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Random Riddle thread

The numbers one:

Had to resort to google as well. I think if you had used the parenthetical notation I might have gotten it, but I didn't think about the +/- 1 possibility. Learned something from this one, thanks smile.

Darrell
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I don't like this thread. It makes me feel stupid. alright

neenerneener
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novice, your post was rather spoilerish for those who want a fair chance a figuring out the riddle shhh However, since I pointed that out in the first place there are no hard feeling here smile

I did solve BRick's number riddle jive
... 104, 108, 138, 150, 180

After recognizing that each number in the series was one greater than a prime number, I dug up an old "primes" module that I created when I was playing around with python. I had a function that would give me the first n prime numbers, so I had it spit out 50. As soon as I wrote them down the pattern became glaringly obvious! Each of the numbers in the series is surrounded by 2 prime numbers; one less than (like I noticed at first) and one greater than the numbers in the series.
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Found one.

Quote:An evil king has 1000 bottles of wine. A neighboring queen plots to kill the bad king, and sends a servant to poison the wine. The king's guards catch the servant after he has only poisoned one bottle. The guards don't know which bottle was poisoned, but they do know that the poison is so potent that even if it was diluted 1,000,000 times, it would still be fatal. Furthermore, the effects of the poison take one month to surface. The king decides he will get some of his prisoners in his vast dungeons to drink the wine. Rather than using 1000 prisoners each assigned to a particular bottle, this king knows that he needs to murder no more than 10 prisoners to figure out what bottle is poisoned, and will still be able to drink the rest of the wine in 5 weeks time. How does he pull this off?

I think I have an answer, but I don't want to work out if it's right. So slaves, I'm farming it out to you lot to see if the general premise is right.

Number each bottle in binary and then label each prisoner 1, 2, 4, etc up to 512 and then give each prisoner some of the wine. So bottle 316 would be given to prisoners labelled 256, 32, 16, 8, 4. After one month, when those prisoners die and the others live you know what bottle is poisoned, because on those prisoners have drank from that specific bottle.
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@Krill

Each poor sod is drinking from ~500 bottles of wine. 2^10 = 1024; use binary:
Bottle #1 = 0000000001 (Pris #10 only)
#2 = 0000000010 (Pris #9 only)
#3 = 0000000011 (Pris #9 and 10)
#127 = 0001111111 (Pris #4-10)
#128 = 0010000000 (Pris #3)
etc. etc.

Based on which ones die, you'll know with certainty which bottle was the bad one.

EDIT after reading Krill's spoiler:
I'd say that's the same solution.
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Ironically, the "give 1000 prisoners a tiny sip" methods A) kills fewer prisoners B) wastes less wine and C) takes less time...
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(February 23rd, 2013, 20:19)AutomatedTeller Wrote:
Ironically, the "give 1000 prisoners a tiny sip" methods A) kills fewer prisoners B) wastes less wine and C) takes less time...

True, but even bad kings rarely kept a thousand prisoners. That's a LOT of useless mouths to feed.
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(February 23rd, 2013, 20:19)AutomatedTeller Wrote:
Ironically, the "give 1000 prisoners a tiny sip" methods A) kills fewer prisoners B) wastes less wine and C) takes less time...

I know, this confuses me too. The question does seem to place a limit on the number of prisoners, it just says "not 1000". I don't know you're supposed to decide the "best" method of the infinite remaining ones.
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@Krill - without reading any spoilers -

Idea here is of course that 2 to the power of 10 is 1024 - which is the number of bottles 10 prisoners can check.
It's easier to imagine 1024 bottles (just take 24 empty bottles, they will not be used except to make the division easier). Divide them in half - first guy tests half, say first half. 1-512. Next guy tests the first half (of each batch), which is 1-256 and 513-768. Third 1-128 (and 3 other I can't be bothered to work out), fourth 1-64, fifth 1-32 sixth 1-16 seventh 1-8 eigth 1-4 ninth 1-2 and tenth 1, 3, 5, ... Each following tests the first half of each batch, ending with last guy who tests every other bottle. After 4 weeks you can work out which bottle is poisoned by looking at who lived. So solution is easy, but I spent quite more time working out how to get out the result, was thinking binary number.

Closest was this, still feels clumpy and unelegant:
Label bottles 1-1000. Label prisoners 1-10. Prisoners test the second half of their interval, Nr 1 tests bottles 513 - end. Nr 2 tests 257 - 512 and 769 - end. Ninth tests 3-4, 7-8 etc. Tenth tests 2, 4, 6, ...
After four weeks, those who die become '1' those who live '0' to form a binary number. Add 1, and this is the number of the poisonous bottle.
So Nr 1, 6, 8 and 10 dies - 1000010101 - is 512+16+4+1=533. 533 + 1 =534. Killer bottle is 534.

...I think, worked it out with 16 bottles and 4 prisoners, just assuming it works out anyway.

EDIT: Yeah suggested solution more elegant and clear.
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It seems to me, if you have at least m*n prisoners and allow m deaths, you can test n^m bottles. Use base n numbering of bottles, and assign n prisoners to each digit. The digits assigned to the m prisoners who die, identify the poisoned bottle.
I have to run.
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