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### 7 to the 1997th power

```
Date: 09/22/97 at 19:50:19
From: Anonymous
Subject: 7 to the 1997th power

How would you find what the 1997th power of 7 is?  It is too long to
put into a calculator.  My eighth grade son got an assignment to
figure out how to solve this equation.
```

```
Date: 09/23/97 at 04:26:26
From: Doctor Pete
Subject: Re: 7 to the 1997th power

Hi,

This is a very odd question to be asked. What is particularly special
about the number 7^1997?  I mean, why not ask what 105^1824 is? It
seems to hold little more than computational value.

I will give you the answer in its exact form. I obtained it by using,
in some sense, a calculator. Well, it's a bit more than a calculator.
It's a program called Mathematica, and it can calculate 7^1997, the
output of which is as follows (with line ends edited out):

Mathematica 3.0 for Solaris
-- Terminal graphics initialized --

In[1]:= 7^1997

Out[1]=

4579160949324087781565673146964641788197506154427481192851907952905019
4309956678395618077471248816909058605574226922612745779509156422105174
7600583919391729526644345966226344259185049512967576291988226580624629
1843143456835498528676559273860051787179476933829484142291803617035612
6372777247241850209351367497982999720860032107213867094079629986736519
8993905459707731623498698573906582035339795136884660098391637477970541
6028843163944040152172111118034048132298313338292594600479696975285690
4544411346085768095457192690072141529116398867015226227667212049466818
5082497451161258972874145856549722452219994292255306630032528073780047
1210666853757217975371039212655984122663483085162332165508117406781727
6482908800183698254500589125167356311226611489244765420861990235649801
0209426132183060436658284967833476974672361803979497730641064739461863
2808348356151668906504852640419389710988650060250936971148671024259770
5944077431085131446729016058265169888115166006316404607609004430719890
9996501464323306565952337310389858655542683659916847355149417697796515
1855338376464646558794955994538160511551527430915749654778913079054722
9126175038660583749586658392484253586283194595185854503295401282550601
8487314672713399676997062146419517527790777603087186409605616819518409
9215287090022011687966750862324137649698220569410799184511727396835772
4366353735346744918365062236491512334061437233827129199387050210285065
3380179824668474792586657024308102265351693956871226571142787671033729
5252315411756476813934785393353151743338954089020017075820757600139251
4761000057317958126219097256108158763171605635331039991494138636068095
6294995946109664921648023520312146267691829406083389145107355160301802
40703207

This is an exact answer, I hope you will take my word for it, because
it looks like a string of random digits to me, too. There are 1688
digits in this number. How did a program compute this value? It uses a
concept called "arbitrary-precision computation."  The basic idea is
that it does what we do with pencil and paper, only it has a *lot*
more scratch paper to work on, and it's *much* faster. You see, most
hand-held calculators have a very limited amount of memory, and the
number of precise digits they can give is determined by their memory
and computing speed (most calculators are *very* slow). However, a
personal computer running a symbolic algebra program such as
Mathematica or Maple has the memory, speed, and algorithms to do
lengthy calculations, often with perfect accuracy.

This is not to say that computers and programs are the mathematician's
salvation. Sure, they do things in milliseconds that would take a
human decades or centuries, but it is the theory behind such
computations that is of interest, not the mere action of calculating.
For instance, Mathematica can give the first 10,000 decimal digits of
pi in a second but it can't tell us if the digits of pi are uniformly
and randomly distributed; that is, each digit from 0 to 9 occurs
statistically 1/10th of the time and in total randomness. This is
something that only mathematicians can prove or disprove, and whether
you can calculate 10,000 digits or 8 trillion, you don't get any

In any case, if you had nothing but a pencil and a lot of paper, you'd
go about finding 7^1997 as follows: Find 7^2, then (7^2)^2 = 7^4, then
((7^2)^2)^2 = (7^4)^2 = 7^8, etc. until you reach 7^1024.  Then you
take 7^1024 and multiply it by 7^512, which you had found just one
step before, to obtain 7^1536. Then multiply by 7^256, 7^128, 7^64, 7^
8, 7^4, and finally 7.  In other words,

7^1997 = (7^1024)(7^512)(7^256)(7^128)(7^64)(7^8)(7^4)(7^1).

This cuts down (a little bit) on the number of multiplications you
need to do, but time-wise, it doesn't really help too much. You'll
notice that the binary representation of 1997 is 11111001101, which
corresponds to the powers of two I have listed above. I leave it as an
exercise to show if you can further reduce the number of
multiplications, and if not, why.

Hope this helps,

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

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