Numbers Theory was the only math course I didn’t take many years ago (I didn’t know that I would end up in philosophy) and just a practical note;
You don’t want to be preferring more than 2 dimensional arrays in Excel, unless they are very small or you are merely using the VBA.

[tab]I thought this one was interesting because it involved 10 prime numbers except for only one of them. Why only prime except for one?!?![/tab]

[tab]No wait, that’s only if he knew that it was possible to know what was a and what was b. But there’s no reason to think that Peter and Sarah knew the question posed to us, let alone before they started talking. Anyway, the problem cannot be solvable with that question, as the numbers can only be 4 and 13; nor is it possible, from the information given, to tell which is which.[/tab]

Both very clever illusions, especially the second.

First:
[tab]The hypotenuse is not quite strait. The visual effect is subtle, but you can calculate the slopes easily: for the red triangle it’s 3/8, and for the green triangle it’s 2/5. The result is the combined hypotenuse bends slightly in for the first triangle, and slightly out for the second, and that bend accounts for a 1x1 unit difference.

I’m sure there’s a more mathematically precise way to put this, but this satisfies me that the laws of mathematics aren’t being locally suspended [/tab]

Second:
[tab]This is a very impressive illusion. Even once I saw it, it’s still hard to believe that someone could come up with it and execute it.

The trick is that, rather than one person being completely removed between one version and the next, different parts of different people are removed by combining and splitting. So e.g., one person’s bottom half of a foot becomes another’s whole foot. It’s hard to identify who’s losing or creating what because it seems like everyone gives a little bit and the result is another full person.

More on #2, which I’ve been thinking about too much:
[tab]A few additional small insights into how it works:

Each person is affected by the swap.

Each person is shorter in the 13 person case than in the 12 person case.

If you trace who ‘gives’ pieces of themselves to who, you will find that there is a chain of A->B, B->C, C->D etc. through every person (start in the lower left, you will end with the 2nd from the left along the top row in the 13 case).

I think a few modifications to this gif might make it clearer what’s happening:

if the transition were done by moving the whole top half of the image to the right, as though it were wrapped around a cylinder.

if the people were replaced with solid blocks or lines.

if different people were different colors, so that you could see the break points more easily.

A hint, responsive to James
[tab]The Master’s reassurance that the problem is not impossible for any true logician constrains the possibilities. If the problem would be impossible to solve with one possible configuration of colors, that configuration is excluded[/tab]

A hint-like clarification, not necessary but constrains the answer space:
[tab]The problem doesn’t have a specific answer, only a general one. We can’t say how many logicians there are, but we can say a lot about minimums and how the numbers affect when the logicians leave.[/tab]

[tab]The only way that this problem can be solved is by assuming that the visual field provides the constraint for the group of colors involved. I can imagine this as if I am one of the logicians and I see that around me all of the colors appear more than once. Therefore, the color of my headband has to be the same as one of the colors I can see. Otherwise the problem is unsolvable since you wouldn’t have a closed group of colors, and the options to chose from are infinite.[/tab]

Alright, someone take it from here. I’m a little busy

Nope:[tab]We know that there must be at least three band colors merely because of the word “many”. The problem is that there are any number of variety of potential colors. If there was only two other logicians, your band could be one of 100 other colors. You might not even know the name of the color. And it doesn’t matter what you see other than to let you know that if there were only two others, you could know that your color is different than theirs. Any more than 3 members and you cannot know that your color is any different.

A logician cannot deduce a color that he perhaps has never even seen before.[/tab]

[tab]You are showing your feminine thinking. In logic puzzles you have to think in terms of what is certainly true, not what might be true, even if almost certain. The fact that all of the others were duplicated doesn’t allow for you to conclude that your color is duplicated.[/tab]

[tab]In the terms of the problem, we are told that it is not impossible. So any assumption that would lead to an impossibility is false.

One can progress from there: If X leads to an impossibility, we know ~X. What can we conclude from ~X?

If assuming that my headband can be any imaginable color makes the problem impossible, then my headband cannot be any imaginable color. The set of possible answers is constrained.[/tab]
I don’t know what “feminine thinking” means (thinking-while-a-woman?), but whatever it is, it’s working.

[tab]It isn’t the fact that all others are duplicated that allows to make that assumption. It is the fact that the problem states that it is solvable.

You said it yourself, you can’t know what your color is, the problem is unsolvable. But because it is specifically stated that the problem is solvable, you have to make the assumption that your color is duplicated at least once because if you can’t see your own color, you don’t have a constrained group of colors to work with.[/tab]

I sort of see what he means…[tab]It is fair to expect that when you are going to work on a problem, whoever gave you the problem must give you all the constraints of the problem. In this problem there is a constraint missing, but there is enough information for you to make an assumption on how to close the problem.
I suppose that you shouldn’t be making any assumptions in a logic problem. In this specific problem, the logic actually begins once you define the last constraint, but the logic itself is not the interesting part of it, it is how you arrive at the conclusion that allows you to define the problem.

Making assumptions or deductions or whatever = feminine thinking?
In any case it is a good skill to exercise, since there are very few problems for which all constraints are known.

I know the solution, I just don’t want to type it out. I already solved it so I want to move on … ok I’ll type it out in a little while, but for a quick and dirty version, read on.[/tab]

Quick and dirty answer:
[tab]color repeated once leaves on 1st bell, color repeated twice leaves on 2nd bell, so on so forth.[/tab]

Yup.
[tab]Once you constrain the problem to require at least two of each headband, it becomes the Blue Eye problem, except that the last color to leave leaves on the bell after the second to last color to leave, not matter how many of them there are.

I do find interesting the logic surrounding the premise that the problem is not impossible. It seems straightforward that the assumption to make is that there’s more than one of each color. But it might be that we’re also implying a least-necessary assumption. Are there other assumptions that would make the problem possible and allow everyone to accurately deduce their band color? I haven’t spent a lot of time on this question, but it’s one I’d like to explore.

This problem reminds me of the concept of meta-programming (with which I’m only passingly familiar), since you need to deduce some of the rules from other rules in order to solve it.[/tab]

[tab]You’ll need to flesh that argument out, James. I never used the word “probably”, and proof by contradiction is not Bayesian.

I take it you don’t like this line of (non-probabilistic) argument:

If the headbands could be any possible color, the problem would be impossible to solve.
The problem is not impossible to solve.
Therefore, the headbands cannot be any possible color.

If the headbands could be any color other than those that each logician can see, the problem would be impossible to solve.
The problem is not impossible to solve.
Therefore, the headbands cannot be any color other than those that each logician can see.

Each logician can see all the possible colors of her own headband
Therefore, for all logicians of headband color X, there must be at least one other logician with headband color X
Therefore, for any color X, there must be either 0 or >1 logicians with that color headband.

So explain what’s wrong with it. Please don’t be sarcastic.[/tab]