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]

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