Weighing coins?

Question

In response to jules09's request, I gave her a tough puzzle. Figured I'd share it with the rest of you. I will select a "best answer":

You have 12 coins which are identical in every way, except that one of them is counterfit in its weight, but you do not know which one it is or whether it is heavier or lighter than the others. (Assume, of course, that the difference is too small to tell just by holding it).

At your disposal you have a balance scale. Using this scale 3 times (no more), you must determine:

a) which coin is the fake
b) whether it is heavier or lighter than the others

Yes, it CAN be done - good luck!

Answer 1

Oh, I love these types of puzzles :)

Quick notation:
V = scale. so ooo v 000 means 3 balls against 3 balls
| = barrier. Basically everything to the right will not be weighed.
n = normal ball (Definitely not the counterfeit)
m = moved ball (See below)
s = stationary ball (See below)

1) Divide the balls into 3 groups of 4:
oooo V oooo | oooo

2) We have 2 interesting possibilities: either the scale says equals, or the left side is lighter. If the left side is heavier, we just turn the scale around and pretend like it was that way all along.

2a) Equals: We know that our suspect ball is in the left out group. We do not know if it is heavier or lighter. We do have a whole slew of normal balls though. So use the following configuration:
n o V o o | o

2b) If the scale says they weigh the same, then we know the ball left over is our bad boy. Just weight that one against one of the other 11, now certain normal balls, to determine if it is heavier or lighter. Finished.

2c) If the scale differs, then we again assume the left side was lighter. Heavier is, of course, exactly the same, just reversed. The leftover ball is definitely normal. We now do some fancy footwork. We take the ball from the left side out. We remove the normal ball, and we move one ball from the right side to the left side:
m V s | o

2d) If the scale is the same, then the ball left over is the bad one, and we know it is lighter from 2a. If the left side is now heavier, we know the "m" ball is the bad one, and it is heavier. If the left side is still lighter, then the "s" ball is bad, and it is lighter.

*small breather before we tackle the big one*

*and now, back at the ranch...*

3) What if step 1 said that the left side was lighter? (Again, if it is heavier, it's the same, just reversed, and this is already very long winded!) We know one of our 8 balls is bad, but we also know that the left 4 are lighter than the right 4. We also know that the 4 left over are normal. Sooo, we move some balls around. 2 balls on the left side and 1 ball on the right come out.

o o V o o o | o o o

2 balls on the right side are moved to the left, and one from the left is moved to the right.

m m o V m o | o o o

Mark the stationary balls as "s"

m m s V m s | o o o

Add a normal ball to the right side to give us ball balance, and weigh:

s m m V m s n | o o o

4) If the scale is balanced, then we know the bad ball is one of the remaining balls. 2 of the remaining balls came from the left side from step 1, and these are the 2 balls we now balance.

o V o | o

4a) If the scale is balanced, the remaining ball is bad, and we know it must be heavier. Finished.

4b) If the scale is not balanced, the lighter ball is bad. Finished.

5) If the scale from 3 says the left side is still lighter, then one of the "s" balls is bad. Either the left s is lighter or the right s is heavier. Pick one, weigh against a normal ball, and we know which one is bad. Finished.

6) If the scale from 3 now says that the left side is heavier, then we know one of the "m" balls is bad. Either one of the 2 "m" balls on the left is heavier, or the one on the right is lighter. Weigh the 2 "m" balls against each other. Balanced means that the one on the right is lighter, otherwise the heavier ball is the bad ball. Finished.

WHEW! DONE!