heavier than other coins. And you are allowed to do 3 weightings. So
what is the maximum number of coins?
And by the way, I found out a way to weight 3 times for 13 coins to
find out the fake coin (without knowing the fake is heavier or
lighter)
Thank you.
--- In Math4u@yahoogroups.com, "Michael S." <M.Suesserott@...> wrote:
>
> There is a large body of literature on this coin weighing problem.
For
> starters, you might look at
> http://www.cut-the-knot.org/blue/OddballProblem1.shtml
> and follow the links given at the end of the article.
>
> This problem can also be attacked via an information theoretical
> approach, using entropy calculations. IMHO, this is one of the
best and
> most elucidating ways to solve the puzzle. For a brief
introduction, see
> http://www.av8n.com/physics/twelve-coins.htm
>
> For those who want to delve deeper into this, there is the
excellent
> book, "Probability and Information" (Theory and Decision Library)
by
> A.M. Yaglom and I.M. Yaglom.
>
> Best wishes,
>
> Michael
>
>
> Mosaad Alabdullatif wrote:
> > So great
> > So magnificent
> >
> > I really love it
> >
> > The question now: if they were 13 coins, we would be forced to
hit the
> > 6-6 route. Would this way leads to such a briliant solution?
> >
> > Thank you Sinichi7kudo
> > Thank you Brian
> >
> > Mosaad
> >
> > */sinichi7kudo <sinichi7kudo@...>/* wrote:
> >
> > Here is how you do it. It certainly can be solved. I'm now
trying to
> > apply the same logic to more than 12 coins, but no successes
so far.
> >
> > Divide the coins into three groups of 4 coins each. First,
weigh two
> > of the groups. If they balance, then the fake coin is in the
third
> > group, so take three of the coins, and weigh them against
three of
> > the coins that we know for sure are real (from the first 8
coins).
> > If they balance, then the remaining coin is the fake one,
but we
> > still don't know if it is lighter or heavier, so weigh it
against a
> > real coin to determine that. But if it didn't balance, then
we know
> > that the fake coin is one of the three, and we also know if
it is
> > lighter or heavier, because we know which side had the real
coins.
> > So weigh two of the coins against each other, if they
balance then
> > the third coin is fake, if they don't balance we know which
coin is
> > fake, because we already know if it is lighter or heavier.
> > Now let's go back to the first weighing; if the two groups
didn't
> > balance, we know that the fake coin is in one of the groups,
but we
> > don't know which group. Let's mark the group that was
heavier as
> > group A, and the group that was lighter group B. For the
second
> > weighing, weigh two coins of group A and one coin of group B,
> > against the other two coins of group A and one coin of group
B. If
> > they balance, the fake coin must be one of the remaining
two, and a
> > weigh between them will tell us which one (they are both
from group
> > B so we know that it's lighter). If the second weigh didn't
balance,
> > look at the side that is heavier: Either the fake coin is
one of the
> > two group A coins, or it can the group B coin on the other
side of
> > the scale. Take the two group A coins, and weigh them
against each
> > other. If they balance, then it is the group B coin that is
fake,
> > and if they don't balance, then the heavier one is the fake
coin.
> > And that's that. We have covered all possibilities.
> >
> > --- In Math4u@yahoogroups.com <mailto:Math4u%
40yahoogroups.com>,
> > BRIAN JENSEN <brianejensen@> wrote:
> > >
> > > My sixth grade teacher gave this problem to us around
1962. It
> > took me 38 years to figure out. He didn't remember me or the
problem.
> > > I have found many solutions to this problem. There is no
regular
> > pattern that I could find. If you can put the coins on the
left pan,
> > right pan, or to the side, then how many different ways are
there to
> > position the coins in three weighings? It certainly can be
solved.
> > > Regards,
> > > Brian
> > >
> > >
> > > Mosaad Alabdullatif <sam362yah@> wrote:
> > > I do not think 3 wieghings will be enough if we do not
> > know if that defected coin is light or heavy!!
> > >
> > > Mosaad
> > >
> > > Brian Edward Jensen <brianejensen@> wrote:
> > > I made a mistake! (Stephen Tavener pointed this out to me.)
> > On the
> > > second problem, I meant to say "sphere," not "cube." I
started
> > > writing this problem thinking that the area of a cube
could be
> > > obtained from derivative of the volume with respect to
half a
> > side.
> > > But this would be too easy because someone could easily
calculate
> > the
> > > area of a side and multiply by 6 sides for the total area.
So then
> > I
> > > decided to use a sphere.
> > >
> > > Here is a resubmission of the two problems and I am
deleting the
> > > original:
> > >
> > > Problem 1:
> > > Here's an old problem, one of my favorites that we haven't
had in
> > a
> > > couple years:
> > > A man has 12 coins and one is bad. So eleven coins have
the same
> > > weight and one has a different weight. The man puts them
on a
> > > balance, some coins on each pan. He can make 3 weighings.
This is
> > an
> > > old-fashioned balance with 2 pans on opposite ends of an
arm. Of
> > > course he must have the same number of coins on each side
but he
> > > doesn't need to weigh all the coins each time.
> > > How can he figure out which coin is bad with 3 weighings
if he
> > knows
> > > the bad coin is heavy? (Actually, he can pick the bad one
out of
> > 27
> > > coins if he knows one is 25% heavy)
> > > Alternately and far more difficult,
> > > How can he figure out which coin is bad with 3 weighings
if it is
> > not
> > > known if the coin is heavy or light? (Actually Stephen
Tavener is
> > > correct in a private communication saying 13 coins would be
> > > possible. If a 13th coin is left on the side without
weighing, and
> > a
> > > bad coin is not identified among the twelve, we would know
it is
> > coin
> > > 13. We would not know if this coin were heavy or light,
however.)
> > >
> > > Problem 2:
> > > Another problem:
> > > If the volume of a sphere is C*r^3
> > > Where
> > > C is a constant
> > > r is the radius
> > > Using simple differentiation instead of looking it up in a
book,
> > what
> > > is the
> > > surface area of the sphere? Show your logic.
> > > Regards,
> > > Brian
> > >
> > >
> > >
> > >
> > >
> > > ---------------------------------
> > > Looking for last minute shopping deals? Find them fast with
> > Yahoo! Search.
> > >
> >
> >
> > -----------------------------------------------------------------
-------
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> > it now.
> >
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> >
>
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