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b-e-a-n-s
Icarian Member

 Posted: Sat Apr 06, 2002 4:55 am    Post subject: 1 What is the fewest number of weighings needed to find the odd ball out of 22,964 balls if all you know is that ONE ball is heavier or lighter than the other 22,963?
robichelli
MI:6 Agent

 Posted: Thu Apr 11, 2002 6:36 pm    Post subject: 2 Well, the largest number would be 22,963. Now all we have to do is work backwards from there
ralphmerridew
Daedalian Member

 Posted: Thu Apr 11, 2002 7:42 pm    Post subject: 3 I think 10 will do.
Heavy Metal

 Posted: Thu Apr 11, 2002 7:47 pm    Post subject: 4 [happy happy joy joy-mode] Uhm, 1 weighing? Hypothetically speaking, you could pick two balls including the right ball, compare them and then you know which ball is heaver than the others . But that a change I wouldn't take. [/happy happy joy joy-mode] If you divide all balls in two groups and compare them, one groups lighter. Take the other group and divide it in two.....and so on until there are 2 left. Or am I overseeing something here? ------------------ There are three kinds of people, those who can count and those who cannot.
Orbiting
very ign-o-rable

 Posted: Thu Apr 11, 2002 7:53 pm    Post subject: 5 ummm... yeah, you might pick the right two balls to start out with, but then how would you know which one was the different one? Seriously though, the half-then-half-again would be my method of choice, except that at some point fairly quickly you get an odd number of balls and have to use a different procedure. I didn't work through the rest of it, so I don't have a good answer. -o-
Zealot
Daedalian Member

 Posted: Thu Apr 11, 2002 8:00 pm    Post subject: 6 Do thirds instead of halves... and you end up with the answer ralphmerridew got.
Quailman
His Postmajesty

 Posted: Thu Apr 11, 2002 8:06 pm    Post subject: 7 araya (anyone remember him?) had put up the formula for figuring this in the discussion of The Counterfeit Coin a couple years ago. I could figure out the method for the counterfeit coin, but I didn't follow the math, nor do I remember it.
ralphmerridew
Daedalian Member

 Posted: Thu Apr 11, 2002 8:15 pm    Post subject: 8 Cadmium: You don't know whether the ball is lighter or heavier.
CrystyB
Misunderstood Guy

CrystyB
Misunderstood Guy

 Posted: Fri Apr 12, 2002 1:41 am    Post subject: 10 incidentally, with 10 weighings, the coins could be as much as 29523. Ow and bah @ mith for posting that AFTER i searched my hard-drive... :P [This message has been edited by CrystyB (edited 04-11-2002 10:10 PM).]
mith
Pitbull of Truth

 Posted: Fri Apr 12, 2002 2:08 am    Post subject: 11 http://www.greylabyrinth.com/Puzzles/answer019.htm
Sumudu2
Daedalian Member

 Posted: Fri Apr 12, 2002 5:03 am    Post subject: 12 what do you do when the number of coins is not divisible by 3?
CrystyB
Misunderstood Guy

 Posted: Fri Apr 12, 2002 12:31 pm    Post subject: 13 Weigh the required number regardless. If the balance doesn't tip, you have 2N known goodies, from which you can add to the leftover pile to make up to N.
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