In fact, it is clear from the riddle that George cannot be a polar bear.

He can't swim because of his injured foot. Sometimes the simplest answers are the correct ones.

"I have a painting that I would like to hang on a wall. The painting has a loose string attached to the upper corners of the back side (though you needn't assume that the string is already attached, we will assume that it is). I also have two nails. I would like to hang the painting in such a way, that if either nail is removed, the painting will necessarily fall. Is this possible? How about for 3 nails?"

This one is a little more complicated... for two i guess you would put one nail on the painting and one on the wall, then you could wraparound the one on the wall and connect the loop around the nail on the painting, if that makes sense.

Still working on 3.

ZaZa,

Not correct, but interesting response.

surely the west thing is not by accident... I guess you are saying he cant be a polar bear since he is shorter than you? Penguin maybe?

He cannot swim because he only has two limbs. Two feet but no arms at all?

@Plastic - my response is "insufficient information to determine". There is no observation indicating George has any motion capabilities other than walking. .

About the cylindrical can:
I think, but haven't actually used any maths to check it, that if you balance the can on the bottom edge so that the opposing side is directly above it and then let go, it will fall back to the bottom if it's less than half full, fall to it's side if it's more than half full and stay where it is if it is exactly half full.

Xapi - your solution for the painting problem is not correct - I even pointed out that your suggestion doesn't work in my problem statement. The painting might fall a few inches, but certainly won't fall completely off the wall. Come on, you can do better! As for your tub solution for the can problem, I feel that it does not account for buoyancy of the material the can is made of.

Draugnar - I really like your solution to the painting riddle, but i would prefer a "generic" solution - i.e. one which is stable under infinitesimal perturbations of the positions involved. If I perturb your system so that the nails aren't tightly together, the painting wouldn't be suspended in the first place. You shouldn't need such a constraint, but your solution is still pretty cool.

nice solution, uclabb - that's a putnam exam question!

Hussar, your question reminds me of the following:

I decided to go on a bear hunt. I put on my boots, hat, and took my gun, and left my cabin. I hiked a mile south. No bears. I hiked a mile east, and shot a bear. It was too heavy to carry, so I hiked the mile north back home. What color was the bear?

Hamish - you're on the money every time - good call!

About the picture questions, it's a little unclear to me if the string has or has not been attached to the corners of the pictureframe. If it has, putting a nail in the pictureframe has no effect. If it hasn't, it becomes pretty easy with 2 nails. For 3 I have an idea, but it's not pretty.

If the rope is attached, it's a whole new problem, and I have no ideas so far.

uclabb has the answer to the hunter riddle, and it is actually a rewrite of the polar bear riddle that PatDragon cites. The reason that the answer is not "polar bear" comes from the date on the note, and the statement that it is pitch black. It cannot be pitch black at the North Pole in July, so the setting has to be the South Pole (because of the directional clues), making George a penguin.

Patdragon: Green. Bear Community representative up in the Arctic doing research into global warming for the U.N. . . . until someone shot him!

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.

Here's a puzzle of my own.
With 6 different paints, how many ways are there of painting a cube so that each of its sides are a different colour? Remember, the cube can be rotated.

@Leo: you said: "@orathaic
Your solution requires more than three weighings. Check out the different options and see that you usually need four. "

no, my solution only required three weighings. The first (three v three) gives you six real coins (either the ones you are weighing or the ones you aren't)

The second gives you both whether the fake is heavier or lighter and which set of three coins it is in.

The Third figures out which of the last three it is.

OMGNSO, you've done the 2-nail problem. can you generalize to more nails, or is such a feat impossible?

6 colour paints, 15 two colour mixes, 30 three colour mixes, 15 four colour mixes, 6 five colour mixes, 1 six colour mix. ~ 73 colours, on six sides of a cube

(6*73) 438 possible choices of a colour plus side, assuming any one side is indistuinguishable from any other, that's still 73 coluors you an paint the first side. The second side has 437 choices * 5 sides you can be chosen, (but four of those can be rotated and match other combinations - so pick the opposite side for simplicity) 436 + 435 + 434 + 433 choices of coulrs for the last four sides, multiplied by the number of ways you can paint those sides uniquely.

