Four 2's Problem

[Anonymous No.3326222]

Hey /sci/, I have a game for you. The point of the game is to represent as many numbers as possible using four 2's and the operations of +,-,x,/, exponentiation, root extraction and factorials. So for example:
1= (2+2)/(2+2)
2=(2/2)+(2/2)
etc...
I've gotten up to 26 with the exception of 7,17 and 19. How well can you do?

[Anonymous No.3326236]

3=(2+2+2)/2
4=2+2+2-2

[Anonymous No.3326253]

3=2*2-(2/2)
4=2+2+2-2
5=2+2+(2/2)
6=2*2*2-2
8=2+2+2+2

[Anonymous No.3326254]

>>3326251
You can't use any numbers besides 2, so the 1 doesn't work.

[Anonymous No.3326260]

9=22/2-2

[Anonymous No.3326266]

10=(2*sqrt(2)) ^2 + 2

[Anonymous No.3326268]

11=[(sqrt(22))^2]/2

[Anonymous No.3326275]

12=(22+2)/2
13=[(2x2)!+2]/2

[Anonymous No.3326278]

12=(22+2)/2

[Anonymous No.3326281]

14=2^(2^2)-2

[Anonymous No.3326282]

you can't use 22 you dumb faggots

lrn2base10notfundamentalffs

[Anonymous No.3326293]

>>3326282
Should have told us
you dumb faggot

[Anonymous No.3326297]

OP here, 22 is fine. If you don't like it go play another game you number fascist.

[EK !EKFQOBUFnQ No.3326299]

>I've gotten up to 26 with the exception of 7,17 and 19.
how did you do 25?

[Anonymous No.3326302]

<span class="math">7 = 2\cdot2\cdot2 - \lim_{n \to \infty }\sqrt[n]{2}[/spoiler]

not sure if it's cheating, since i'm not using extra numbers

[Anonymous No.3326307]

>>3326299
25=(2x2)!+(2/2)

[Josef !!nUf2NflSAyw No.3326308]

>>3326302
>and the operations of +,-,x,/, exponentiation, root extraction and factorials
I guess the limit counts as an operation.

[Anonymous No.3326309]

>>3326302
You just made my day, that is awesome :D

[Anonymous No.3326314]

>>3326297


7 = 22 - 2/2 in base 3
17 = 22 - 2/2 in base 8

this thread is now lame.

[Anonymous No.3326320]

>>3326314
You can't say base 3 or 8 since they aren't 2's :P

[Anonymous No.3326323]

>>3326308
true, but there's no limit to the amount of operations

I could've done an infinite number of <span class="math">\sqrt\sqrt\sqrt\ldots[/spoiler] I guess

[Anonymous No.3326326]

>>3326323
argh, I meant ... \sqrt\sqrt\sqrt ...

[Anonymous No.3326330]

17=((2?)!)?-2-2
>face it

[Anonymous No.3326339]

>>3326330
oops

[Anonymous No.3326340]

>>3326320
>missing the point

you can't say base 10 either, which you are doing tacitly.

[Josef !!nUf2NflSAyw No.3326347]

>>3326323 there's no limit to the amount of operations
Then I've found a way of representing
2601218943565795100204903227081043611191521875016945785727541837850835631156947382240678577958130457
0826199205758922472595366415651620520158737919845877408325291052446903888118841237643411919510455053
4665861624327194019711390984553672727853709934562985558671936977407000370043078375899742067678401696
7207846280629229032107161669867260548988445514257193985499448939594496064045132362140265986193073249
3697704776060676806701764916694030348199618814556251955925669188308255149429475965372748456246288242
3452659778973774089646655399243592878621251596748322097602950569669992728467056374713753301924831358
7076125412683415860129447566011455420749589952563543068288634631084965650682771552996256790845235702
5521862223581300167008345234432368219357931847019565107297818043541738905607274280485839959197290217
2661229129842051606757903623233769945396419147517556755769539223380305682530859997744167578435281591
3461340394604901269542028838347101363733824484506660093348484440711931292537694657354337375724772230
1815340326471775319845373414786743270484579837866187032574059389242157096959946305575210632032634932
0922073832092335630992326750440170176057202601082928804233560664308988871029738079757801305604957634
2838683057190662205291174822510536697756603029574043387983471518552602805333866357139101046336419769
0973974322859942198370469791099563033896046758898657957111765666700391567481531159439800436253993997
3120306649060132531130471902889849185620376666916446879112524919375442584589500031156168297430464114
2538074897281723375955380661719801404677935614793635266265683339509760000000000000000000000000000000
0000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000
00000000000000000000000000000000000000000000000
(It's 3!!!)

