coins. From a single problem, which seems to have no systematic way
of solving, it was generalized into a mathematical method.
Thank you Brian,
Thank you other people who participated
Now I wonder if anyone can propose another difficult problem about
coins.
Sinichi7kudo
--- In Math4u@yahoogroups.com, "Brian Edward Jensen"
<brianejensen@...> wrote:
>
> To distribute the coins, the coin labeled 120 in the base 3 number
> 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
> > > > > > >
> > > > > > >
> > > > > > >
> > > > > > >
> > > > > > >
> > > > > > > ---------------------------------
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