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Truly lost here, i know abba could look anything like 1221 or even 9999 $$\\overline{abba}$$ so it is equal to $$1001a+110b$$ and $1001$ and $110$ are However how do i prove 11 divides all of the possiblities?
Use the fact that matrices commute under determinants For example a palindrome of length $4$ is always divisible by $11$ because palindromes of length $4$ are in the form of 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
In digits the number is $abba$ with $2 (a+b)$ divisible by $3$.
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.
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. There must be something missing since taking $b$ to be the zero matrix will work for any $a$.
The commutator [x, y] of two matrices is defined by the equation $$\begin {align} [x, y] = xy − yx
This is nice work and an interesting enrichment 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.
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