Ahh crap. You are indeed.

Yes.

Take the rectangle, cover it with big circles, and then scale everything to 1/4 size (divide side lengths by 2, change circle radii to 1) and now you have 1/4 of the rectangle completely covered by 25 1-inch circles. If 1/4 of the rectangle is covered by 25 circles, the whole rectangle can be covered by 100 of them.

Am I missing something?

^This.

Edges. Size of rectangle. Amount of overlap. These factors need to be specified or accounted for in the solution.

so obvious... now :)

@Mujus, nope, doesn't matter. It just has to work for *any* rectangle that can be covered by the 25 circles of radius 2. The amount of overlap is entirely dependent on the rectangle.

A magician's assistant gets five audience members to draw a card each from a standard 52 card deck. He looks at the five cards and then hands four of them, one at a time, to the magician, who immediately identifies the fifth card. How is this possible?

Ok, so maybe this is cheating, but he is handing the magician 4 cards, so there are 4! = 24 ways he can pass them on. In order to get another *2, he can either pass them all upside down or right side up (the audience doesn't know whether he was planning on passing them right side up or upside down). This gives 48 units of information. Any function from these 48 choices to the 48 remaining cards would work.

Yeah?

The only thing that can be changed is the order of the cards, so no, that doesn't work.

So he chooses which one is the card to be guessed?

Yes.

Ok, I have an idea:

Since there are 5 cards, there are some 2 of the same suit. So we should for sure choose one of those 2 as the card to be guessed. That way, we can put the other one first, to give suit, and then still have 3 cards of freedom. That is still 4!= 24 units of information, but we still have choice between those two cards (which is first and which is to be guessed). I haven't yet figured out how to choose between them, and I am getting distracted by my Chargers getting taken apart.

Ok, I think I have it.

I'm pretty sure that the first card shown can narrow the secret card down to 6 cards as follows: There are certainly 2 cards of the same suit. One of those is going to be the secret card and one of them is going to be the first card. Also, one of these cards is 6 or less "behind" the other. I.E. if you have a 4 of spades and a king of spades, the king is 4 (ace, 2, 3, 4) behind the 4. So if you make the first card the one that is "behind", then you hve narrowed it to the six cards after that card, and can encode the choice from there with the last 3 cards using permutations.

Yes?

uclabb, you are right (or at least, your method works)

Yep.