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?

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: …

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

Mar 11, 2025 · Suppose $A$ and $B$ are complex linear operators of some finite-dimensional vector space $X$. In my definition, positive definite operators are only for self-adjoint ...

Aug 29, 2013 · 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 ...

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

The commutator [X, Y] of two matrices is defined by the equation $$\begin {align} [X, Y] = XY − YX. \end {align}$$ Two anti-commuting matrices A and B satisfy $$\begin {align} A^2=I \qu...

May 13, 2016 · This is nice work and an interesting enrichment. (+1). 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 …

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

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