It's been 6 days without a solution.. should I post it now or give you more time?

I could give you some clues..
the answer is 3 in all cases.
Start with 3 groups of 4 balls.



Posted on 2006-03-19 11:32:43 by pdixon
To the Ineffable All,
好bsp; 好bsp; 好bsp;With the clues given by pdixon, I was able to solve the 12 ball problem.好bsp; Naturally I am kicking myself for not seeing it sooner, because I did consider 3 groups of 4 balls.

好bsp; 好bsp; 好bsp;Let's define some conditions first. The solutions to the conditions are so simple that I need not bother to explain them.

Condition one:  If one has 2 balls, then the odd ball can be found with one weighing.
Condition two:  If one has 3 balls, and it is known that the odd ball is heavy/light, then the odd ball can be found in one weighing.
Condition three: If one has 4 balls, then the odd ball can be found in two weighs.

    Weigh any two groups of 4 balls. If the scales weighs even, then condition three applies to the group that did not get weighed. Otherwise, remove three balls from one side of the scale and set aside.好bsp; Replace the three balls that were set aside with three balls from the other side of the scale.好bsp; Replace the three balls from the other side of the scale with three balls from the unweighted 4 ball group which will consist of only even balls.好bsp; If the scale does not change, then the odd balls are one of the two balls that did not get switch or replaced on the scale.好bsp; Use condition one to find the odd ball.好bsp; If the scale goes even, then the three ball set aside group contains the odd ball.  If the scale reverses its tilt, then the 3 ball group that was switched contains the odd ball.  It can easily be determined from the even or reverse tilt whether the odd ball was heavy or light.好bsp; Apply condition two to find the odd ball.

好bsp; 好bsp; 好bsp;Now I will give you an easy problem and a hard one.

1) Factor the polynomial x^4+64 .好bsp; In other words factor the term x to the 4th power plus 64.好bsp; Do it algebraically.

2) With the possible exception of the last problem and the earlier problem involving a hollow sphere, explain to me how developing skill in solving these kinds of problems is of any practical use in the real world.好bsp; Carry on. Ratch

Posted on 2006-03-20 07:55:07 by Ratch
Ratch,
Replace the three balls that were set aside with three balls from the other side of the scale.  Replace the three balls from the other side of the scale with three balls from the unweighted 4 ball group which will consist of only even balls.


That's the trick along with knowing to start with 3 groups of 4 balls.
Well done!.. When I first had that one to solve it took me a week, but I didn't know what the solution was so I could've been looking for a 3 weigh solutuion that didn't exist.

            Now for your 2 questions.

Factorise x^4 + 64
 
Consider a^2 + b^2
The general solution can been seen by inspection as (a+jb)(a-jb) where j= square root of -1
We can verify this by expanding:
(a+jb)(a-jb) = a^2 + ajb - ajb -(j^2)(b^2) = a^2-(-1)b^2 = a^2 + b^2, as expected.

Notice that x^4 + 64 = (x^2)^2 + 8^2
Now substitute x^2=a and 8^2=b and we get:
x^4 + 64 = (x^2+j8)(x^2-j8)


Now sort out these 2 factors:
(x^2+j8) = (x+ sqrt(-j8))(x- sqrt(-j8))
(x^2-j8) = (x+ sqrt( j8))(x- sqrt( j8))

Note that sqrt(j)=((1+j)/sqrt(2))
We can verify this by expanding:
((1+j)/sqrt(2)) ((1+j)/sqrt(2)) = (1+j)(1+j)/2 = (1 +2j -1)/2 = j

Use this to simplify the 2 factors above:
(x^2+j8) = (x+ sqrt(-j8))(x- sqrt(-j8))
= (x+ sqrt(-1)  sqrt(j)      sqrt(8))(x- sqrt(-1)  sqrt(j)    sqrt(8))
= (x+    j  (1+j)/sqrt(2) 2sqrt(2))(x-      j  (1+j)/sqrt(2) 2sqrt(2))
= (x+    j  (1+j)        2      )(x-    j      (1+j)        2    )
= (x+ 2j-2)(x -2j+2)

