[Math4u] Fun Problems

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@gmail.com> 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
>
>
>
>
>
> ---------------------------------
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Yahoo! Search.
>

---

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