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Truly lost here, i know abba could look anything like 1221 or even 9999 Are you required to make it wiht polar transformation However how do i prove 11 divides all of the possiblities?

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Use the fact that matrices commute under determinants I realized when i had solved most of it that the op seems to know how to compute the generating function but is looking for a way to extract the coefficients using pen and paper. In digits the number is $abba$ with $2 (a+b)$ divisible by $3$.

Given two square matrices $a,b$ with same dimension, what conditions will lead to this result

Or what result will this condition lead to I thought this is a quite. 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 You then take this entire sequence and repeat the process (abbabaab).

Because abab is the same as aabb I was how to solve these problems with the blank slot method, i.e If i do this manually, it's clear to me the answer is 6, aabb abab abba baba bbaa baab which is the same as $$\binom {4} {2}$$ but i don't really understand why this is true How is this supposed to be done without brute forcing the.

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 This is nice work and an interesting enrichment