A doctor in Michigan has a sister who lives in California but the sister does not have any brother that lives in Michigan, how come??
:P

if he sees other two people both wearing red cap, he knows that he is the third one and the only one with black cap.

Typo: "Now, the are given..."

To be read: "Now, they are given..."

Ritthe, there is no square of negative number in Galaab's question. He just expanded -6 to 4 - 10 and 9 - 15. And, you can manipulate those numbers to be in (a-b)^2 = a^2 + 2.a.b + b^2.

Well, no, Ritthe.

The question says: They know that either they are given red or black. It could be all red or all black, but very true one can see two others' caps.

Also, one is able to tell he is wearing RED cap, not black.

oops...I meant (a-b)^2 = a^2 - 2.a.b + b^2.

Ok guys, see you tomorrow. I am off to bed now. Good night y'all! :)

since they were told to raise the hand when they see red, X and Y had red and Z had black, X realised he had red cap because Y was raising hand when Z had black cap, he was the one left with red cap which Y saw.

2 - 5/2 is a negative number it is like

4 = 4
(-2)^2 = 2^2
-2 = 2

Going back to Republican's milk and water problem... hopefully you can follow it:



X cups of milk------>xm
X cups of wate----->xw

Container 1: xm Container 2: xw


Step 1: xm-1m, xw+1m
(x-1)m xw+1m

Step 2: (x-1)m+ (xw+1m)/(x+1) xw+1m-(xw+1m)/(x+1)
{(x^2-1)m+xw+1m}/(x+1) {x(x+1)w+(x+1)m-xw-1m}/(x+1}

Final: (x^2m+xw)/(x+1) (x^2w+xm)/(x+1)

Since, coefficient of m in container 1= coefficient of w in container 2;
They are equal...

Correction

(2 - 5/2)^2 = (3 - 5/2)^2
=> 2 - 5/2 = -(3 - 5/2)
=> -1/2 = -1/2

Ignore last post:

going back to Republican's milk and water problem... hopefully you can follow it:

Nicely done, hetterika.

Another one:

There is a big, perfectly round table. You and I each have a heap of quarters (let's say unlimited amount to avoid any confusion). Each of us have to take turns placing one quarter on the table. The person who places the last coin (when there is no space left anywhere on the table), wins the game.

How would you ensure that you win the game?

What is the comment for this?

1 = 1

=> 1^1 = 1^0

=> 1 = 0

Let the other guy place the first quarter?

republican,
Is it you calculate the serface area of table and coin. divide the surface area of the table by the surface area of the coin. if the answer is odd number I start if first and if the answer is even number let the other person start first. Is this the solution?? I am confused coz there will be some spaces unused inbetween.

Pretty, you could be right, but the answer is a little more specific : )

ritthe, the size of the table and coin is unknown, the solution would be applicable to any table size.

preety could be wrong. what if the size of the table is equal to the size of the coin, other person will win in first move.

we can calculate the area by measuring the diameter of coin and table. What is the solution then if we don't know the area????

this is logical more than mathematical.. also let's not assume that the size of a table could be the same as the size of a quarter! we're making reasonable assumptions here..a table that is perfectly round and perfectly round coins..eg quarters.

hint: you've already taken the first step to the solution by explaining how preety could be wrong.

republic, I think it always takes odd number of smaller circles to cover bigger circle, like, 1, 3, 5. So we should start first to win. Is it the answer??? I am not sure though ......

i am wrong ...... republic I give up ...... give me the answer coz I am at work and don't want to be thinking abt it all the time ..........hehe