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### Hard Powers

```
Date: 12/11/97 at 20:41:58
From: Justin Pearce
Subject: POWERS, HARD POWERS

There is a challenge in our math class. There will be one hundred
dollars to the person who can do this problem. It is due by
March 20, 1998:

nine to the ninth to the ninth power

9
9
9

It has to be written out on paper.

Thank you.
```

```
Date: 12/12/97 at 18:06:28
From: Doctor Tom
Subject: Re: POWERS, HARD POWERS

Hi Justin,

Well, 9^9 = 387420489, so 9^(9^9) will have about 360 million digits.
If you can get 5000 to a page, you're talking about approximately 8000
pages of digits.

-Doctor Tom,  The Math Forum
Check out our web site!  http://mathforum.org/dr.math/
```

```
Date: 01/05/98 at 14:01:35
From: Doctor Sonya
Subject: Re: POWERS, HARD POWERS

Dear Justin,

I hope you have a lot of free time, and write really fast, since this
number has about 387,000,000 digits. Even if you wrote 10 digits every
second, it would take you over a year just to write the number down,
so to do it by 4 months from now, you would have to write about 30 per
second. Clearly, this is unrealistic. In fact, you would *lose* money
on the bet!

If you assume about 2000 digits to a page of ordinary notebook paper,
it would take you about 200,000 sheets of paper to write the number
down. At \$6 for a ream (480 sheets) of paper, the paper alone would
cost about \$2400, and you would only get \$100 as a prize. Sounds like
a ripoff to me!

The reason that this is such a big number has to do with a certain
convention that you might not be aware of.  When you write two
exponents, as you did above or as in a^b^c, this is (by definition)
equal to a taken to the power (b^c). If, on the other hand, you want
to write down what (a^b) taken to the power c is, you *have* to use
parentheses and write it as (a^b)^c. This, of course, is just a^(bc),
if you use the laws of exponents.

So 9^9^9 (the number you wrote down) is equal to 9 to the power 9^9.
Since 9^9 is about 387,000,000 (it's exactly equal to 387,420,489),
that means that the number you wrote is equal to approximately
9^(387000000).  Since 9 is about 10, that means that this number is on
the order of 10^(387000000), which is a 1 followed by 387000000 zeros.
(If you were to work it out exactly, you would find that it has about
369 million digits; since you are 14, that would put you in about 9th
grade, I am guessing, so you probably don't know about logarithms yet;
you'll get that in precalculus, and once you do, you will be able to
figure out about how many digits the number has by using logarithms
and their properties. That's how I did it: I certainly did
*not* actually multiply it out).

Incidentally, there are a few other things that you can find out about
this number, but it takes some fairly advanced math to do so. For
instance, it is possible to show that the last two digits of this
humungous number are 89!

This is a much bigger number than the number (9^9) taken to the 9th
power, i.e., (387420489)^9, which is the same thing as 9^(9*9) = 9^81.
This is "only" approximately a 1 followed by 81 zeros (if you work it
out exactly, it starts out 19, then has 75 other digits, and ends with
a 9).  That is a number you *could* write down by March 20th.

I think that whoever devised this problem is either playing a trick on
you, or is confused about what he or she is asking.  I would check to
see what it is that is meant.

-Doctor Mark,  The Math Forum
Check out our web site!  http://mathforum.org/dr.math/
```
Associated Topics:
Middle School Exponents

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