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
- 1986: David Wells: Curious and Interesting Numbers ... (previous) ... (next): $142,857$
- 1997: David Wells: Curious and Interesting Numbers (2nd ed.) ... (previous) ... (next): $142,857$