TITLE: Proving you *can't* make $2011$ out of $1,2,3,4$: nice twist on the usual
QUESTION [224 upvotes]: An undergraduate was telling me about a puzzle he'd found: the idea was to make $2011$ out of the numbers $1, 2, 3, 4, \ldots, n$ with the following rules/constraints: the numbers must stay in order, and you can only use $+$, $-$, $\times$, $/$, ^ and $!$. In words, "plus minus times divide, exponentiation and factorial". The game was to construct $2011$ with $n$ as small as possible.
To my amazement, he had managed to do $n=5$: indeed
$$((1+2)!)!+(3!)^4-5=2011.$$
He now wanted to solve the puzzle completely by proving that $n=1,2,3,4$ are impossible.
So there's where it gets interesting. My first thought was "computer search -- done". But it's not as easy as that, because factorial is only a unary operator. For example one has to rule out any possible amazing cancellations between very large numbers, e.g. one has to check
$$(1+2)!!!!!!!!!!!!!-3^{(4!!!!!!!!!!!)} \neq  2011.$$
This one in particular is not hard to rule out, because, for example, the left hand side is a multiple of $3$ and the right hand side isn't. In general, this approach can perhaps be used to bound the number of factorials that can occur in any presentation of $2011$ using $1, 2, 3, 4$ only -- but making this rigorous seemed a bit delicate and I wondered if I was missing something. Anyone any ideas?
EDIT: Ron Maimon's heroic attempt to deal with the problem by brute force has led him to the interesting case of $2011=(1+2)!!!...!!!/(3!!!...!!!/4!!!!...!!!)$, which seems to be tougher to rule out than the others: the game here is to prove that $2011$ cannot be written as $3!!!...!!!*4!!!!...!!!/3!!!!...!!!$ for any choices of numbers of factorials.
EDIT: Size considerations seem to deal with the above case. Ultimately it seems that the question I asked can be answered using a rather lengthy, but finite(!), procedure. Thanks Ron for your efforts.

REPLY [49 votes]: If square roots and percentages (and unary minus) are permitted, then there are a number of solutions, the simplest probably being
$$2011 = (-1 + (2 + 3!)!\%) \div \sqrt{4\%}$$
Many years ago, I wrote a simple Windows program called $200$ Up (link to Dropbox folder) for solving problems like these:
$\hspace{1.75in}$ 
The domain of calculation is $\mathbb{Z}/d$ for some specified denominator $d\in\mathbb{N}$ (and with limits on the magnitude of the numerator).
Note that, in the image above, the solution for 2018 is
$$
\large 2018 \;\;=\;\; \big( \sqrt[\sqrt{\cdot \dot{1}}]{\Large 2} \; + \; 3!!\%\% \big) \; \div \; \cdot 4\%.
$$