How did Hermite calculate $e^{\pi\sqrt{163}}$ in 1859? Pretend you are in 1859. What is a fast, efficient, and accurate way to numerically evaluate constants like that to, say, 20 decimal places, using ONLY pen and paper?
 A: Often you can find some expression for a number that lends itself well to pencil and paper computation. For instance, suppose you were trying to compute $\sqrt{3}$. You can use this expression:
$$\sqrt{3}=\frac{1732}{1000}\left(1-\frac{176}{3000000}\right)^{-1/2}$$
If you use the binomial series for $(1-x)^{-1/2}$ you can compute that efficiently. $176/3000000$ is small enough that the series converges rapidly. This particular example is a problem from Apostol's Calculus, volume I, and it asks for 15 places, which if I recall correctly requires out to the $x^5$ term of the binomial series.
Doing this kind of calculation used to be part of a decent mathematical education, before calculators were common, and any working mathematician then would be quite agile at this (or would have an assistant whose job was to do calculations for the mathematician). Now we have pocket calculators, and computers that will quickly do these things for us to hundreds of decimal places, and so becoming adept at pencil and paper calculation is simply not a skill we are required to develop.
A: This is a bibliographical complement to Raymond wonderful answer. Since it contains no mathematics, I used the community wiki mode.
Hermite wrote a series of five articles under the title Sur la théorie des équations modulaires:
Hermite, C. Sur la théorie des équations modulaires. Comptes Rendus Acad. Sci. Paris 48, 1079-1084 and 1095-1102, 1859.
Hermite, C. Sur la théorie des équations modulaires. Comptes Rendus Acad. Sci. Paris 49, 16-24, 110-118, and 141-144, 1859.
The five articles can be freely and legally retrieved in pdf format by following the appropriate links given on this French National Library page.
These articles have been reprinted in Volume 2 of Hermite's Oeuvres Complètes, which can also be freely and legally retrieved in pdf format from this University of Michigan page.
The retrieval being somewhat tedious in both cases, I've put the various pdf files here.
The files corresponding to the five articles (from the French National Library) are named hermite_1.pdf to hermite_5.pdf here.
University of Michigan divided Volume 2 of Hermite's Oeuvres Complètes into twenty page files. The relevant article is scattered through three such files, named hermite_a.pdf, hermite_b.pdf, and hermite_c.pdf here. The article goes from p. 38 to p. 82.
The page, containing the first digits of $\exp(\pi\sqrt{43})$, Raymond points to at the beginning of his answer is page 8 of this pdf file (p. 1101 of the scanned text), or page 15 of this pdf file (p. 60 of the scanned text).
There is a fact I find very surprising about this famous page: In the Comptes Rendus version, Hermite thanks C.-J. Serret for having done the computation:

... on trouve (*) $$e^{\pi\sqrt{43}}=884736743.9997775\dots$$
(*) Je dois ce calcul à l'obligeance de M. C.-J. Serret.

Again, I find strange that this acknowledgment has been suppressed from the reprinted version.
EDIT. Life is sometimes funny:
Tito asked "How did Hermite calculate $e^{\pi\sqrt{163}}$?"
Raymond answered "Hermite calculated $e^{\pi\sqrt{43}}$, not $e^{\pi\sqrt{163}}$".
Then it turns out that Hermite did not calculate $e^{\pi\sqrt{43}}$, but C.-J. Serret did.
Note that C.-J. Serret (not to be confused with Joseph-Alfred Serret) was not a mathematician, but an astronomer. This suggests (I think) that, probably, methods similar to the ones described by Raymond were used.
A: Reducing everything to calculating something that is easier to manage
$$e^{\pi\sqrt{163}}=x$$
In order to simplify the calculation take $x=\frac{p}{10^{18}}$
$$\ln(p)^2-36\ln(p)\ln(10)+324\ln(10)^2-163\pi^2$$
Now this one has two zeros and one is substantially smaller than another so much easier to find
$$y^2-36y\ln(10)+324\ln(10)^2-163\pi^2=0$$
You calculate this by either replacing all constants or just calculating the final result first.
$$y_1=1.33736168276030...$$
It is simpler to find and handle this smaller number
$$e^{y_1}=3.808980937007642...$$
Now it is simply
$$e^{\pi\sqrt{163}}=\frac{1}{3.808980937007642...}10^{18}$$
Three required operations can be done with any precision available at the time. $\pi$ was known to hundred decimal places which is more than sufficient for our calculations.
