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

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The discussion proves that a repunit \( R_n \) is divisible by 9 if and only if \( n \) is divisible by 9. It starts by showing that if \( 9 \mid R_n \), then the sum of its digits, which consists of \( n \) ones, must also be divisible by 9, leading to \( 9 \mid n \). Conversely, if \( 9 \mid n \), the sum of the digits in \( R_n \) equals \( n \), ensuring \( 9 \mid R_n \). The proof concludes that the divisibility condition holds in both directions. This establishes a clear relationship between the divisibility of repunits and their indices.
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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 ##.
 
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Almost.

Math100 said:
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 ##.

Math100 said:
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.)
Math100 said:
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 ##.
 
10^m=999..99+1 \equiv 1(\mod 9)
R_n =\sum_{m=0}^{n-1}10^m \equiv n(\mod 9)
 
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