Rainman

You and your friend are standing on a road which goes on forever in both directions. Your friend suggests a game of chase. Since you are faster, your friend will get a head start. Also, to make it more interesting than a simple Achilles vs Tortoise race, you will not know in which direction your friend takes off. You move at constant speed x>1 m/s and your friend moves at constant speed 1 m/s. Your friend gets a 1 hour head start. Visibility is limited so you will not be able to see your friend from a distance. You must not stray from the road, same goes for your friend. Can you catch up to your friend? Does the answer depend on your speed x, and if so, what is the minimum x for which you can catch up? What is your strategy?

You're missing that they are not just using one template sequence. They define equivalence classes by how a sequence ends, so that two sequences belong to the same class if and only if they are different in only finitely many places. Then they choose one template sequence from each equivalence class.

And that's why you should never go by intuition in mathematics (or at least have better intuition than I do)

The central almost empty area of a polygon is the set of internal points which are more than halfway out from every corner. At most one such point exists except in triangles, and that is the midpoint of the polygon. The midpoint can be almost empty and give rise to infinitely many other almost empty points. However, since they are discrete points, they don't form any noticable pattern. So triangles are the only polygons with these patterns. (Proofs omitted because I'm just going by intuition here )

The almost empty area consists of triangles of different sizes. You can only get to one of these almost empty triangles from a larger almost empty triangle. So no almost empty triangle can contain more than one dot. You can get out of the almost empty area if you started inside it, but you can't get back inside it once you've left.

Suppose coins 1,2, and 3 are gold, silver, and bronze respectively. - Put coin 1 in pile A. - Weigh coin 2 against coin 1, not equal. (Both coins have been weighed once) - Put coin 2 in pile B. - Re-order piles so coin 2 is in pile A, coin 1 in pile B. - Weigh coin 3 against coin 2, not equal. - Weigh coin 3 against coin 1, not equal. (All coins have been weighed twice) I think this will only work if you can see the result of the third weighing of a coin before the coin magically disappears.

I think uncountably countably is all that's needed. The boxes themselves would only contain countable amounts of information, but I think the information the mathematicians need to bring into their rooms (i.e. their strategy) is uncountably infinite. Unless I'm missing a simpler solution.

No form of communication is necessary once they have entered their rooms. They can not see, hear, or otherwise perceive anything another mathematician does. We do need to allow some other wild stuff though.

Actually we have the same amount of apples used, but he was more efficient in terms of answering first

This is a "famous" unsolved problem. The conjecture that every positive integer eventually leads back to 1 is called the Collatz conjecture. We won't be able to prove or disprove it here, sorry