If a book can hold 720000 characters and there are 37 characters how
many different books are there? (I think this is 37 to the power of
720000 but I'm not sure). If these books are 17.5 cm long, 11cm wide,
and 1.9cm deep, and they are all packed into a box (a cube) how long
are the sides of the box? Please express this answer in light years.

Hi ioawnat-ga,

You are correct that the number of different books would be 37 to the
power of 720000. Using the KCalc calculator (on Linux), I found that
37^720000 approximates to 1.7431238 times 10 to the power of 1129105.

Using the Linux "units" program, I found that there were 2.3152062
times ten to the power of 51 books per cubic lightyear.

Dividing the number of books by the number of books per cubic
lightyear gives 7.5290219 times 10 to the power of 1129053 cubic
lightyears for all the books, so each side of the box will each be the
cube root of this number, or 1.9599554 times 10 to the power of 376351
lightyears.

That is an incredibly big number - way bigger than a Googol or the
number of particles in the known universe - so I think we can safely
assume that not all possible books will ever be written.

(For this answer, I have had to assume that KCalc works correctly for
these huge numbers. Before you build your box, I suggest you
double-check my calculations on a different calculator.)

Regards,
eiffel-ga

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Request for Answer Clarification by ioawnat-ga on 10 May 2006 21:34 PDT

Sorry, also, does anyone know the size of the universe? I know it's
expanding but at any particular time it must have a size. Heard any
estimations? Would my box fit in the universe?

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Clarification of Answer by eiffel-ga on 11 May 2006 03:12 PDT

Hi ioawnat-ga,


According to the following article, the universe is estimated to be
156 billion lightyears across:

"Universe Measured"
http://www.space.com/scienceastronomy/mystery_monday_040524.html

A billion is ten to the power of nine, which is enormously smaller
than the ten to the power of 376351 lightyears required to hold your
books.

Regards,
eiffel-ga

Many of those "books" are going to be just different by just one or more typos.

What about books that do not use the full compliment of 720000?  Would
a book with just a single character be considered in your calculation?

to add books with fewer characters consider the sum

1 (book with no characters) + 37 (books with 1 character) + 37^2 +
37^3 + ... + 37^720000

this sums to (37^720,001 - 1)/36, which of course is roughly
37^720,000 * 37/36, around 3% more than the books with exaclty 720,000
characters.

(I leave the light years calculation to you).

Of course, depending on the nature of the characters, these shorter
books are already included (assume 26 letters, 10 numerals, and one
blank).  The descrepancy is nonexistant if one character corresponds
to blank space (as in most languages I use).

The characters include a space- so one character and the rest of the
book blank is just one character and many spaces- so the second part
of the equation is not needed. Thanks though, i.