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Jan 26, 2026 · That is, there seems to be fairly strong symbolic evidence that for $n=4$, if $ABBA-BAAB = A-B$ and $A$ is nilpotent, then $B^4 = \lambda I$ for some $\lambda$. Nov 21, 2013 · Truly lost here, I know abba could look anything like 1221 or even 9999. However how do I prove 11 divides all of the possiblities? 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.

Apr 19, 2022 · 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:.