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#1 MSGMOTG

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Posted 18 July 2007 - 10:04 PM

This one took me days to solve. Here's the setup: an ungraduated balance scale and 12 apparently identical balls. Only, one of the 12 balls is of a slightly different density than the other 11. Using the scale only 3 times, how can you discern without any uncertainty, which is the different ball? Sorry, but I actually have to resolve this one myself again. Work on it, and I'll post the solution soon.
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#2 unreality

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Posted 18 July 2007 - 10:56 PM

i saw this in a book.... it was like a year ago though, i dont remember any more than that, this solution is totally mine. In my solution i also determine if the ball is hevier or lighter:


so you have 12 balls: 1,2,3,4,5,6,7,8,9,10,11,12

first weighing: 1234 vs 5678



root 1: if weighing 1 balances

if weighing 1 balances, it means the ball is in 9-12, so do 1,2,3 vs 9,10,11


if 9,10,11 side is lighter, the ball is lighter, if it is heavier, the ball is heavier. If they balance it means 12 is the pesky ball, and in the third weighing weigh it against a normal ball to see if 12 is heavier or lighter

but if they didnt balance, and its 9,10 or 11 (and you know if the ball is heavier or lighter):
so now weigh 8,9 vs 7,10 in your last weighing. Since you know if the ball is heavier or lighter, whichever one goes in the direction you know the special ball will go is your ball (either 9 or 11, since 8 and 7 are normal)

If 8,9 vs 7,10 balances, you know the ball is 11. (and u already know if its lighter or heavier)


root 2: if weighing 1 DOESNT balance
i need to think about this..
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#3 rookie1ja

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Posted 18 July 2007 - 11:04 PM

I have already posted this puzzle a long time ago (and other similar water and weighing puzzles as well) - it's one of my absolute favorites ... check Weighing IV. and add your posts there

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