Some definitions from the story

• Each line has 80 characters

• There are 40 lines on each page and 410 pages on each book

We can calculate the number of characters on each book.

With the computer

\[ \begin{aligned} pages\_per\_book &= 410\\ char\_per\_book &= 80 \cdot 40 \cdot pages\_per\_book \end{aligned} \]

How many books

Each character can be any of 25 symbols

The total number of books is \(25^{1312000}\) which is not easy to calculate without the proper tools

Back of the envelope

A quick “back-of-the-envelop” calculation can confirm this result. We have \[25^n = 5^{2n} ≈ 4^{2n}=2^{4n}\] Thus we are speaking about a little more than 5.2 million bits

Using the rule of thumb \(2^{10}≈10^3\) we get \(10^{1.6\text{ million}}\) books.

Trying harder

The difference between \(5^{2n}\) and \(4^{2n}\) is important, so we will try better. We have \[25^n = 5^{2n} = 4^{2n\log_4(5)}\]

We have that \(\log_4(5)≈1.16\) so to change from base 5 to base 4 we need to increase the exponent by 16%. So 1.3 millions become 1.5 millions, and so we have near \(4^{3\text{ million}}=2^{6\text{ million}}\) books.

Another way

In fact it may be easier to see that \(\log_2(5)\) is a little more than 2, in fact near 2.32, thus \[25^n = 2^{2n\log_2(5)}≈ 2^{4.64 n}\] When \(n\) is 1.3 million, then \(4.64 n\) is about 6 million. We need 6 megabits just to count the number of books in The Library. Now, \(2^n ≈10^{3n/10},\) therefore there are near 10^1800000 books.

We need to work with orders-of-magnitude of orders-of-magnitude.

How many rooms?

Using more data from the story, we can calculate how many books fit on each room, and therefore how many rooms are in the Library.

\[ \begin{aligned} books\_per\_shelves =& 32\\ shelves\_per\_wall=& 5\\ walls\_per\_room =& 4\\ books\_per\_room =& books\_per\_shelves \cdot shelves\_per\_wall\\ & \cdot walls\_per\_room\\ rooms =& books / books\_per\_room\\ \end{aligned} \]

Area of each room

The area of a hexagon of side \(s\) is \[\frac{3\sqrt{3}·s²}{2}\] Thus, we need to know the width of each wall

We know that there are 32 books on each shelf, so we need to estimate the width of each book.

We need to make some assumptions

• Paper thickness is around 0.0001m, that is, 10 pages per millimeter. This is based on the fact that a pack of 500 page takes more or less 5cm.
• Each sheet has 2 pages
• The covers are 1 mm each

\[ \begin{aligned} paper\_width &= 1⋅ 10^{-4} m\\ book\_width &= paper\_width \cdot \frac{pages\_per\_book}{2} + 2e-3 m \end{aligned} \]

Therefore, the wall width is

\[ wall\_width = book\_width \cdot books\_per\_shelves \]

Now we calculate the area of one room

\[ room\_area = \frac{3 \sqrt 3}{2} \cdot wall\_width^2 \]

and then the area of the whole library

\[ library\_area = room\_area \cdot rooms \]

Compare this area to the surface of Earth

\[ earth\_radius = 6371 ⋅ 10^3 m\\ earth\_area = 4 \pi \cdot earth\_radius^2 \]

Book size

\[ \begin{aligned} room\_height &= 1.6 m\\ book\_height &= room\_height / shelves\_per\_wall\\ aspect\_ratio &= 3/4\\ page\_area &= aspect\_ratio \cdot book\_height^2 \end{aligned} \]

Weight of book

\[ \begin{aligned} paper\_weight &= 50⋅ 10^{-3} kg/m^2\\ page\_weight &= page\_area \cdot paper\_weight\\ book\_weight &= pages\_per\_book \cdot page\_weight \end{aligned} \]

Weight of room

\[ room\_weight = books\_per\_room \cdot book\_weight \]