= 277,219,690 * the number of unique ways to paint the four sides...
</great answer done badly>

@ two nail problem. that depends on the friction of the rope and the weight of the painting.

try it if you don't believe me.

>i mean, try a piece of rough string with a weightless painting (let's say no painting for the lower limit of possible painting weights) with one nail and wrapped clockwise and anticlockwise.

see if the string CAN stay. (because the problem requested a guarentee that the painting would fall)

6 colors on a cube (without mixing colors):

there are certainly 6! arrangements possible on the cube, but this does not take into account the symmetries of the cube. allowing only rigid rotations (no reflections to turn the cube "inside out"), we know the rigid symmetries of the cube. they are described by a subgroup of a "Weyl group" - but all that's important for our purposes is how many there are. this is easy to count, playing with a six-sided die: if you roll the die, there are 6 possibilities for the top, which consequently forces the bottom face's value. there are four vertical sides now, one of which can be called the "front" - but this forces the remaining three faces. hence, there are 6*4 rigid symmetries of the cube. taking our 6! colorings and factorizing with respect to these symmetries, (if you can rotate a cube, this doesn't really change its coloring), we get (6!) / (6*4) = 5*3 = 15 colorings. there is perhaps a more elegant interpretation in terms of pairings:

15 counts the number of ways you can sort 6 objects into pairs. all we've done with our colors is pair them (on opposite faces of the cube). we don't care which face they color, only the relationships between them.

unfortunately, this "pairing argument" does not generalize to other shapes - we can check that the tetrahedron allows 2 distinct colorings, but there are 3 ways to pair 4 objects. we can also show that the octahedron allows (8!) / (6*4) distinct colorings, but there are 7*5*3 ways to pair 8 objects. I haven't checked the dodecahedron or icosahedron, but I conjecture that the cube is the only platonic solid that this can work for. anyone care to explain why?

An old one and probably easy for you gus:

You are on your way and face 2 doors... one is the correct way the other one is not. You need to know which one you must choose.
Near the doors there are 2 guys who know which one is the correct door... One of the guys always speaks the truth the other one always lies. You know this but you don't know which one speaks what.

You have 1 question and only to one of the guys to get the conclusion to which door choose... What is that question?

@ orthaic
If you weight A,B,C against D,E,F
say they are equal, that means that the fake is one of G,H,I,J,K,L

Continue from there and show me how you find in two weighings which is the fake and whether it is heavier or lighter... I am curious as to how to do it with only two weighings, as far as I was able to work out, I sometimes needed three.

orthaic, if you want to allow mixing of colors, you have to count the number of ways to mix them correctly - you're right that there are 6 pure colors, and 15 secondary colors. we can also argue by symmetry that you're right about the muddiest mixes of colors. but you're wrong about tertiary (mixes of three) colors,

there should be "six choose three" = (6!) / ((3!)((6-3)!)) = 20 tertiary colors.

after making this small correction, there isn't much of a way to salvage your argument - there is no reason you should be multiplying the number of sides by the number of colors. and then your subtracting one from this product instead of its factors is just crazy! what we should do is the following:

first, choose six of your 63 colors (note, this should really be 64, if you admit the existence of an "empty color") - there are (63!) / ((6!)((63-6)!)) such choices. then color the cube exactly as i described in my previous post (15 distinct ways). thus, we would get a total of "(63 choose 6) times 15" possible colorings. it comes out to be over a million. so you're only off by a factor of about 5... ;)

@pat dragon: did i get my colour combination calc wrong? I wasn't sure how to add the three colour mixes...

@Troodonte: simple binary type problem which seems much simpler than the weighing scales problems above (less info to be gained - though a different way of getting it hmm... and one question rather than multiple weighings) I'm not going to answer it to allow someone else try.

@ Troodonte
You ask a predetermined question you know how the answer will come out. For example:
"If I ask your friend which door leads out, what would he (or she) answer?" From the response, you take the opposite door every time.