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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?
- $A^2=AB+BA$. Prove that $\det (AB-BA)=0$ [duplicate]
I get the trick Use the fact that matrices "commute under determinants" +1
- matrices - When will $AB=BA$? - Mathematics Stack Exchange
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
- 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,
- How many $4$-digit palindromes are divisible by $3$?
Hint: in digits the number is $abba$ with $2 (a+b)$ divisible by $3$
- 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
- elementary number theory - Common factors for all palindromes . . .
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
- 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 $A$
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