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Googolplex: Sheets of Paper

Date: 11/18/97 at 00:02:53
From: Eich
Subject: Googolplex

Can you tell me how many sheets of paper it will take to make a
googolplex if you can have 20,000 zeros on each page? 


Date: 11/18/97 at 12:11:47
From: Doctor Rob
Subject: Re: Googolplex

Since a googolplex is N = 10^(10^100), there are 10^100 zeroes in
its decimal form. If 20,000 = 2*10^4 fit on one page, you will need
(10^100)/(2*10^4) = 5*10^95 pages.

-Doctor Rob,  The Math Forum
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Date: 01/05/98 at 13:55:39
From: Doctor Mark
Subject: Re: Googolplex


Well, first let's make sure that you mean a googolplex, and not some
smaller number, like a googol. You will remember that a googol is the
number that is written by putting down a 1, then following that on the
right with 100 zeros, i.e., it is the number 10^100. (Just so you 
know: computers are not good at writing exponents.  So when I write 
10^100, the 100 is the exponent.) 

This is a really big number, of course. For instance, the number of 
seconds since the beginning of time is only about a 1 followed by 
18 zeros, and the number of atoms in the entire universe is
estimated to be only about 10^80, a 1 followed by 80 zeros, so you'd 
need 10^20 (written out: 100000000000000000000 [that's a 1 followed by 
20 zeros]) universes to have a googol of atoms. (Still with me?  This 
stuff gets pretty spacey after a while...).

Now a googolplex is a 1 followed by a *googol* of zeros. That's a 
truly humungous number. In fact, if you took all the atoms in the 
*entire universe* and lined them up, put a l on the first one, and 0's 
on all the rest, you still would not have been able to write down a 
googolplex, since there are only 10^80 atoms, and you need to write 
10^100 zeros.

So now back to your original question.  If you put 20000 zeros on a 
page (it would have to be a pretty big page: more like a newspaper 
page than one in a book, unless the zeros are really small: in type 
like this, you can only fit about 3500 zeros on a normal page), then 
to write a googol, you would need to N pages, where

   (number of zeros in total) = 
   (number of zeros on one page)*(number of pages), i.e.,

   10^100 = (20000)N  ---> N = (10^100)/(20000).

If you work this out using the laws of exponents (writing 10^100 as 
(10) X (10^99), and 20000 as (2 X 10^4), then dividing), you find that 
N is 5 X 10^95, i.e., a 5 followed by 95 zeros.

If in fact you really needed to find the answer for a googolplex, 
then you would write down the same kind of equation, except instead 
of 10^100, you would need 10^(googol):

   10^(googol) = (20000)N ---N = (10^googol)/(20000).

Now if you work *this* out using the laws of exponents, you'll find 
that the answer is

   N = 5 X (10^(googol - 5)),

i.e., a 5 followed by a (googol - 5) zeros, i.e., by

99999999999999999999999999999 - 5 zeros (there are 99 nines there).

As before, there aren't enough atoms in the univere to write this 
number down, so aren't you glad that we have scientific notation?

Dr. Mark,  The Math Forum
Check out our web site!   
Associated Topics:
Elementary Large Numbers

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