# 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 = 2

^{3}16 = 2·2·2·2 = 2

^{4}and the

**Powers of ten**:

1000 = 10·10·10 = 10

^{3}100 = 10·10 = 10

^{2}

**Powers of ten and comma shift**:

1,23456·1000 = 1 234,56 1,23456·10

^{4}= 12 345,6

**scientific / technical notation**/

**Floating point number**:

10^{0} | = | 1 (strange but understandable) | 10^{3} | = | 1 000 | = | thousand | |||

10^{6} | = | 1 000 000 | = | million | 10^{9} | = | 1 000 000 000 | = | billion | |

10^{12} | = | 1 000 000 000 000 | = | trillion ^{*}) | 10^{15} | = | 1 000 000 000 000 000 | = | Billiards | |

10^{18} | = | 1 000 000 000 000 000 000 | = | Trillion | 10^{21} | = | 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**:

99^{9} | = | 99·99·99·99·99·99·99·99·99 | = | 9,1352� ·10^{17} |

9 times the factor 99 | Comma by 17 places To the right! |

**much bigger**:

9^{99} | = | 9·9·9·9·9·9·9·�·9·9·9·9·9·9 | = | 2,9513� ·10^{94} |

99 times the factor 9 | 94 places To the right! |

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

"**lousy big**"(as we shall see):**9 ^{99}** ((also written as hyperpotency 9 ^ 9 ^ 9 = 9 ^^ 3)) auxiliary number: 9

^{9}= 3,8742� ·10

^{8}

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?
^{9})·(9^{9})·(9^{9})·(9^{9})·...·(9^{9}) = 9 ^{9·9} = 9^{81}= (3,8742...·10 ^{8})^{9} = 3,8742...^{9}·(10^{8}) ^{9}= 1,9663...·10 ^{5}·10^{72} = 1,9663...·10^{77}So not particularly big, see above. | or so?
^{3,8742� ·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 · 10^{6}) per working year!

So there are over (369 x 10^{6})/(2·10^{6}) 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 = 10

^{4}Digits / leaf.

10 sheets are approx. 1 mm thick, i.e. 10 sheets / mm = 10,000 sheets / 1,000 mm = 10

^{4}Leaves / m = 10

^{4}·10

^{4}Digits / m = 10

^{8}Digits / m = 100 x 10

^{6}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 digits*number*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 · 10^{3}·10^{3} m = 380 x 10^{6} times *almost 400 million m*

The distance *SMG m* would then be the (SMG m / 380 · 10^{6} 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 · 10^{6}·10^{3} m = 150 x 10^{9} m = ** 150 billion m** (just!).

Diameter of **Milky Way** (our galaxy) approx. 100,000 light years = 10^{5} Light years.

**Speed of Light**: 300,000 km / s = 3 · 10^{5} km / s = 3 10^{5}60 60 24 365 km / year = 9.4 ... 10^{12} km / year,

so it applies **1 light year = 9.4 ... 10 ^{12} km**i.e.

**.**

*almost 10 quadrillion m* Thus the diameter of the Milky Way: 10^{5}·9,4·10^{12} = 9,4·10^{17} km = 9.4 x 10^{17}·10^{3} m = 9.4 x 10^{20} m,

This is *almost 1 trilliard m*.

Diameter of **Universe**: almost 100 billion light years = 10^{11} Light years = 10^{11}·9,4·10^{12}·10^{3} m =

9,4·10^{26} 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 10^{79} Particle (Einstein / Eddington) and is therefore pretty empty.

Dietrich Tilp 2007 (2018)

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