Value of 9^(9^9)?

Date: 10/21/2002 at 20:59:03
From: Stephen W.
Subject: Value of 9^(9^9)?

A kid in my class came in one day and asked my teacher, "What is the 
largest possible number that can be made with 3 digits?" I answered 
9 to the 9 to the 9. I was correct. 

Today the teacher asked me to figure out 9 to the 9 to the 9. I'm 
going to tell her tomorrow that it is 1966270505 with 68 0's after it. 
(My dad's calculator says it is approximately 1.9662705065 with 68 0's 
after it.) 

Do you know what the answer is? I was going to try to do it by hand 
but my dad said I shouldn't because it would be a ton of writing.

All I did to get the following number is add 68 0's to 1966270505. 
Is that the real anwer or just approximate? I think the answer is 
approximately 

19,662,705,050,000,000,000,000,000,000,000,000,000,000,000,000,000,
000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000.

What is the exact number? I know it is HUGE!
Thanks,
Stephen W.

Date: 10/21/2002 at 23:21:35
From: Doctor Peterson
Subject: Re: Value of 9^(9^9)?

Hi, Stephen.

Your answer is actually much too small. You seem to have found (9^9)^9 
rather than 9^(9^9), which is much larger: it has 369693100 digits! 
Your number is only 9^81, while this is 9^387420489.

The number you got is only an approximation to (9^9)^9, since you have 
put zeroes in many digits whose value you don't really know. There are 
calculators that could give the exact value of that number (and in 
fact it would not be terribly hard to do by hand, if you know some 
tricks); but the number you really want would take thousands of pages 
to print out. In any case, with huge numbers like this, the important 
thing is usually their size, not knowing every digit, so an 
approximation is good enough.

We have answered questions about this number before, which you can 
find by searching our archives for "9^9^9". Here is one:

   9^9^9
   http://mathforum.org/library/drmath/view/59172.html 

Here is another:

   Hard Powers
   http://mathforum.org/library/drmath/view/58235.html 

If you have any further questions, feel free to write back.

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

Date: 10/22/2002 at 00:06:58
From: Doctor Jeremiah
Subject: Re: Value of 9^(9^9)?

Hi Stephen,

It is definately approximate.  I know this without doing any 
calculations because something that ends in zero is divisible by 10 
and 10 is 2 times 5. But 9^9^9 is the same as 9 times itself a bunch 
of times and 9 is not divisible by 2 or 5 so the answer can't be 
divisible by 10, which means it can't end with zero.

Now, how do you calculate 9^9^9?

Well it depends. Do you mean (9^9)^9, which is 387420489^9, or do you 
mean 9^(9^9), which is 9^387420489 (much bigger)?

This is one way of looking at it:

(9^9)^9 = 9^(9*9)
(9^9)^9 = 9^81
(9^9)^9 = 1966270504.7555291361807590852... x 10^68

9^(9^9) = 9^387420489
9^(9^9) = 9^(9*43046721)
9^(9^9) = (9^9)^43046721
9^(9^9) = (9^9)^(9*4782969)
9^(9^9) = ((9^9)^9)^4782969
9^(9^9) = (1966270504.7555291361807590852... x 10^68)^4782969

And here is another:

(9^9)^9 =        (9^9)^9
9^(9^9) = ((((((((9^9)^9)^9)^9)^9)^9)^9)^9)^9

Clearly I cannot tell you what 9^(9^9) is because 
(1966270504.7555291361807590852... x 10^68)^4782969 is an absolutely 
huge number!

But this is the value of (9^9)^9:
 
196,627,050,475,552,913,618,075,908,526,912,116,283,103,450,944,214,76
6,927,315,415,537,966,391,196,809

I used an "arbitrary precision" calculator for this. If you need to do 
this often search the Internet for "arbitrary precision" to find a 
calculator that does this.

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