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Truly lost here, i know abba could look anything like 1221 or even 9999 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 ways However how do i prove 11 divides all of the possiblities?
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You'll need to complete a few actions and gain 15 reputation points before being able to upvote You then take this entire sequence and repeat the process (abbabaab). Upvoting indicates when questions and answers are useful
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Instead, you can save this post to reference later. I'm trying to figure this one out I know that if a number is divisible by $3$, then the sum of its digits is divisible by $3$ 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 $1001$ and $110$ are
