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visits member for 3 years, 2 months
seen Apr 18 at 15:55

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


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 $2^r\ln \frac{2^r}{2^r-r}$
Oct
24
comment Probability that a sequence repeats itself
@joriki: Disjoint events are independent (but not necessarily the other way around). I was using the contrapositive: that because they are not independent (which agrees with the statement in your answer) they are not disjoint, so you cannot simply add them. Do you agree?
Oct
23
comment Probability that a sequence repeats itself
At first blush, 1/9 suggests itself: 1/10+1/100+1/1000+... but this isn't right because the probabilities are not independent. So it appears the answer is between 1/10 and 1/9.
Oct
4
comment Approximation for $2^r\ln \frac{2^r}{2^r-r}$
@Mercy: By "about linear" I mean the function is $\tilde{O}(r)$
Oct
4
asked Approximation for $2^r\ln \frac{2^r}{2^r-r}$