plasmid

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Ah, thanks Rainman and Bonanova. That actually makes sense (which is sort of frightening). Suppose we want to make things easier for the poor mathematicians who don't have infinite memories. For each equivalence class, we choose the representative sequence in that class such that x1 = 1, x2 = 2, x3 = 3 .... until we reach a point N out in infinity-land where it starts matching every subsequent term in the class. We know that such a point must lie infinitely far out since for any N if your representative sequence stops following x(N) = N and starts following the equivalence class' sequence, you can define a sequence which has x(N) = N and then starts matching the equivalence class for all subsequent terms. Since there is thus no finite limit to how large N can be, each representative sequence will be x1 = 1, x2 = 2, x3 = 3 ... for an infinite number of terms. If the proposed solution using the axiom of choice works, then it should work using these representative sequences. So if a mathematician guesses the value in any box numbered M in his subsequence, he must guess that it contains the value M. Now I'm going to set up the boxes so they all contain negative numbers.

I finally had enough and went ahead and googled the OP. I've got to say, after looking at the solution they proposed, I would definitely place it in the category of "shenanigans". But maybe I'm missing something.

I'll be the judge of that That is, if I decide I've had enough of this problem and just ask for the solution you two seem to have in mind. Which is not yet.

This reminds me of the "Team of 15" problem, although it doesn't look like it can really be approached the same way. Are you working under the premise that all boxes must contain a different number? It doesn't look like that's the case.

I get the feeling this will depend on how exactly you define communication. Since this is one of the best puzzles you've seen, your definition of communication would probably prohibit such a straightforward approach. But if you take an ultimately strict interpretation of communication, like "no mathematician can leave any evidence of their existence to any other mathematician after the game starts", then this should be indistinguishable from sending each mathematician into the game completely alone and there would clearly be no solution. So there must be some form of communication. We'll need to know what's allowed and what isn't: whether mathematicians can rearrange boxes or see the boxes that other mathematicians have rearranged, whether the other mathematicians hear a guess that any other mathematician makes or whether or not it was correct, etc.

I still can't recognize anything that stands out as FF. Forgive and forget? Far flung? Fast and furious? Freedom fighters? 255?

Now we just need seven more of those to make the whole tongue-twister.

[spoiler= ]nn-1/2 = 1; n = 2 sqrt(n)-1 = 2; n = 9 sqrt(n)+1 = 3; n = 4 n2 = 4; n = 2 (3n)!-n = 5; n = 1 (3(n!))!+n = 6; n = 0 2srqt(n)+1 = 7; n = 9 n*1n (=n if n<infinity) = 8; n = 8 2+n = 9; n = 7 n2+1 = 10; n = 3 n2+(n-1)! = 11; n = 3 2n+|n| (=3n, n>0, or =n, n<0) = 12; n = 4 (n3+n)/10 = 13; n = 5 2n = 14; n = 7 10(n+1)/n = 15; n = 2 2n-(sqrt(n)-1)! = 16; n = 9 2n+1 = 17; n = 8 (2n)n+n = 18; n = 2 (n-1)!-n = 19; n = 5 (2n)!-n2 = 20; n = 2 n*sqrt(9) = 21; n = 7 n+20 = 22; n = 2 [10/n]n/2*n-sqrt(n) = 23; n = 4 (11-n)! = 24; n = 7 n2 = 25; n = 5 a(b·cdsqrt{ef-g+sqrt[(((h·i-j)·k·l+m)·n+o)·p] ·q(r·s)t} -u+v/sqrt(w)-x)y 2(9·42sqrt{10-9+sqrt[(((8·7-3)·3·4+5)·7+2)·9] ·8(2·5)2} -7+2/sqrt(4)-7)5 2(144 sqrt{ 1 + sqrt[(( 641 )·7+2)·9] · 800 } - 12 )5 2(144 sqrt{ 1 + 201 · 800 } - 12 )5 2(144 401 - 12 )5 577320

With that setup, I don't think you can call the width of the moat equal to the hypotenuse of the triangle formed by half of the first board plus all of the second board. A line from one end of the first board to the end of the second board that's touching the opposite shore would not be perpendicular to the moat. You would need to solve for the length that's perpendicular to the moat in order to find out how wide of a moat you could cross.

Yoruichi has this one. Definitely on the right track with the previous guesses, just not quite hitting the actual thing that I was trying to describe.

I don't think 25 is the answer -- the first expression would evaluate to 2524/2 which is not an integer.

Something quite close to those answers, but think of a union that's held together by a thread of steel that's curved. And the "you" in "teeth brush you" really is referring to the "you" and not the "I" of this riddle.

Not a padlock, but you're so close it hurts.

If you try to compact the top row like this, at some point during compaction the circles on the top row must lie directly above the circles on the bottom row at which point such compaction cannot continue. Or if you propose that the compaction traverses a zone where circle N-1 is just a tiny bit to the right of circle M-1 below it and circle N is just a tiny bit to the left of circle M below it, all without ever having the top row of circles exactly above the bottom row, then if the lower circles M-1 and M are touching then it would require you to somehow pack more efficiently in that block of four circles than rectangular packing. Edit: although now that I understand TSLF's answer, I think that it should work in principle. Haven't verified the calculations myself yet though. Edit2:

Not a vending machine. I can see where you're coming from with that answer; clues about maintaining a union with a thread of steel, not letting a fraudulent claim through, and (maybe) teeth brushing you could maybe be made to fit it, but I think the answer I have in mind is enough of a more natural fit that most people could recognize it as the right answer when re-reading the riddle.

It might be easier with a rephrasing where the clues are more clearly clues. I have a hollow skull, but am not "dead" because of it I have come from or been sent from some sort of "burrow" I either am a part of or am a union kept together with steel And that steel is not straight The "you" in the riddle should recognize/inspect me Any liars or impostors won't get through I will, with a handshake, pick something to "reject" Teeth brush you

I take that back, it probably is impossible. The three equations with three unknowns I could set up turned out to be degenerate.