Skunked by Leo, again.

orthaic - think of mixing three colors this way: you have to select three from 6, one at a time (there are 6*5*4 ways to do this), but the order you chose them in is irrelevant. there are 3*2*1 possible orders that the three you chose could be arranged into. hence, we should have (6*5*4) / (3*2*1) possible mixes. you should read wikipedia's article on binomial coefficients for a more complete explanation.

@PatDragon: thanks for he correction to my colour mix.

and i didn't think of the empty colour, it makes the symmery up, but i feel the problem requires that it be excluded. I like your better answer.

@Leo: I think you're right.
Without knowing if the fake is heavier/ligher you can only get down to one fake among two. dang...

It's something I studied in combinatorics... it's all about distributing balls into bins. It's if you can distinguish the balls, and if you can distinguish the bins... and all the possibilities that arise from all these options.

Orathaic and Leo,

The solution of Leo is correct.

There is a easy trick to see if you are on the right track.
With one weighing, you can distinguish 3 possibilities (going left, right or balance).

With two weighing you can distinguish 9 possibilities. With 3, 27 etc.

If you do 3 to 3, as orathaic suggested, then if they balance, you have 6 coins which could be heavier or lighter.

So, that makes 12 possibilities. But with 2 weighing you can only distinguish between 9. So, that is the wrong start.

Lucas

I already answered the nail riddle. It's easier than that. Put the nails close together, tie a knot in the string at the center (you don't have to cut a really loose string to tie a knot in it, just use the slack), then put the string between the nails with the know above it. I guarantee if you remove one of the nails it will fall. I answered this a while back.

As far as the penguin/north pole/south pole. Because you have the explorer go south, then west, then north to arrive at his origin, he can't be at the south pole, but because it is dark on June whatever, he can't be at the north pole. This problem is unsolvable because it has a serious paradox in it. Either correct the date or the order of the directions.

Oh, and George can swim either way. Both polar bears and penguins can swim.

*know = knot

I can get down to two coins and knowing which is heavier, counterfeit or real. But I need a fourth weighing to get which coin is the fake.

Do 4 against 4.
If they balance, pull 4 off and do the other 4. If the other 4 are lighter, the fake is lighter, conversely, if they are heavier the fake is heavier.
If the first reading didn't balance, pull off the heavier four and put on the known good four (weren't weighed yet, but we know the fake was on the sacle in weighing #1). If they are also heavier, the fake is lighter, if they balance the fake is heavier.

No matter what, you now have it down to 4 coins and know if the fake is heavier or lighter. But you only have one measurement left.

You have a 50/50 shot at picking one of the coins if you pick two and weigh them against each other, or you could weigh two against two to determine which stack, but still need a 4th to determine the exact coin.

Draugnar, if they balance, and you do the 4 remaining against 4 coins that are certainly real, then they can't balance. Because this outcome is not possible, you waste it. It gives you two possibly answers, and your possibilities go from 8 to 4. And you should come to 3 or lower.

The trick is to shuffle something, in such way that they can also balance.

Lucas

Draugnar:

Re: the south pole. The Explorer is not exactly at the south pole (and nobody says he was). Instead he is at a more than x miles north of the south pole, such that when he travels south x miles he is at the edge of an imaginary circle that has the south pole at its centre and is x miles in circumfrence. By moving directly west or east on that line, he will return to where he stands, when he has travelled x miles..

Actually, re-reading my riddle, none of the distances are given, so you don't need to worry about "x." All you need to imagine is that he is somewhere close to the South pole, but not actually at the pole itself.

@lkruijsw - the second weighing was to determine, using a known factor, whether the counterfeit was heavier or lighter. But I agree, there has to be a shuffle in there somewhere. That is, if it is solvable...

@Plastic Hussar - Then he didn't travel x miles south. No matter what x is, know or unknown, he only travelled a part of x south. The rest he travelled north. The moment he crosses the south pole he is no longer traveling south. So this is *still* a bogus riddle. As it stands the riddle is a paradox.

Draugnar. Imagine he is ten miles north of the south pole when he starts. He travels seven miles south. Then he stops, before hitting the south pole. At that point, he starts travelling due west. If he keeps travelling due west, he will eventually come back to the point he started from (this is true of any point on the globe, it is just faster when you are near a pole). When he gets back to that point, he knows he is due south of his camp, so he just has to travel seven miles north and he is back in camp.
No paradox, not unsolvable, and not bogus.

