work out such a solution. It appears that his solution covers all the
possibilities. It is also possible to decide the positions of the 12
coins on the three weighings before starting and to analyze the
results afterwards.
Regards,
Brian
--- In Math4u@yahoogroups.com, "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, 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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