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Re: cotpi 41 - Counting locally prime numbers
Posted:
Feb 28, 2012 9:59 PM
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"Tonico" <Tonicopm@yahoo.com> wrote in message
news:c72e20a5-44b3-4fe7-b07b-50beb86195cf@w4g2000vbc.googlegroups.com...
> On Feb 27, 9:07 pm, "Mike Terry"
> <news.dead.person.sto...@darjeeling.plus.com> wrote:
> > "Tonico" <Tonic...@yahoo.com> wrote in message
> >
> >
news:f3c955d7-13fd-4f4a-bdcd-411c06f2c280@do4g2000vbb.googlegroups.com...
> >
> >
> >
> >
> >
> > > On Feb 27, 8:50 am, cotpi <puzz...@cotpi.com> wrote:
> > ...
> >
> > > > Let A be a set of 10 consecutive integers. Let B be a subset of A
such
> > > > that every element in A that is coprime to every other element in A
is
> > > > present in B. What are the possible values for the cardinality of B?
> >
> > > > I hope this helps.
> >
> > > > cotpi
> >
> > > Take the set A:= {10, 11,..., 19} . As with any other of the given
> > > characteristics, it has exactly 5 even numbers and 5 odd numbers, and
> > > according to the presented condition NONE of the even numbers can be
> > > present in the wanted subset B (as any of these ones is NOT coprime
> > > with any other even element in A).
> >
> > No, you're misreading (or I am...)
> >
> > For A:= {10, 11,..., 19} we can have B:= A. (In fact, we can do this for
> > any A.)
> >
> > Mike.-
>
>
>
> I wasn't addressing your post, but now I am your last one: what do you
> mean by "we can have"?!? Either we have B = A (and I can't understand
> why you use the mathematical sign of "definition" here) or we don't.
No, B is not fixed. First A is chosen, then we choose a subset B which
satisfies the OP's criterion above. I repeat that criterion here:
Let B be a subset of A such that every element in A that is
coprime to every other element in A is present in B.
Note "let B be *a* subset of A such that...". For a given set A there are
dozens of possible subsets B meeting the criterion. When I said "we can
have" B:= A I was just saying that if we take the set B to be the set A,
then the set B satisfies the criterion above.
>
> Is is true that ANY element in A = {10, 11,...,19} is coprime with ANY
> other element in A? No. For example, 10 is not coprime with 12 and,
> thus, 10 cannot belong to B so, I don't really understand how can you
> argue that IN THIS CASE we [can] have B = A...
Yes, 10 is not coprime to 12, but that doesn't exclude 10 from being in B.
That's not what the criterion says - just read it carefully! :)
>
> And there can't be any set A of 10 consecutive integers s.t. B = A
> since any of the 5 even ones is NOT coprime with any of the other even
> ones, as I explained before.
Sure it can - just read it carefully and check the criterion with A:= {10,
11,..., 19} and B:= {10, 11,..., 19}. (I.e. check that every element of A
which [blah blah blah..] is present in B.)
But this is sort of irrelevant (other than answering your question) because
the OP has already said that the criterion we're discussing here wasn't the
one he intended for the problem, and he's posted a corrected version.
(Where B=A does not meet the new criterion.)
Mike.
>
> Tonio
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