What is 9 9 9

What is the greatest number
which can be written with three digits?

1. First attempt at a solution

Naturally 999 or 9·9·9 = 729 or 9·99 = 891
but all numbers are less than a thousand.
It can be much bigger!

2. Mathematical tools:

Potencies using the example of Powers of two: (see also hyperpotencies)
8 = 2·2·2 = 23     16 = 2·2·2·2 = 24 and the Powers of ten:
1000 = 10·10·10 = 103     100 = 10·10 = 102Powers of ten and comma shift:
1,23456·1000 = 1 234,56     1,23456·104 = 12 345,6 scientific / technical notation / Floating point number:
100 = 1 (strange but understandable)     103 = 1 000 = thousand
106 = 1 000 000 = million     109 = 1 000 000 000 = billion
1012 = 1 000 000 000 000 = trillion *)     1015 = 1 000 000 000 000 000  =  Billiards
1018 = 1 000 000 000 000 000 000 = Trillion     1021 = 1 000 000 000 000 000 000 000 = Trillion
and so on*) note the difference between the USA and continental Europe !!

3. New attempted solution (completely "without" arithmetic symbols, only the three digits nine):

little big:
999

very large:

999=99·99·99·99·99·99·99·99·99=9,1352� ·1017
9 times the factor 99Comma by 17 places
To the right!

much bigger:

999=9·9·9·9·9·9·9·�·9·9·9·9·9·9=2,9513� ·1094
99 times the factor 994 places
To the right!

Until then, the calculator can easily calculate, but it can be even bigger:

"lousy big"(as we shall see):
999 ((also written as hyperpotency 9 ^ 9 ^ 9 = 9 ^^ 3)) auxiliary number: 99 = 3,8742� ·108
However, the order in which the exponentiation should take place must still be determined here
(Extension of the "right of way"):

"Right of way" like that?

(99)9 = (99)·(99)·(99)·(99)·...·(99)
= 99·9 = 981
= (3,8742...·108)9 = 3,8742...9·(108) 9
= 1,9663...·105·1072 = 1,9663...·1077
So not particularly big, see above.

or so?

9(99) = 93,8742� ·108
That's over 387 million nine factors.
If there were so many tens factors, if the number had over 387 million digits, it would only have over 369 million digits, as we can unfortunately only calculate in class 10 (logarithms!). And that is very souchmmoderate GHorse!! So let's call this number SMG.

 

How big is SMG practical

1. If a person had to write down all digits of this number SMG, how long would he be busy?

Calculate normal "working years".

20 digits per minute, 1,200 digits per hour, 9,600 digits per working day (8 hours), 10,000 digits rounded up.
Per working year (365 - 104 - 30 - 13 = 218) with approx. 220 working days, this results in 2,200,000 digits, i.e. approx. 2 million digits (= 2 · 106) per working year!
So there are over (369 x 106)/(2·106) Years = 184.5 years.

2. How high would the stack of A4 paper be if the number SMG were printed using a laser printer?

60 lines of 80 digits per A4 page, i.e. 4,800 digits, rounded to 5,000 digits.
So 10,000 digits per sheet = 104 Digits / leaf.
10 sheets are approx. 1 mm thick, i.e. 10 sheets / mm = 10,000 sheets / 1,000 mm = 104 Leaves / m = 104·104 Digits / m = 108 Digits / m = 100 x 106 Digits / m = 100 million digits / m. So the pile is 3.69 m high.

3. How big is SMG really

So far we have only dealt with the digitsnumber played by SMG. And SMG itself?
Let us consider the stretch of length SMG meter.

Compare with distance earth-sun, diameter of the Milky Way, diameter of the universe!

distance Earth - moon approx. 380,000 km = 380 · 103·103 m = 380 x 106 times almost 400 million m
The distance SMG m would then be the (SMG m / 380 · 106 m) times the earth-moon distance.
However, this number is only approx. 8 digits shorter than SMG, i.e. still over 369 million digits long!

distance Earth - sun approx. 150 million km:
that's 150 · 106·103 m = 150 x 109 m = 150 billion m (just!).

Diameter of Milky Way (our galaxy) approx. 100,000 light years = 105 Light years.

      Speed ​​of Light: 300,000 km / s = 3 · 105 km / s = 3 10560 60 24 365 km / year = 9.4 ... 1012 km / year,
so it applies 1 light year = 9.4 ... 1012 kmi.e. almost 10 quadrillion m.

Thus the diameter of the Milky Way: 105·9,4·1012 = 9,4·1017 km = 9.4 x 1017·103 m = 9.4 x 1020 m,
This is almost 1 trilliard m.

Diameter of Universe: almost 100 billion light years = 1011 Light years = 1011·9,4·1012·103 m =
9,4·1026 times almost 1 quadrillion m.

How many universes (next to each other) do you then need for the whole route ???

 

From a cosmological point of view, the meter is a quite arbitrary unit. Let us therefore rather consider the length of a chain of SMG atoms placed one behind the other. An atom is on average about 10-10 m tall (with atomic shell), the core (i.e. without shell) is approx. 10-14 m tall (the space in between is empty).

It is just stupid that there are not enough atoms in the entire universe for our chain, because:

The universe consists of only 1079 Particle (Einstein / Eddington) and is therefore pretty empty.


Dietrich Tilp 2007 (2018)