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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?
- How many $4$-digit palindromes are divisible by $3$?
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 have done is listed out lots of numbers that work I haven't developed a nice technique for this yet
- 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
- Matrices - Conditions for $AB+BA=0$ - Mathematics Stack Exchange
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
- 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 to show that $A^2=AB+BA$ implies $\det (AB-BA)=0$ for $3\times3 . . .
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
- 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
- sequences and series - The Perfect Sharing Algorithm (ABBABAAB . . .
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)
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