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A mediocre professor at a top-notch university making mediocre contributions to this top-notch community


Nov
22
comment How to explain that division by $0$ yields infinity to a 2nd grader
@Marcus My kid came home talking about his friends having a competition about "who can name the biggest number". One said "infinity!" Another said "infinity + 1!" My kid said "infinity to the infinity!" I gave him a time-out for that. Infinity is not a number.
Nov
18
answered Language over $\{0,1\}$
Nov
16
comment Possible values of $N$
@RossMillikan: He edited it: he had the condition "$(n+7)\ |\ 36m$ for integer $m$" as a condition.
Nov
16
comment How many chess games of a suspect does one have to analyze to have a reliable answer to the question whether the suspect cheats?
Don't you get into trouble when there is a long sequence of exchanges? Then every move is obvious to a human and computer: would they be flagged as cheaters?
Nov
16
comment Careers in Math
Well, you almost certainly won't fail to not be rich.
Nov
16
comment Possible values of $N$
I don't think this is right; for example $n=65$ satisfies your criterion, but not the equation required by the asker.
Nov
16
answered How many subsets does $S_{n+1}$ have?
Nov
4
comment Ensuring that a graph has odd number of hamiltonian paths
I think if $G$ is undirected, and if we define a hamiltonian path as a permutation of the vertices, then there are always an even number of such permutations if the number of vertices > 1.
Nov
4
comment Ensuring that a graph has odd number of hamiltonian paths
Is $G$ a digraph?
Oct
31
answered Relations between the maximum matching, minimum vertex cover, maximum independent set, and maximum vertex biclique for a bipartite graph
Oct
31
answered Cards and numbers
Oct
27
comment Why does $1234^{1234^{1234}\ \bmod\ 10^{10}}$ = $1234^{1234^{1234}\ \bmod \ \phi(10^{10})}$
It's true that, if $a\in 10^{10}$ then you can say $a^b = a^{b \bmod\ \phi(10^{10})}$. It's also true you can instead say $a^b = a^{b\ \bmod\ \phi(10^{10})/4}$ since I believe $\phi(10^{10})/4$ is the exponent of the group. For $a\not\in 10^{10}$ the assertion in the question above isn't faulty unless $a$ is a multiple of 10, and 1234 is not. You have given a proof, however, that doesn't provide for this.
Oct
27
comment Why does $1234^{1234^{1234}\ \bmod\ 10^{10}}$ = $1234^{1234^{1234}\ \bmod \ \phi(10^{10})}$
Also, the LCM of all group orders (for periodic groups) is called the "exponent" of the group. Although this seems to not be very well-known (for example, Wolfram Alpha appears to not know the term).
Oct
27
comment Why does $1234^{1234^{1234}\ \bmod\ 10^{10}}$ = $1234^{1234^{1234}\ \bmod \ \phi(10^{10})}$
Does it affect your answer at all that 1234 $\not\in R_{10^{10}}$?
Oct
26
comment Why does $1234^{1234^{1234}\ \bmod\ 10^{10}}$ = $1234^{1234^{1234}\ \bmod \ \phi(10^{10})}$
I just tried it for 1000 values other than 1234, and it always works. Assuming your argument above is correct, there is only a 1 in $2^{1000}$ chance that was just luck, so there is something missing here still.
Oct
26
comment Why does $1234^{1234^{1234}\ \bmod\ 10^{10}}$ = $1234^{1234^{1234}\ \bmod \ \phi(10^{10})}$
I mean why are they equal for these particular constants even though the approach, in general, is flawed. Is this an artifact of base 10? Does 1234 have a particular property? Or does this always work for any base and any constant?
Oct
26
comment Why does $1234^{1234^{1234}\ \bmod\ 10^{10}}$ = $1234^{1234^{1234}\ \bmod \ \phi(10^{10})}$
$\phi(10^{10}) = \phi(2^{10})\phi(5^{10}) = 2^9 \cdot 4\cdot 5^9$.
Oct
26
asked Why does $1234^{1234^{1234}\ \bmod\ 10^{10}}$ = $1234^{1234^{1234}\ \bmod \ \phi(10^{10})}$
Oct
25
revised Expected tail and head length of $\rho$ for a finite random function
added 222 characters in body
Oct
24
accepted Approximation for $2^r\ln \frac{2^r}{2^r-r}$