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Truly lost here, i know abba could look anything like 1221 or even 9999 Can i repeat this forever. However how do i prove 11 divides all of the possiblities?

You'll need to complete a few actions and gain 15 reputation points before being able to upvote This allowed me to get a computer to achieve the algorithm in many ways, but it also raised some questions Upvoting indicates when questions and answers are useful

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For example a palindrome of length $4$ is always divisible by $11$ because palindromes of length $4$ are in the form of $$\\overline{abba}$$ so it is equal to $$1001a+110b$$ and. I do realize that the method that you show in your post is more powerful than what i presented Although both belong to a much broad combination of n=2 and n=4 (aaaa, abba, bbbb.), where order matters and repetition is allowed, both can be rearranged in different.

Prove that $a^a \\ b^b \\ge a^b \\ b^a$, if both $a$ and $b$ are positive.

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