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07-28-2017 , 01:36 AM
Biggest number you can form with 1,2,3,4,5,6,7,8,9,0,+,-,*,/,(,),^,. used only once each (all must be used).

Can you beat that if you give up any number of digits for same number of latin characters that can form a particular well known function name like like log, sin, cos, tan, exp etc (name it, it ought to be seen often in calculators ) (they all come with automatic parentheses free) etc you see on a regular scientific calculator and/or any number of / symbols? (eg replace 4 7 8 with Tan or 4 7 8 and 0 with tan and /)

(natural units are assumed in any of these functions)


( i mean just to make the point that you can always keep busy when you have lost other options to have fun lol in some poor setting or situation)
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07-28-2017 , 02:57 AM
It'd take some programming.
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07-28-2017 , 03:25 AM
is this how john nash went insane
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07-28-2017 , 03:37 AM
Ok let me start.

Please advise rules improvement if you find a richer structure can emerge without making it too ridiculously big to explore in a competitive creative spirit.

1,2,3,4,5,6,7,8,9,0,+,-,*,/,(,),^,.


Spoiler:
8^(9*76543/.01+-2)

about 10^(6.22127*10^7)

Last edited by masque de Z; 07-28-2017 at 03:56 AM.
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07-28-2017 , 05:40 AM
Spoiler:
1,2,3,4,5,6,7,8,9,0,+,-,*,/,(,),^,.

sacrifice 234 for Exp()

6^(Exp(9*87/.01)+-5)


about 10^(1.40928*10^34005)
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07-28-2017 , 05:51 AM
Quote:
Originally Posted by masque de Z

Spoiler:
8^(9*76543/.01+-2)
Reading this inadvertently might spoil it for you.
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07-28-2017 , 08:19 AM
Quote:
Originally Posted by masque de Z
Ok let me start.

Please advise rules improvement if you find a richer structure can emerge without making it too ridiculously big to explore in a competitive creative spirit.

1,2,3,4,5,6,7,8,9,0,+,-,*,/,(,),^,.


Spoiler:
8^(9*76543/.01+-2)

about 10^(6.22127*10^7)
Spoiler:
Wouldn't 8^(9*76543/.01)+-2 be bigger?
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07-28-2017 , 02:03 PM
Quote:
Originally Posted by darksideofthewal
Spoiler:
Wouldn't 8^(9*76543/.01)+-2 be bigger?
Yes unless the+- is like the quadratic solution +-ie suggesting 2 solutions. This is why on the second one i keep 5 in and not in its place eg 2.

But if this interpretation is not allowed we can do what you said too.
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07-28-2017 , 02:38 PM
Here is another question too. How many different numbers can one represent with these symbols only used once each?
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07-29-2017 , 07:22 AM
Quote:
Originally Posted by darksideofthewal
Spoiler:
Wouldn't 8^(9*76543/.01)+-2 be bigger?
Could also do new record about 5.119847*10^(6.5823587*10^7) (come on guys nobody wants to challenge me?)

Spoiler:
7^(9*86543/.01)+-2 without the +- possible ambiguity rule.

That takes it higher to about 5.119847*10^(6.5823587*10^7)

smaller base for bigger exponent wins most times it looks close

If -base is allowed as well then -7^(9*86543/.01+2)

Last edited by masque de Z; 07-29-2017 at 07:45 AM.
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07-29-2017 , 07:18 PM
8^(9 * xxx/xxx etc) immediately came to mind without looking at the spoilers. Beyond that I lost interest because beer is more important than numbers.
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07-30-2017 , 12:58 AM
Quote:
Originally Posted by Zeno
8^(9 * xxx/xxx etc) immediately came to mind without looking at the spoilers. Beyond that I lost interest because beer is more important than numbers.
You should have stayed longer to it and see that 7^ is better lol.
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07-30-2017 , 01:14 AM
Seven is a holy number, so I discounted it immediately as the main prop for number bigness. Silly prejudges cause many to stumble about. It's best to let beer do the stumbling for you. Life is easier that way.

Fun number game nonetheless.
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08-01-2017 , 02:36 PM
Spoiler:


1^(987*64/.5) + 2 - 3



what do I win?
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08-02-2017 , 03:16 AM
Quote:
Originally Posted by Alobar
Spoiler:


1^(987*64/.5) + 2 - 3



what do I win?
You forgot the 0 but its fixable instantly, also we want the maximum possible number not the smallest lol. But its good to have it though thanks. The smallest in absolute value nonzero could also be a good target to obtain.
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08-02-2017 , 10:31 PM
This made me think of the video on Graham's number on numberphile. I still don't feel like I've read a good explanation of tree(3)
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10-16-2017 , 03:42 AM
New Record!

5.00182656311013843993950036687042132613972×10^(1. 488041806*10^9)

Spoiler:
According to rules above sacrifice 3 digits like 236 to get cos()

Now try 7^.8/(1+Cos(40*9-5))


or

[IMG]http://latex.codecogs.com/gif.latex?%5Clarge%207%5E%7B.8/%281+Cos%2840*9-5%29%29%7D[/IMG]

That takes it higher way higher to 7^(1.761338617*10^9) or

5.00182656311013843993950036687042132613972 × 10^(1.488041806*10^9)
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10-16-2017 , 06:01 AM
Basically it is

Spoiler:



Where i tried to avoid abusing the . notation as in not having 8. but .8
or using +-.

Still the Exp solution of #5 is bigger but this uses Cos as an alternative so its bigger in its subclass. Also it uses rads not deg.


Last edited by masque de Z; 10-16-2017 at 06:22 AM.
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10-16-2017 , 06:34 AM
The important little devil here is Cos(355)=-0.999999999545658980165935841692754081123824951499 9282448 which makes 1+cos(355)= 4.543410198340641583072459188761750485000717552 × 10^-10.

Now that is a tiny number for such low argument. It has the nasty little value that if you add it as a parameter in a little program very few people would know its such a tiny number and you can use it to do things that are not immediately obvious to others especially if lazy lol.
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