\) \(\ds 10^5 - 1\)


Sources

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