Divisors of Repunit with Composite Index

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Theorem

Let $R_n$ be a repunit number with $n$ digits.

Let $n$ be composite such that $n = r s$ where $1 < r < n$ and $1 < s < n$.

Then $R_r$ and $R_s$ are both divisors of $R_n$.

Proof

Let $n = r s$.

Then:

\(\ds R_n\)

\(=\)

\(\ds \sum_{k \mathop = 0}^{n - 1} 10^k\)

Basis Representation Theorem

\(\ds \)

\(=\)

\(\ds \sum_{j \mathop = 0}^{s - 1} \paren {\sum_{k \mathop = 0}^{r - 1} 10^k} 10^{r j}\)

\(\ds \)

\(=\)

\(\ds \paren {\sum_{k \mathop = 0}^{r - 1} 10^k} \paren {\sum_{j \mathop = 0}^{s - 1} 10^{r j} }\)

\(\ds \)

\(=\)

\(\ds R_r \paren {\sum_{j \mathop = 0}^{s - 1} 10^{r j} }\)

Similarly:

$R_n = R_s \paren {\ds \sum_{j \mathop = 0}^{r - 1} 10^{s j} }$

$\blacksquare$

Thus, for example:

\(\ds R_6 = 111 \, 111\)

\(=\)

\(\ds 11 \times 10 \, 101\)

\(\ds \)

\(=\)

\(\ds 111 \times 1001\)

\(\ds R_{10} = 1 \, 111 \, 111 \, 111\)

\(=\)

\(\ds 11 \times 101 \, 010 \, 101\)

\(\ds \)

\(=\)

\(\ds 11111 \times 100 \, 001\)

The pattern is clear.