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Title: CRIMINAL CUPBEARERS (answer) Post by gavnook on Nov 7th, 2002, 7:51pm 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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Title: Re: CRIMINAL CUPBEARERS (answer) Post by towr on Nov 8th, 2002, 12:07am 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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