Since there's all these riddles going on, I do have a few (but I'll tell em one at a time).

Here's one:

Suppose you have a total of nine metal balls. 8 of these balls are exactly the same weight and 1 is slightly heavier. You have a balancing scale which you could put any number of balls on either side to determine which balls are heavier. You also can not tell which ball is heavier with just your hands (so you must use the scale). You can ONLY use the scale 2 times to figure out which is the oddball. How do you do it?

 

put 3 balls on each side of the scale, either 1 side will drop, or it will stay even. if one side drops take those 3 balls, and if its even take the 3 you didnt use. then put 1 ball of the 3 you chose on each side, either 1 side will drop or it will be even. if it is even then the last ball is the heavyist, if one side drops, then that is the heaviest.

 

at first i misread and thought it said one ball just weighed different, but it made it a lot easier knowing that the oddball was only heavier, and not possibly lighter.

 

why does that make it easier?

 

because if the oddball could be heavier or lighter, and took 3 balls, put one on each side of the scale and left one off, then even if one side went down, you couldnt determine the oddball because the one you didnt put on could be either the same as the lighter one or the same as the heavier one. but with the oddball being only heavier, you know that every time the scale goes down, then the oddball MUST be on that side of the scale.

i hope my ramblings make sense to you, its kinda hard for me to put it into words. i see it in my mind, but cant really explain it.

 

You can do it with twelve balls one of which is either lighter or heavier with only three weighings. It's, to say the least, more complicated.

 

[QUOTE]Originally posted by brerspidur
put 3 balls on each side of the scale, either 1 side will drop, or it will stay even.

 

I bet Magician could figure out any riddle put on this board.... he's just TOO INTELLIGENT!!!!

 

[QUOTE]Originally posted by magician
You can do it with twelve balls one of which is either lighter or heavier with only three weighings.

 

Let me see if I got it right:
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