# 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.