Similarly for the other factor
(x^2-j8) = (x+ sqrt(8j)) (x- sqrt(8j))
= (x+ sqrt( 8 )    sqrt(j)    )(x- sqrt( 8 )    sqrt(j))
= (x+ 2sqrt(2) (1+j)/sqrt(2) )(x- 2sqrt(2) (1+j)/sqrt(2))
= (x+ 2+2j)(x-2-2j)

So the answer is:
x^4 + 64 = (x+2j-2)(x-2j+2)(x+2+2j)(x-2-2j)


That took longer to figure out how to type in without a sqrt symbol and without smileys appearing everywhere than it did to figure out the answer!


On the second question, anything which gets you used to thinking differently is good as it expands the potential range of thought at your disposal for solving future real world problems.

Paul.
Posted on 2006-03-20 16:35:41 by pdixon
pdixon,

On the second question, anything which gets you used to thinking differently is good as it expands the potential range of thought at your disposal for solving future real world problems.


    I am not so sure about that.  I think probs like that are more of a test than a mind builder.  But I surely could be wrong.

    That first problem occurred in my high school algebra textbook.  I never forgot it.  Your answer is quite correct, but a more succinct method is as follows.

    x^4+64 = x^4 + 16x^2 + 8^2 - 16x^2 = (x^2+8)^2 - (4x)^2 = (x^2 + 4x +8)(x^2 - 4x +8) , from which the quadradic formula can take over.  I am sure the authors of that textbook wanted the students to complete the square and then factor the difference of two squares.  Ratch
Posted on 2006-03-20 17:06:43 by Ratch
Ratch,
    "completing the square" was not taught in my class at school, but other classes in the same school did learn it. For some reason, it never occurs to me to use that method even though I came across it later, but there's always an alternative.

Paul.
Posted on 2006-03-20 17:21:47 by pdixon
Two simple ones :P.



Objetive, dont modify the look of the lines and get the trash "x" outside of the "|_|".
Rules, only move 2 lines.

|x|
-
|




Objetive, get 4 box.

Rules, only move 2 short lines.

For the figure

_ _ _
|_|_|_|_
    |_|_|

Posted on 2006-03-29 15:30:19 by rea
I don't really get what you mean by the first one, do you mean like

|x
-
| |


or do you mean that it can't change at all (ie it can't be upside down). For the second one..


_ _ _ _
|_|_|_  |
    |_|_|


You didn't say the boxes couldn't overlap or could not contain other boxes.
Posted on 2006-03-29 15:40:06 by Synfire
lol, sorry, I dont know exactly how to explain :P.

The first one is the correct answer ;).

For the second, the boxes should not overlap, the big square in this case isn't correct ;).
But without taking this last, is OK.
Posted on 2006-03-30 11:04:02 by rea
Do they all have to be square boxes? Or can two be rectangular boxes and two be square.. also for future reference, do they all have to be the same size?


_ _ _
|_| |_|_
  |_|_ _|
or
_ _ _ _
|_|_|_ _|
    |_ _|


And can you have any latent lines like


|_ _ _
|_|_|_|_
    |_|
    |


I'm sorry for all the questions, I tend to think very openly and honestly I see quite a few ways to get four boxes...
Posted on 2006-03-30 12:11:17 by Synfire
Ops, yes.

1) A hint is: all of the same size.
2) the conjuntion of each set of lines do a square, then isn't valid have lines that dosent close a square (I see the term, latent lines aren't valid ;)).
3) Rectangles aren't valid.


Ops, my bad, I have some of those questions when I do it, the good thing about, is that I have at "hand" the person that put it for clear the questions out ;).
Posted on 2006-03-30 15:22:57 by rea