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Mytical
Responsible
Undefeatable Hero
Chaos seeking Harmony
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posted November 09, 2009 11:33 AM |
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Simple, take the bottles back to the pharmacy, explain the situation. Exchange them for more of both A and B, and take them 
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ihor
Supreme Hero
Accidental Hero
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posted November 09, 2009 11:46 AM |
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OK The pills are very expensive, no... the secret to prepare them knows only druid who lives in the Antarctic ... and that druid could only communicate with xerox.
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Mytical
Responsible
Undefeatable Hero
Chaos seeking Harmony
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posted November 09, 2009 12:05 PM |
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Although it doesn't solve the issue in the long run, here is what you do. Set aside those three pills, grab a new one from both A and B. You'd have till the 30th day to figure out what to do then.
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ihor
Supreme Hero
Accidental Hero
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posted November 09, 2009 12:24 PM |
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Aha and you must try to die heroically(!) before 30th day 
Any other ideas? TheDeath?
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Adrius
Honorable
Undefeatable Hero
Stand and fight!
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posted November 09, 2009 07:22 PM |
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Edited by Adrius at 20:07, 09 Nov 2009.
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Argh, umm... *thinks til veins start bursting*
Maybe... cutting them in half?
Like, there are two A pills and 1 B pill, we divide them all by half which makes:
Pill number 1:
½ A
½ A
Pill number 2:
½ A
½ A
Pill number 3:
½ B
½ B
I then eat ½ from each pill and then pull out another B pill from the box and cut it in half, I then eat one half. Thus only half a B pill is lost.*
ERGO:
½ A + ½ A + ½ B + ½ B = Survival
My brain hurts.
EDIT: *No wait, in fact no pills are lost, since I can add the other ½ B to the remaining halved bits and make another day's supply of medication.
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TheDeath
Responsible
Undefeatable Hero
with serious business
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posted November 09, 2009 07:52 PM |
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Are you allowed to cut them?
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Adrius
Honorable
Undefeatable Hero
Stand and fight!
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posted November 09, 2009 07:54 PM |
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Hope so
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TheDeath
Responsible
Undefeatable Hero
with serious business
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posted November 09, 2009 08:05 PM |
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Quote:
Hope so
Me too, I don't see another solution.
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VokialBG
Honorable
Legendary Hero
First in line
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posted November 09, 2009 10:01 PM |
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@Adrius: Mmmm, ok so you cut them and you have 4 halfs of A and 2 halfs of B. The problem is that even if you cut them you don't know if the half you take is A or B half, since you do not know what exacly you cutted - A or B pill. So, you can take 1 half A by cutting 1 pill from the box and saving the other half. You can do the same with one B pill from the other box. But after that when you need to take one more half from both you still can't know if you cutted A or B first so you still can take 1 half of the 3 pill, it doesn't matter ifts A or B, but this is only of the first one you can't say if the second half is not A again.
So basicly if you cut them you get 6 x 1/2 Y pills (mixed halfs) you can split them in 2 -since you know the two parts of every cutted pill and you acctualy have 3x 1/2 Y.
So, uh you have 50% to die this way when taking the last half. Or I miss something.
On the other way even if you take the 3 Y halfs you can die... So uh... I have no idea
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veco
Legendary Hero
who am I?
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posted November 09, 2009 10:08 PM |
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Edited by veco at 22:09, 09 Nov 2009.
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No, you take a half from every pill - that way you are sure to get a full A and half B so you take another half B from the box.
On the nest day you take al the other half which are 2 A halfs and 2 B halfs.
I can't make it any more clear
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none of my business.
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TheDeath
Responsible
Undefeatable Hero
with serious business
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posted November 09, 2009 10:10 PM |
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Adrius is right, at least if you are allowed to cut them.
Let's name the pills 1, 2 and 3. You don't know which one is A and which one is B.
Cut Pill 1, you have Pill 1a and Pill 1b.
Cut Pill 2, you have Pill 2a and Pill 2b.
Cut Pill 3, you have Pill 3a and Pill 3b.
Now you know that Pill 1a, Pill 2a and Pill 3a in TOTAL, if you take them all, will get you a full Pill A and half of a Pill B.
So you take a Pill B from the box and cut it in half, then take them all.
Next day you take Pill 1b, Pill 2b and Pill 3b, and the other half from the Pill B box.
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Adrius
Honorable
Undefeatable Hero
Stand and fight!
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posted November 09, 2009 11:07 PM |
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Yeah, can't explain it any better than that.
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Kareeah_Indaga
Responsible
Undefeatable Hero
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posted November 10, 2009 01:27 AM |
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I recall a simple one from the OotS forums.
You have a 5 gallon pail and a 3 gallon pail, and an unlimited water source.
Assuming you have no way of measuring the water other than knowing the pail sizes (i.e. no measuring marks denoting the 1 gallon mark, 2 gallon mark, etc...) get EXACTLY four gallons total into one or both containers.
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TheDeath
Responsible
Undefeatable Hero
with serious business
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posted November 10, 2009 01:37 AM |
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Fill the 5 gallon pail, then use that to fill the 3 gallon pail, leaving the 5 gallon pail with 2 gallons of water. Empty the 3 gallon pail. Empty the 5 gallon pail into the 3 gallon pail. Fill the 5 gallon pail again. Now empty the 5 gallon pail into the 3 gallon pail. Because the 3 gallon pail was filled with 2 gallons, you can only fill 1 more gallon. 5 - 1 = 4, so there are 4 gallons in the 5 gallon pail now.
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ihor
Supreme Hero
Accidental Hero
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posted November 10, 2009 08:41 AM |
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Yes, Adrius is right.
That was correct solution. Good job 
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Mytical
Responsible
Undefeatable Hero
Chaos seeking Harmony
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posted November 10, 2009 08:49 AM |
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That water riddle was in one of the Die Hard movies  .
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Message received.
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ihor
Supreme Hero
Accidental Hero
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posted November 10, 2009 08:54 AM |
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Edited by ihor at 08:54, 10 Nov 2009.
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Aha, I also noticed that - the third.
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mvassilev
Responsible
Undefeatable Hero
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posted November 10, 2009 05:18 PM |
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Okay, new riddle. For this one, you can ask me "yes or no" questions.
The snow saved her life.
Figure out what happened.
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Eccentric Opinion
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Geny
Responsible
Undefeatable Hero
What if Elvin was female?
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posted November 10, 2009 06:14 PM |
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Edited by Geny at 18:15, 10 Nov 2009.
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Is that a trick question that utilizes the HC censorship system?
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mvassilev
Responsible
Undefeatable Hero
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posted November 10, 2009 06:35 PM |
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This thread is pages long: 
