Found this via Reddit. Here's my writeup of the solution.
The strategy involves the axiom of choice like so: the mathematicians group sequences of real numbers such that two sequences are in the same group if and only if they agree on all but the first few terms. For example, (pi,e,sqrt(2),2^(4/3),1,4,1,5,9...) and (ln(2),phi,-7.8,3,1,4,1,5,9,...) are in the same group, assuming that they both keep repeating the final digits of pi in their sequence.
The axiom of choice is required to choose an arbitrary representative from each group. For example, I can choose (1.49,3,-cos(4),3,1,4,1,5,9,...) to represent the group I described above, but since there's infinitely many groups and I only have a short amount of time, I need a special rule that says I can pick those infinitely many representatives. (That's a clumsy short version of how the axiom works, anyway.)
Alright, so now I'll describe the plan. Let's say I'm mathematician #1. I'm going to open every box except 1, 101, 201, 301, etc. My buddy one room over (mathematician #2) will open every box except 2, 102, 202, 302, etc. And so on.
Back to me. I know what's inside the boxes that my buddy #2 didn't open. Let's pretend the numbers are:
- 2 -> 1739218.33
- 102 -> sqrt(5)-sqrt(2)
- 202 -> Arctan(37.238)
- 302 -> 382
- 402 -> -832.019
- 502 -> 4
- 602 -> 1
- 702 -> 5
- 802 -> 9
Okay, I know the group that falls in. (Coincidentally, it's the group I talked about above.) I'm going to write a note that it started matching the representative from the sixth box onward. Let's write that note like this "x(2)=6".
I'll repeat that process for #3, noting x(3)=5, and for #4 perhaps x(4)=8, and so on. I know x(m) for m=2,...,100, but I don't know x(1) since those are the boxes I haven't opened yet. I'm going to take the biggest of those numbers, add one, and call that y(1), that is: y(1) = max(x(2),x(3),...)+1. It just so happened that x(3)=8 was the biggest, so y(1)=9.
It's finally time to open most of the remaining boxes. I'll open all the boxes in my sequence 1, 101, 201, etc., starting with the y(1)=9^(th) box. (Since every x(m)>=1, y(1) had to be at least 2, so in general there's always some unopened boxes.) Let's see what I got:
- 1 -> ???
- 101 -> ???
- 701 -> ???
- 801 -> 7pi+sqrt(3)
- 901 -> 2
- 1001 -> 8
- 1101 -> 4
Seeing those last digits, I know enough to figure out which group it belongs to: the group with representative (e,2,7,1,8,2,8,1,8,2,8,4,6,...).
I now know enough to make my guess. I'm going to use the representative, and pick the y(1)-1=8^(th) member, which is the maximum value of the x(m)s. In this case, the 8^th member happens to be "1", and I'm going to guess the 8^th unopened box, box #701, holds number "1" inside.
Now, I need to prove to you that this strategy (and similar strategies for the other mathematicians) works. It's okay to assume that I, mathematician #1, am wrong. I just need to prove that if I'm wrong, then everyone else is right! (99/100 ain't bad, according to the rules.)
Okay, I'm wrong, so what happened. Well, obviously, the y(1)-1^(th) term didn't match the representative. That means that $x(1)>y(1)-1$, since x(1) is the first number where it and all later boxes in my sequence match the representative.
With this information, I realize something. $y(1)-1>=x(m)$ for every other number m, since I defined it to be the biggest x(m) plus one! So here's what I now know:
Oh man! This is great news. Everyone else defined their own y(m) based on the other x, and x(1) is the biggest x out there. So:
So, for each mathematician #m, all boxes x(1) and onward match the representative. Every mathematician #m opened x(1)+1 onward, and guessed the representative's x(1) entry for the x(1) box, which has to match the representative.
Thus, if I'm wrong, everyone else has to be right at least. And that beats the game.