Draug, it is correct. Saying that, I'm now interested to know that this means geometrically. I think I'll ask my geometry lecturer. He has travelled in a closed loop on a sphere, which included 3 straight elements and so you would expect the area enclosed by his travels to be a spherical triangle. Yet this in itself seem rather strange as the area of the triangle is equal to the area of the section (I'm not quite sure what you'd call this) within the circle walked, including the south pole. Anyone able to explain that to me? :S

Still, the problem & solution are valid Draug - its not a paradox

figles - there's a bigger ambiguity you should worry about, in terms of computing areas of sections of a sphere: take any closed path on the sphere which does not intersect itself (no figure-eights) and try to define an "inside" of the path - this is nontrivial! On a plane, any closed path divides the plane into a bounded part, and an infinite part. On the sphere, both parts are bounded. You might try to call the smaller one the "inside", or use an orientation on your path to define an "inside". In any case, it's probably nicest to think of south- and north-bound paths that are actually just very close to one-another, than exactly overlapping. Then your "inside" might just be a circular section around the pole, plus a spike between the N/S paths. Still, it might be bothersome that this triangle is not convex...

is it not usual (at least in physics) that pick the handedness of the system?

though this is really used to determine which side of a surface is up (rather than which is inside) a left-hand rule based on the curl of your fingers and the up in which your thumb points...

@PH - I see that possibility now. And because there are no distances given, I accept it as not being a prardox after all.

has anyone successfully done the one with 12 coins of which one is heavier or lighter... with only 3 weighings?

@dave
Yes, the solution was given. And counting the possibilities is a big hint in solving it.

Lucas

Draugnar - I know, I meant what would be thought of as the smaller section if this example was applied to the earth

But yes obviously calculating area's in spherical geometry is much harder than in planar geometry, although arguably much more interesting

The coins solution wasn't ever actually given. Orathaic's solution had a seriosu flaw in that there was a set of three unweighed after the second weighing that he claims determined heavier or lighter. If the two sets of three match and the third set of three also matched, you have it narrowed to the final set of three but still don't know if it is heavier or lighter, thereby requiring two reading to truly determine which coin it is for certainty.

I did finally find a solution (googled, not solved).

1 2 3 10 against 4 5 6 11
1 2 3 11 against 7 8 9 10
1 4 7 10 against 2 5 8 12

We give the table, with only those outcomes for these weightings, that start with `l r', `r l' and `b b' (as all other cases already have been dealt with):
outcome fake coin:
-----------------------
l r l 10 too heavy
l r b 11 too light
l r r (not possible)
b b l 12 too light
b b b no fake coin
b b r 12 too heavy
r l l (not possible)
r l b 11 too heavy
r l r 10 too light

Not trying to sound stupid (although I know it will) but how do you know its not coin 1 to 9

l l l coin 1 is heavy, conversly r r r coin 1 is light
l l r coin 2 is heavy...
l l b coin 3 is heavy...
r b l coin 4 is heavy...
r b r coin 5 is heavy...
r b b coin 6 is heavy...
b r l coin 7 is heavy...
b r r coin 8 is heavy...
b r b coin 9 is heavy...

Interesting Wiki article on the coin problem that explains the mathematics behind the solution.

http://en.wikipedia.org/wiki/Counterfeit_coin_problem

crazy... I think I just about get it...

"I've seen the 2 nails one before. Essentially if you have two nails A and B you wrap the string in this order. Clockwise around A, Clockwise around B, AntiClockwise around A, AntiClockwise around B. Removing either nail leaves you with a clockwise and an anticlockwise turn around the remaining nail, which cancel out."

Unless I am reading this wrong, doesnt this require you to untie one side of the string from the picture frame and then reattach it? I thought that the string was attached and always attached.

Oh, and for the area of triangles on a sphere thing, this is how you do it:
the area of a triangle on a sphere is (the sum of the angles (in radians)- pi)*r^2.

If this is hard to visualize, consider cases you can visualize:
for a very very thin triangle, the angles will be about pi, 0, and 0 for an area of close to zero.
for an eighth section of a sphere, the angles would all be pi/2 (90 degrees), making the are pi*r^2/2, which is an eighth of the surface area of a sphere (4pi r^2).
etc.