Show that ## 9\mid R_{n} ## if and only if ## 9\mid n ##

[Math100]

Homework Statement

Given a repunit $R_{n}$, show that $9\mid R_{n}$ if and only if $9\mid n$.

Relevant Equations

None.

Proof:

Suppose $9\mid R_{n}$, given a repunit $R_{n}$.
Then $R_{n}=111\dotsb 1$.
Observe that $9\mid R_{n}\implies 9\mid (1+1+1+\dotsb +1)$.
Thus $9\mid n$.
Conversely, suppose $9\mid n$.
Then $R_{n}=1+1+1+\dotsb +1=n$, where $n$ is the sum of digits in $R_{n}$.
Since the sum of digits in $R_{n}$ is divisible by $9$, it follows that $9\mid R_{n}$.
Therefore, $9\mid R_{n}$ if and only if $9\mid n$.

 

[fresh_42]

Almost.

Homework Statement:: Given a repunit $R_{n}$, show that $9\mid R_{n}$ if and only if $9\mid n$.
Relevant Equations:: None.

Proof:

Suppose $9\mid R_{n}$, given a repunit $R_{n}$.
Then $R_{n}=111\dotsb 1$.
Observe that $9\mid R_{n}\implies 9\mid (1+1+1+\dotsb +1)$.
Thus $9\mid n$.
Conversely, suppose $9\mid n$.

Then $R_{n}=1+1+1+\dotsb +1=n$, where $n$ is the sum of digits in $R_{n}$.

Then $1+1+1+\dotsb +1=n$, where $n$ is the sum of digits in $R_{n}$.
(It does not equal $R_n$ and it is not necessary.)

Since the sum of digits in $R_{n}$ is divisible by $9$, it follows that $9\mid R_{n}$.
Therefore, $9\mid R_{n}$ if and only if $9\mid n$.

 

[anuttarasammyak]

$$10^m=999..99+1 \equiv 1(\mod 9)$$
$$R_n =\sum_{m=0}^{n-1}10^m \equiv n(\mod 9)$$