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 …

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?

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

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: …

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 …

Oct 4, 2016 · The algorithm is normally created by taking AB, then inverting each 2-state 'digit' and sticking it on the end (ABBA). You then take this entire sequence and repeat the process (ABBABAAB).

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

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 30, 2017 · Maybe you can assume SVD (Singular Value Decomposition) of A and B and see where that leads.

The 4 4 -digit palindrome abba a b b a is divisible by 101 iff a = b a = b. The 5 5 -digit palindrome abcba a b c b a is divisible by 101 iff c = 2a c = 2 a. The 6 6 -digit palindrome abccba a b c c b a is divisible …