# Show that ## 11\mid R_{n} ## if and only if ## n ## is even

• Math100
In summary, this proof shows that for a repunit ## R_{n} ##, ## 11\mid R_{n} ## if and only if ## n ## is even. This is based on the criterion that a number is divisible by 11 if and only if its alternating sum of digits is, which is used to show that ## T=0 ## and thus ## n ## must be even. The second part of the proof shows that if ## n ## is even, then ## R_{n} ## is divisible by 11. Therefore, we can conclude that ## 11\mid R_{n} ## if and only if ## n ## is even.
Math100
Homework Statement
Given a repunit ## R_{n} ##, show that ## 11\mid R_{n} ## if and only if ## n ## is even.
Relevant Equations
None.
Proof:

Suppose ## 11\mid R_{n} ##, given a repunit ## R_{n} ##.
Let ## R_{n}=1\cdot 10^{m}+\dotsb +1\cdot 10+1 ## and ## T=(a_{0}-a_{1})+(a_{2}-a_{3})+\dotsb +(-1)^{m}a_{m} ##.
Then ## T=(1-1)+(1-1)+\dotsb +(-1)^{m}a_{m}=0 ##.
This means ## 11\mid R_{n}\implies T=0 ##.
Thus, ## n ## is even.
Conversely, suppose ## n ## is even.
Then ## 1-1+1-1+\dotsb -1+1=0 ## and ## 11\mid 0 ##.
Thus ## 11\mid R_{n} ##.
Therefore, ## 11\mid R_{n} ## if and only if ## n ## is even.

Math100 said:
Homework Statement:: Given a repunit ## R_{n} ##, show that ## 11\mid R_{n} ## if and only if ## n ## is even.
Relevant Equations:: None.

Proof:

Suppose ## 11\mid R_{n} ##, given a repunit ## R_{n} ##.
Let ## R_{n}=1\cdot 10^{m}+\dotsb +1\cdot 10+1 ## and ## T=(a_{0}-a_{1})+(a_{2}-a_{3})+\dotsb +(-1)^{m}a_{m} ##.
Then ## T=(1-1)+(1-1)+\dotsb +(-1)^{m}a_{m}=0 ##.
This means ## 11\mid R_{n}\implies T=0 ##.
Thus, ## n ## is even.
Conversely, suppose ## n ## is even.
Then ## 1-1+1-1+\dotsb -1+1=0 ## and ## 11\mid 0 ##.
Thus ## 11\mid R_{n} ##.
Therefore, ## 11\mid R_{n} ## if and only if ## n ## is even.
The first part is a bit clumsy and a link to the thread with that criterion that you use would have been nice, meaning: I couldn't find it.

I would write the first part as:

Suppose ## 11\mid R_{m} ##, given a repunit ## R_{m} ##.
Let ## R_{m}=1\cdot 10^{m}+\dotsb +1\cdot 10+1 ## and ## T=(a_{0}-a_{1})+(a_{2}-a_{3})+\dotsb +(-1)^{m}a_{m} ##.
Then (insert link) ##T=0## and ##m## has to be even.

The second part is also shown in
https://www.physicsforums.com/threa...-number-of-digits-is-divisible-by-11.1017287/

fresh_42 said:
The first part is a bit clumsy and a link to the thread with that criterion that you use would have been nice, meaning: I couldn't find it.

I would write the first part as:

Suppose ## 11\mid R_{m} ##, given a repunit ## R_{m} ##.
Let ## R_{m}=1\cdot 10^{m}+\dotsb +1\cdot 10+1 ## and ## T=(a_{0}-a_{1})+(a_{2}-a_{3})+\dotsb +(-1)^{m}a_{m} ##.
Then (insert link) ##T=0## and ##m## has to be even.

The second part is also shown in
https://www.physicsforums.com/threa...-number-of-digits-is-divisible-by-11.1017287/
But what specifically, should I insert in the first part of the proof, where "Then (insert link) ## T=0 ##..."?

Math100 said:
But what specifically, should I insert in the first part of the proof, where "Then (insert link) ## T=0 ##..."?
Sorry, my fault. You just use the divisibility by 11 criterion. I thought it was an earlier thread. I haven't had it in mind. So maybe a reminder for people like me would be nice: A number is divisible by 11 if and only if its alternating sum of digits is.

Math100

## What does it mean for a number to be divisible by 11?

A number is divisible by 11 if it can be divided evenly by 11 without any remainder.

## What is the significance of the number 11 in this statement?

The number 11 is significant because it is the divisor in the statement. In order for a number to be divisible by 11, it must be a multiple of 11.

## Why does the statement only apply to even numbers?

This statement only applies to even numbers because when an even number is divided by 11, the result will always be an integer. This is because an even number can be expressed as 2 multiplied by another integer, and when divided by 11, the 2 can be factored out leaving only the other integer.

## How can I prove that a number is divisible by 11?

To prove that a number is divisible by 11, you can use the divisibility rule for 11 which states that if the alternating sum of the digits is divisible by 11, then the original number is also divisible by 11.

## Can this statement be generalized to other numbers besides 11?

Yes, this statement can be generalized to any prime number. In general, if a prime number p divides a number n, then n must be a multiple of p. This can be written as "if p|n, then n is a multiple of p".

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