system, for example, would be in pan 1 for the first weighing, pan 2
for the second weighing, and off to the side for the third weighing,
hence the "120."
Yes, there are many solutions because we must make an arbitrary
choice between label 112 and 221 which are opposites. We must choose
one from each pair so that for all the coins, we have the same number
of coins on each pan. But with thirteen coins, we wind up with 4
coins on one pan and 5 coins on the other pan so we must borrow a
genuine coin.
I don't know about other methods. You came up with a brilliant
alternative method for 12 coins.
Regards,
Brian
--- In Math4u@yahoogroups.com, "sinichi7kudo" <sinichi7kudo@...>
wrote:
>
> To Brian,
> The coins problem became so beautiful.
> Surprisingly, I solved it the same way as you did, and I also
> figured out that 14 coins can be solved with certain restrictions.
> But I didn't know how to distribute the coins on the balance and
how
> to move them. So according to your solution, are there a lot of
ways
> to arrange the coins on the balance?
> Further more, for the 13 coins, I wonder if we have any solution to
> distinguish the fake (don't know if it's lighter or heavier),
> besides using the method that you did.
>
> Regards,
> Sinichi7kudo
> --- In Math4u@yahoogroups.com, "Brian Edward Jensen"
> <brianejensen@> wrote:
> >
> > To Sinichi7kudo
> > I will study your 13 coin solution when you give it. You caused
> me
> > to study my solution further. The basic solution can be done for
> 12
> > coins. We can do it for an additional coin is we leave one coin
> off
> > to the side for all three weighings but we won't know is this
coin
> is
> > heavy or light. We can do an additional coin if we add a genuine
> > coin on pan 2 for each of the three weighings so that we test 5
> coins
> > on pan 1 and 4 coins on pan 2 each time. This gets us up to 14
> coins
> > but we will not know if the unweighed coin is heavy or light.
> >
> > Let's say we call the two pans on our balance pan 1 and pan 2.
> > IDEA: We could have a balance with 3 pans arranged in a
> triangle.
> > Group 3 is off to the side and not weighed.
> >
> > Let's write all the three digit numbers 0-26 in the base 3 number
> > system:
> > 000
> > 001
> > 002
> > 010
> > 011
> > 012
> > 020
> > etceteras
> >
> > Let these 27 numbers be the individual ID numbers of the coins.
> > Let the first digit be the first weighing, the second digit be
the
> > second weighing and the third digit be the third weighing.
> > Let the "0" mean the coin is not on the balance.
> > Let the "1" mean the coin is on pan 1.
> > Let the "2" mean the coin is on pan 2.
> > If we know that one of the coins is 25% heavier, we can identify
> > which of the 27 coins is the heavy coin. If the balance
balances,
> > then we have a zero. If pan 1 goes down, then we have a "1." If
> pan
> > 2 goes down, then we have a "2."
> >
> > If we don't know if the bad coin is heavy or light, we will not
be
> > able to distinguish between 001 and 002 or 012 and 021 for
> examples.
> > So in this case we must pair up the indistinguishable pairs of
> > numbers as follows:
> > 000
> > 001, 002
> > 010, 020
> > 011, 022
> > 012, 021
> > 100, 200
> > 101, 202
> > 102, 201
> > 110, 220
> > 111, 222
> > 112, 221
> > 120, 210
> > 121, 212
> > 122, 211
> >
> > What we need to do next is to choose one of each of the two
> numbers
> > so that we have some coins on each pan. There are many solutions
> so
> > just try some choices at random and then make a few switches:
> >
> > 000
> > 001, 002 choose 001
> > 010, 020 choose 010
> > 011, 022 choose 011
> > 012, 021 choose 012
> > 100, 200 choose 200
> > 101, 202 choose 101
> > 102, 201 choose 102
> > 110, 220 choose 220
> > 111, 222 choose 111
> > 112, 221 choose 221
> > 120, 210 choose 120
> > 121, 212 choose 212
> > 122, 211 choose 122
> >
> > If we count the 1's in the first column of the 13 choices, we
find
> we
> > have 5 coins in pan 1 and 4 coins in pan 2. The same is true for
> the
> > second and third weighings.
> >
> > Options:
> > If we delete coin 111, then we will have 4 coins in each pan.
For
> 12
> > coins, we can tell if the bad coin is heavy or light.
> > We can add a thirteenth coin 000 which will be indicated as the
> bad
> > coin if the balance balances all three weighings but we won't
know
> if
> > it is heavy or light.
> > We can add a genuine coin so that we have 5 coins on each side
> > instead of 4 on one side and 5 on the other. Now we can tell
> whether
> > the bad coin is heavy or light out of 13 coins.
> > We can add a fourteenth coin 000 which will be indicated as the
> bad
> > coin is the balance balances all three weighings but we won't
know
> if
> > it is heavy or light.
> >
> > Regards,
> > Brian
> >
> > --- In Math4u@yahoogroups.com, "sinichi7kudo" <sinichi7kudo@>
> > wrote:
> > >
> > > I just have another question: If you know that the fake coin is
> > > 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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> ---
> > -
> > > -------
> > > > > Be a better friend, newshound, and know-it-all with Yahoo!
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> > > >
> > >
> >
>
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