Re: Maths problems (interesting ones)
 

sory i been a bit busy lately.
some new problems coming up later

Re: Maths problems (interesting ones)
 

Ok, I know this is a little late, but here goes...

The whole 1/2 1/3 debate was bugging me, so I e-mailed her about this with links to a couple of different arguments (in addition to the thread it self). Here's a copy:

She is a profesor at a local college.

Re: Maths problems (interesting ones)
 

I just wrote my own computer program.


IF you choose a random child to open the door then it's an outcome of 50%

IF you choose a girl to answer the door (I still say the question provides that as a given), then the outcome is 33%


But I don't know... I honestly don't agree with your teacher's logic... yes I agree that the gender of one doesn't influence the gender of the second... but she seems to ignore the fact that there are twice as many families with a boy and a girl than there are with two girls. The gender of the second kid isn't generated when you meet them (i.e. flip the coin); it's already been made... and fact is twice as many "tails" get flipped than "heads."


Also, why does opening the door and getting a girl eliminate 500 doors? Shouldn't it only eliminate 250 doors (b/b) as even in the b/g & g/b families, either one could have answered the door? And wouldn't that leave 250 doors with GG and 500 with some combination of GB?

Re: Maths problems (interesting ones)
 

and that, boys and girls, is the sound of my head exploding.

Re: Maths problems (interesting ones)
 

Of course there is. GG, GB, BG, BB. But that assumes you can identify A and B. But in this instance, you can only identify them by which one opened the door, so you have to account for both instances. A or B may open the door!

A opens:
+GG
+GB
-BG
-BB
B opens:
+GG
-GB
+BG
-BB
It's still 1/2, 50-50.

Also, That list is generated from the combinations of 2 mutually exclusive chances, a table:The bottom line is: THE RESULT OF ONE DOES NOT AFFECT THE RESULT OF THE OTHER.

Re: Maths problems (interesting ones)
 

I don't understand what your + and - signs are denoting.

Re: Maths problems (interesting ones)
 

the + seems to represent a possible combination and the - shows that it does not work. That post aims to prove that there are 2 ways that the Girl-Girl combination works (because either one could open the door and then the other child would be unknown), which would then make the odds 50-50.

Re: Maths problems (interesting ones)
 

Right, i remembered this thread from a while ago, so i decided to post some new problems up
Number 1

given that the rectangle is 2x4, find the radius

Number 2
Divide $261 (in whole $ increments) into a number of bags so that I can ask for any amount between $1 and $261, and you can give me the proper amount by giving me a certain number of these bags without opening them. What is the minimum number of bags you will require?

When i get around to it i'll add a couple more

Re: Maths problems (interesting ones)
 

See attached drawing. (The original diagram looked something like the top left picture.)

Re: Maths problems (interesting ones)
 

AHHHHH my brain hurts ... i need the solution!

Re: Maths problems (interesting ones)
 

You don't need to use the scale at all.

First, if you need a small graduated cylinder. 11 coins should be exactly the same, so they all have the same density. The fake, although not necessarily weighing the same, will have a different density than the others. Simply put in X mL of water in the cylinder and calculate the displacement of the water for each coin dropped, and the fake coin will be different.

Re: Maths problems (interesting ones)
 

Well yea i think that is definitly an option if i had to do this in reality... but we're not in "reality", we're in Steve howlands reality... and he descided to give us merley a stupid scale with a bunch of stupid coins... AHHH i need to know the answer dangednabbit! lol

Re: Maths problems (interesting ones)
 

ok, here got it from a book some place and i figured out the solution (actually there are two but only two)

--so here it goes: a 9-digit number (123456789; the digits only represent the position for further labeling),
--now each position has to be divideable (english, gramar, I sorry.) by the #of the position (like 1/1=n; 12/2=n; 123/3=n; 1234/4=n and so on; n representing a natural number)
--and finaly, the number consist of the digits 1 through 9, but they can be used only once

So get the solution, and I'm sorry for my terrible english

Re: Maths problems (interesting ones)
 

Right,

For the coins one you can't do it unless you know whether its heavier or lighter.
if you know whether it is heavier or lighter:
put 6 on either side of the scales and weigh
One side will be lighter than the other so you take the side witth the heaveier or lighter one and weigh 3 on 3. One side will be heavier/lighter. Then weigh 1 on 1 of and leave one off. If it is even then the 3rd is fake. If it is not even then the heavier/lighter side is fake

Coin Problem
 

I got the solutions for the coins! However I need 4 steps. Maybe I can get it with 3 (I need some time on that)

I'll post the solution with 4 later today (School now).
I know it doesn't count but at least it's something and maybe then someone else can do it then with three...