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, let $\pi_i$ denote the $i$-th digit of $\pi$ (i.e. $\pi_0=3,\pi_1 = 1,\pi_2=4,\dots$).

Then the sequences $(\gamma,e,\sqrt 2,2^{4/3},\pi_1,\pi_2,\pi_3,\dots)$ and $(\ln(2),\gamma,-7.8,\pi_0,\pi_1,\pi_2,\pi_3,...) $
are in the same group, because they both are equal on all but the first 4 elements.

The Axiom of Choice is required to *choose* an arbitrary representative from each group. For example, I can choose $(1.49,3,-\cos(4),\pi_0,\pi_1,\pi_2,...)$ to represent the group I described above, but since there's infinitely many groups and I only have a finite amount of space to describe my strategy, I must appeal to the Axiom of Choice to produce a "choice function" that tells me which representative should be chosen from each of the groups.

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, and so on. Meanwhile mathematician #2 will open every box except 2, 102, 202, 302, etc., and in general mathematician #$n$ will open all boxes except those labeled $n,100+n,200+n,300+n,...$.

Back to me. I know what's inside the boxes that my buddy in room 2 didn't open. Let's suppose the numbers are:

- 2 -> 1739218.33
- 102 -> sqrt(5)-sqrt(2)
- 202 -> Arctan(37.238)
- 302 -> 382
- 402 -> -832.019
- 502 -> 4
- 602 -> $\pi_3$
- 702 -> $\pi_4$
- 802 -> $\pi_5$
- ...

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 box $602$ onward. Let's write that note like this: "x(2)=6". I'll repeat that process for #3, noting "x(3)=5" for box $503$, and for #4, perhaps I note that "x(4)=7" for box $704$, and so on.

What I've done is defined $x(m)$ for $m\in\{2,\dots,100\}$ to be the first box that disagrees with the associated representative given by the Axiom of Choice. However, I don't know the value of $x(1)$ since I haven't opened boxes $1,101,201,\dots$ yet.

What I can do though is let $y(1) = \max(x(2),...,x(100))+1$ be larger than all the observed numbers. It just so happened that $x(4)=7$ was the biggest, so $y(1)=8$.

It's finally time to open most of the remaining boxes. I'll open all the boxes in my sequence $1, 101, 201, \dots$, starting with the box given by $y(1)=8$: box $801$. (Since $y(1)$ has to be at least $1$, this strategy always leaves at least box $001$ closed.) Let's see what I got (let $e_i$ denote the $i$-th digit of $e$):

- 1 -> ???
- 101 -> ???
- ...
- 701 -> ???
- 801 -> 7pi+sqrt(3)
- 901 -> $e_{9}$
- 1001 -> $e_{10}$
- 1101 -> $e_{11}$
- ...

Seeing those last digits, I know enough to figure out which group it belongs to: the group with representative $(\sqrt 2,e_1,e_2,e_3,...)$.

I now know enough to make my guess. I'm going to use the representative, and pick the box given by $y(1)-1=7$, which is the maximum value of $\{x(2),\dots,x(100)\}$ (which we said was $x(4)$ in this example). In this case, the seventh entry is $e_7=1$ (note that we're zero-indexing so that the first entry goes with box $001$). So I'll guess that box $701$ contains $1$.

Of course, mathematician #$n$ will do exactly the same thing by considering the values of $\{x(1),\dots,x(n-1),x(n+1),\dots,x(100)\}$,
computing $y(n)$, and so on. Now, I need to prove to you that these strategies work.

To do this, 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? Obviously, the box given by $y(1)-1$ didn't match the Axiom of Choice's representative. That means that $x(1)>y(1)-1$, since $x(1)$ is the number for which every other mathematician knew the boxes from that point forward matched the Axiom of Choice's representative.

With this information, I realize something. $y(1)-1\geq x(n)$ for every other number n, since $y(1)$ is defined to be the maximum of the $x(n)$ plus one! So here's what I now know:

$x(n)\leq y(1)-1<x(1)$

This is great news. Everyone else defined $y(n)$ knowing $x(1)$, and $x(1)$ has just been shown to be bigger than the other $x(n)$. So:

$x(n)\leq y(1)-1<x(1)<x(1)+1=y(n)$

So, for each mathematician #$n$, all boxes $x(1)$ and onward match the representative given by the Axiom of Choice. Every mathematician #$n$ opened $x(1)+1$ onward, and guessed the choice function's $x(1)$ entry for the $x(1)$ box, which has to match up!

Thus, if I'm wrong, everyone else has to be right. And that beats the game.