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 $\lambda$.

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 …

Feb 28, 2018 · Hint: in digits the number is $abba$ with $2 (a+b)$ divisible by $3$.

Apr 5, 2026 · I’ve been trying to build some intuition around convolution in signal processing and I thought of an analogy with carry in decimal arithmetic, but I’m not sure if it’s valid. When we add …

I get the trick. Use the fact that matrices "commute under determinants". +1

I've found and proven the following extensions to palindromes of the usual divisibility rules for 3 and 9: A palindrome is divisible by 27 if and only if its digit sum is. A palindrome is divisible by 81 if and only if …

There must be something missing since taking $B$ to be the zero matrix will work for any $A$.

Sep 19, 2023 · The first condition holds for all $b$ or for some particular $b$?

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: …

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