Re: [Math4u] Fun Problems

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@prodigy.net> 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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