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Topic: CRIMINAL CUPBEARERS (answer) (Read 1450 times) |
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gavnook
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CRIMINAL CUPBEARERS (answer)
« on: Nov 7th, 2002, 7:51pm » |
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Here's the answer:
You need use 20 prisoners in order to save all 999 good bottles
Each one will drink a small portion of each of about half the bottles.
Prisoner #1 gets the a cup filled with a drip from each of the 1st 500 and #2 gets the 2nd 500
#3 gets the 1st and 3rd 250 while #4 gets the 2nd and 4th 250
#5 and 6 get every 125
#7 and 8 get every 64
#9 and 10 get every 32
#11 and 12 - 16
#13 and 14 - 8
#15 and 16 - 4
#17 and 18 - 2
and finally #19 gets the odd numbered bottles and #20 gets the even numbered bottles
After they drink, wait 5 weeks and see who died. If prisoners #1, #4, #5, #8, #10, #11, #13, #16, #17, and #20 die you throw away bottle 0+250+0+64+32+0+0+4+0+1=351 and drink the rest.
If you change the total bottles to 1024 (or just start prisoner #1 with 512 instead of 500) you can write a more elegant representation of the solution.
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towr
wu::riddles Moderator
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Re: CRIMINAL CUPBEARERS (answer)
« Reply #1 on: Nov 8th, 2002, 12:07am » |
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you only need 10, 2^10 = 1024 (give every prisoner a drink whose bit is 1 in the binary representation of the bottle number)
There was allready a thread about this riddle though..
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Wikipedia, Google, Mathworld, Integer sequence DB
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