How Big is a Googol?

Date: 03/02/2003 at 18:15:37
From: Anna
Subject: Googol

Is there any way to hold a googol in your hands? I read once on a 
Web site that if you did something with cards the chances would be a 
googol or more.

Date: 03/20/2003 at 11:00:11
From: Doctor Ian
Subject: Re: Googol

Hi Anna,

A googol is a 1 followed by 100 zeros.  That's a pretty big number! 
For example, a trillion is a 1 followed by 12 zeros.  So 

 1 googol = 10 thousand trillion trillion trillion trillion
               trillion trillion trillion trillion

A much easier way to write this is 10^100. 

Now, just how big is that?  Well, consider that a nickel weighs about
5 grams.  A diamond that weighs as much as a nickel would contain
about 10^23 carbon atoms.  That's also a big number, but it's
insignificant compared to a googol.  

The earth weighs about half of 10^28 grams.  So if you imagine a
diamond that weighs as much as the earth, it would have about 10^50
carbon atoms.  Still pretty small compared to a googol.  

A billion diamonds that big would have about 10^60 carbon atoms. So
clearly you're not going to be able to hold a googol of things in your
hands.  

But can you hold a googol of _possibilities_?  That's where things
like cards come into the picture. If you have a deck of cards, how
many different ways can you arrange the deck? You have 52 choices for
the first card, 51 for the second, 50 for the third, and so on. So
that works out to 

  52! = 52 * 51 * 50 * ... * 3 * 2 * 1

      = about 10^67

Which is still a lot less than a googol. But also, it's a big number
to come from something as small as a deck of cards. And in fact, if
we add more cards, so that we have at least 70, the number of possible
arrangements is greater than a googol.  

So in fact, to 'hold a google in your hand', all you'd have to do is
write the numbers from 1 to 70 on individual index cards, and hold
that 'deck' in your hand... if what you're counting is possible ways
to arrange the deck.  

Does this help? 

- Doctor Ian, The Math Forum
  http://mathforum.org/dr.math/