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# Bottle

Mon 15 Feb, 2010 12:06 pm
There are 15 bottles arranged in chronological order. There is poisous in one of the bottle. We have to form a group of minimum people so that if a person consumes from any of the bottle, he dies on the 7th day. In which bottle is the poisous?
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fresco

0
Tue 16 Feb, 2010 01:27 am
@Megha ,
oolongteasup

2
Mon 29 Dec, 2014 10:06 pm
@fresco,
Megha said:

Quote:
There are 15 bottles arranged in chronological order. There is poisous in one of the bottle. We have to form a group of minimum people so that if a person consumes from any of the bottle, he dies on the 7th day. In which bottle is the poisous?

fresco said:

Quote:

I say:

My friendly local apothecary has offered me at least 15 assorted bottles, phials and flacons containing his finest. One of the vials contains the vilest deadly poison. I have amazed a group of 4 rats. How do I find the bottle containing the poison with their inadvertent assistance?

Clue: The pellet with the poison's in the flagon with the dragon; the vessel with the pestle has the brew that is true.

chai2

1
Mon 29 Dec, 2014 10:28 pm
hehehe....

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oolongteasup

1
Tue 30 Dec, 2014 07:28 pm
@oolongteasup,
Is anyone here able to answer the question?

15 bottles, 4 rats, which bottle contains the poison?
fresco

1
Wed 31 Dec, 2014 12:54 am
@oolongteasup,
It looks like a variation of the minimum number of scale pan weighings problem for 12 balls. I suppose you set up samples of mixture from 5 bottles at a time and on finding the lethal sample do 2 against 2 etc. (The twelve ball problem is a lot harder with scale pan switching to find the odd one out)
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markr

1
Fri 16 Jan, 2015 03:42 pm
@oolongteasup,
Number the rats 1, 2, 4, 8. Number the bottles 1-15. On each bottle, write the bottle number as the sum of unique powers of two (e.g. 11 = 1+2+8).

Each rat will get a mixture which consists of liquid from all of the bottles with the rat's number in the sum expression (e.g. rat 4 will get liquid from bottles 4, 5, 6, 7, 12, 13, 14, 15).

After the rat(s) has/have died, add their numbers to get the bottle number that contains the poison.
fresco

1
Sat 17 Jan, 2015 02:39 am
@markr,
Give first rat mixture from bottles 1 - 5
If rat dies, give second rat mixture 1+2. (If it lives give 3+4... etc)
If rat dies give third rat bottle 1 hence answer 1 or 2

If bottles 1-5 keep rat alive, give first rat mixture 6-10 ( or 11-15)
and repeat logic of "halving" as above.

This method has the "advantage" of economy of 3 rat maximum usage .

(This method is an adaptation of the "12 ball minimum balance pan weighings" problem in which the balls are identical except for the weight of one of them)

oolongteasup

1
Sat 17 Jan, 2015 03:05 am
@fresco,
Rats. For one wonderful moment I wondered whether or not the rats had a solution.

I prefer the elegance of painting the rats with a non-toxic solution al a markr pencilled fresco. The same number of rats are imperilled in either scenario if you are patient.

My money is on the mystery bottle 16.

Thank you chai. Tea for 2?

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markr

1
Sat 17 Jan, 2015 04:38 am
@fresco,
I believe the original problem I'm familiar with had a time constraint, and it took a while for the poison to kill. In that case a sequential procedure would not be completed in time. It can be solved with a single rat if you feed it from one bottle at a time.
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