

Subject:
Large number Computation
Category: Science > Math Asked by: ioawnatga List Price: $2.00 
Posted:
05 May 2006 06:19 PDT
Expires: 04 Jun 2006 06:19 PDT Question ID: 725701 
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. 

Subject:
Re: Large number Computation
Answered By: eiffelga on 05 May 2006 07:38 PDT Rated: 
Hi ioawnatga, 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 doublecheck my calculations on a different calculator.) Regards, eiffelga  
 

ioawnatga rated this answer: 

Subject:
Re: Large number Computation
From: myoaringa on 05 May 2006 08:25 PDT 
Many of those "books" are going to be just different by just one or more typos. 
Subject:
Re: Large number Computation
From: herkdrvrga on 06 May 2006 05:47 PDT 
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? 
Subject:
Re: Large number Computation
From: mpchandlerga on 08 May 2006 05:27 PDT 
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). 
Subject:
Re: Large number Computation
From: thebigbrainiacga on 09 May 2006 23:05 PDT 
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). 
Subject:
Re: Large number Computation
From: ioawnatga on 10 May 2006 21:29 PDT 
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. 
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