Can You Solve the Real Challenge?

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can you give the next two rows?

1
11
21
1211
111221
312211
13112221
1113213211
31131211131221
-------------------
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maybe the next row is 12212311321211?
 
The next rows are:

13211311123113112211
and then
11131221133112132113212221
(each line tells you what was in the previous line)
but this is not number theory; it should be in the brain teaser section.
 
Also, it's been done already in the Brain Teaser subforum...but is an fun puzzle, nevertheless - requires some out-of-the-box thinking.
 
Thread 'Determine whether ##125## is a unit in ##\mathbb{Z_471}##'

Thread 'Determine whether ##125## is a unit in ##\mathbb{Z_471}##'

This is the question, I understand the concept, in ##\mathbb{Z_n}## an element is a is a unit if and only if gcd( a,n) =1. My understanding of backwards substitution, ... i have using Euclidean algorithm, ##471 = 3⋅121 + 108## ##121 = 1⋅108 + 13## ##108 =8⋅13+4## ##13=3⋅4+1## ##4=4⋅1+0## using back-substitution, ##1=13-3⋅4## ##=(121-1⋅108)-3(108-8⋅13)## ... ##= 121-(471-3⋅121)-3⋅471+9⋅121+24⋅121-24(471-3⋅121## ##=121-471+3⋅121-3⋅471+9⋅121+24⋅121-24⋅471+72⋅121##...
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Thread 'Trouble understanding an online solution to an exercise in Dummit & Foote'

##\textbf{Exercise 10}:## I came across the following solution online: Questions: 1. When the author states in "that ring (not sure if he is referring to ##R## or ##R/\mathfrak{p}##, but I am guessing the later) ##x_n x_{n+1}=0## for all odd $n$ and ##x_{n+1}## is invertible, so that ##x_n=0##" 2. How does ##x_nx_{n+1}=0## implies that ##x_{n+1}## is invertible and ##x_n=0##. I mean if the quotient ring ##R/\mathfrak{p}## is an integral domain, and ##x_{n+1}## is invertible then...
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Thread 'Questions about non existence of GCDs vs (coimages, cokernels)'

The following are taken from the two sources, 1) from this online page and the book An Introduction to Module Theory by: Ibrahim Assem, Flavio U. Coelho. In the Abelian Categories chapter in the module theory text on page 157, right after presenting IV.2.21 Definition, the authors states "Image and coimage may or may not exist, but if they do, then they are unique up to isomorphism (because so are kernels and cokernels). Also in the reference url page above, the authors present two...
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