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

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 …

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 …

As long as a, b, c a, b, c are not all 0 0, this has rank 2 2, so there is a 2 2 -dimensional linear space of B B for which AB + BA = 0 A B + B A = 0 in this case. On ...

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?

Jun 23, 2022 · Are you required to make it wiht polar transformation? Because with the change $x=uv$ $y=u (1-v)$ it's easier

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...

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 …

Jan 26, 2015 · Let $A$ and $B$ be two $3\times 3$ matrices with complex entries such that $A^2=AB+BA$. Prove that $\det (AB-BA)=0$. (Is the above result true for matrices with real ...

Dec 26, 2012 · A different function with zero spacing pattern ABAB might have period $\pi$. If extended, the ABBA pattern is ABBAABBAABBA.., and there are no periods of less than four …