安裝中文字典英文字典辭典工具!
安裝中文字典英文字典辭典工具!
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- 11 | abba, where a and b are the digits in a 4 digit number.
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?
- If $ABBA-BAAB=A-B$, show $\operatorname {tr} (A^2)=\operatorname {tr . . .
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 $\lambda$
- How many ways can we get 2 as and 2 bs from aabb?
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 the same as $$\binom {4} {2}$$ But I don't really understand why this is true? How is this supposed to be done without brute forcing the
- How to calculate total combinations for AABB and ABBB sets?
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: First one: AABB, BBAA,
- Matrices - Conditions for $AB+BA=0$ - Mathematics Stack Exchange
There must be something missing since taking $B$ to be the zero matrix will work for any $A$
- How many $4$-digit palindromes are divisible by $3$?
Hint: in digits the number is $abba$ with $2 (a+b)$ divisible by $3$
- prove $\\Gamma(a)\\Gamma(b) = \\Gamma(a+b)B(a,b)$ using polar . . .
Are you required to make it wiht polar transformation? Because with the change $x=uv$ $y=u (1-v)$ it's easier
- $A^2=AB+BA$. Prove that $\det (AB-BA)=0$ [duplicate]
I get the trick Use the fact that matrices "commute under determinants" +1
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