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?

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 …

Feb 28, 2018 · How many $4$-digit palindromes are divisible by $3$? I'm trying to figure this one out. I know that if a number is divisible by $3$, then the sum of its digits is divisible by $3$. All I …

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Let $A,B$ be two $3\times 3$ matrices with complex entries, such that $A^2=AB+BA$. Prove that $\det (AB-BA)=0$

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 …

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 …

Jan 11, 2015 · there is a similar thread here Coordinate-free proof of $\operatorname {Tr} (AB)=\operatorname {Tr} (BA)$?, but I'm only looking for a simple linear algebra proof.

May 13, 2016 · I do realize that the method that you show in your post is more powerful than what I presented. Otherwise with non-trivial words you have to construct a DFA from the prefixes of …

Dec 11, 2014 · Prove that $a^a \\ b^b \\ge a^b \\ b^a$, if both $a$ and $b$ are positive.