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21866
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location Mexico City
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visits member for 2 years, 6 months
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Hello, I am a freshman at The National Autonomous University of Mexico (UNAM).

My interests are contest problems,elementary number theory, combinatorics, graph theory, linear algebra, group theory, abstract algebra and I have been meaning to learn category theory and matroids.


1d
answered Set with distinct subset sums
1d
comment How to solve this kind of problem?
I solved it using integers, I think it becomes harder if we can only use naturals.
1d
revised How to solve this kind of problem?
added 267 characters in body
1d
comment Check if Sequence is Graphic: 8, 8, 7, 7, 6, 6, 4, 3, 2, 1, 1, 1
I mean, you should count the degrees in the graph to make sure, but I can't find more conclusive evidence it is graphic
1d
comment Check if Sequence is Graphic: 8, 8, 7, 7, 6, 6, 4, 3, 2, 1, 1, 1
Well, I counted the degrees in the graph in the solution and it is correct, are you sure you copied the sequence correctly?
1d
answered How to solve this kind of problem?
1d
comment How to solve this kind of problem?
The number in the bottom right corner has to be $1,-1,11$ or $-11$
1d
answered Check if Sequence is Graphic: 8, 8, 7, 7, 6, 6, 4, 3, 2, 1, 1, 1
1d
comment Is it possible to permute an unknown binary sequence so that two particular bits are equal?
I editted it, is it convincing now?
1d
revised Is it possible to permute an unknown binary sequence so that two particular bits are equal?
added 886 characters in body
1d
awarded  abstract-algebra
2d
comment An abelian subgroup of symmetric group
Wow, that was pretty sweet. Using abelians are clt was rather ingenious.
2d
comment Normal subgroups in groups of odd order
I don't see why $H$ needs to be normal for the first proof. Couldn't we use the same argument without normality?That is: the order of $N_G(H)/C_G(H)$ divides the order of $N_G(H)$ and therefore the order of $G$. But it also divides the order of $Aut(H)$? since the order of $Aut(H)$ is relatively prime to the order of $G$ we get$C_G(N)=N_G(N)$. Oh, I see why you need it to be normal lol.
2d
comment Normal subgroups in groups of odd order
proposition $2$ can be proven using the thing about normal subgroups being a union of conjugacy classes.the smallest non trivial ones have size $p$. Although prop 2 is not generalization.
2d
comment Normal subgroups in groups of odd order
The proof of the proposition is the same isn't it? You get the order of $N_G(H)/C_G(H)$ divides the order of $\Aut(H)=p-1$ and the order of $N(G)$. Why does $H$ have to be normal though? Can't you get the same results without $H$ being normal?
2d
comment Is it possible to permute an unknown binary sequence so that two particular bits are equal?
My point is that he can safely assume the swap was done when $r>k$ and he can safely assume the swap was not done when $r<k$. Because the sequence will be the same. (Of course it is possible the swap didn't actually occur, but for all intents and purposes we can assume it did, since the outcome would be the same).
2d
comment Is it possible to permute an unknown binary sequence so that two particular bits are equal?
By induction the sequence will always be a permutation of the $b$'s. So if after a number of steps he can determine the bit at position $m$ is the same as the one in position $n$ he knows that those bits correspond to $b_k$ and $b_r$ for example(wlog $k<r$). and then he knows that the number of zeroes is not $k$ (since otherwise they would be different).
2d
comment Is it possible to permute an unknown binary sequence so that two particular bits are equal?
Hmm, what I mean is the following: The mathematician starts with the ordered sequence $b_1,b_2,b_3\dots b_{2015}$. In the first move if he sends an electron from $b_r$ to $b_k$ with $r<k$ he knows that the sequence will stay the same. so the sequence is still $b_1,b_2\dots b_{2015}$. if $r>k$ he knows that the sequence is going to be the same as the sequence when $b_r$ and $b_k$ are switched (because the only possibilities are $b_r=b_k=1$ or $b_r=1,b_k=0$. In both cases the string becomes the string when $b_r$ is switched with $b_k$ and everything stays the same.
2d
accepted Is it possible to permute an unknown binary sequence so that two particular bits are equal?
2d
comment Is it possible to permute an unknown binary sequence so that two particular bits are equal?
well yes, but in this case the mathematician knows which where the "coordinates" of the diodes he did, what I mean is that he can reverse his own process.