[Anonymous No.3326360]

>>3326330
>>3326339
19=((2?)!)?-2-2+2

[Anonymous No.3326368]

>>3326323
in my opinion, you can't do an infinite number of square roots, since that's just the power of a half

[Anonymous No.3326369]

>The point of the game is to represent as many numbers as possible using four 2's and the operations of +,-,x,/, exponentiation, root extraction, factorials and concatenation while using only base 10.
There, happy?

[Anonymous No.3326388]

13 = log_2 ( ( (2^(2+2/2))*((2^10)-(2*((2^(2+2)-(2^2)-2))+2^2) ) ) + [ (2^(2^2+2+(2/2)))-(2^(2^2+(2/2)))+(2^2) ] + ( (2^(2^2+2+(2/2)))-(2^(2^2+(2/2)))-(2+2) ) )

check it on wolframalpha if you don't believe me

[Josef !!nUf2NflSAyw No.3326413]

>>3326388
I'm not good at counting, but I think you used more than 4 "2"s there.

[Anonymous No.3326425]

7 = (2^2+2^2) - (2/2) ?

[Anonymous No.3326434]

>>3326413
nope but i forgot do disolve 2^10 to twos :|

<-- here to correct my mistake

[Anonymous No.3326436]

>>3326425

Shit, wait. The four 2s includes the exponents doesn't it?

[Anonymous No.3326455]

7=(2?)*2+2/2

[Anonymous No.3326456]

<div class="math">7=\; !(2+2) - \sqrt{2^2}</div>

[Anonymous No.3326463]

>>3326456

fullretard.jpg

[Anonymous No.3326478]

>>3326463
What?
!(2+2)-sqrt(2^2)=!4-sqrt(4)=9-2=7

[Anonymous No.3326486]

1 = (2+2)/(2+2)
2 = 2
3 = 2 + 2/2
4 = 2+2
5 = (2+2) + 2/2
6 = 2+2+2
7 = 2+2+2 + 2/2
8 = 2+2+2+2
9 = 2+2+2+2 + 2/2
And so on

[Anonymous No.3326490]

>>3326468
nope
You HAVE to use FOUR "2"

[Anonymous No.3326496]

>>3326486

I did not realize it was just four 2's. Sorry.

[Anonymous No.3326581]

-62 = <span class="math">2-\sqrt{\sqrt{(2^{(2+2)!})}}[/spoiler]
-46 = 2-((2+2)!*2)
-44 = (2-(2+2)!)*2
-42 = 2-(22*2)
-40 = (2-22)*2
-24 = 2-((2+2)!+2)
-23 = (2-2)!-(2+2)!
-22 = <span class="math">2-(\sqrt{(2+2)}+2)![/spoiler]
-21 = (2-2)!-22
-20 = 2-((2+2)!-2)
-18 = 2-(22-2)
-14 = <span class="math">2-((2+2)^{2})[/spoiler]
-11 = (2-(2+2)!)/2
-10 = 2-((2+2)!/2)
-9 = 2-(22/2)
-6 = 2-((2+2)*2)
-4 = 2-((2+2)+2)
-3 = ((2-2)!-2)-2
-2 = <span class="math">2-(\sqrt{(2+2)}+2)[/spoiler]
-1 = <span class="math">(\sqrt{(2+2)}-2)!-2[/spoiler]
0 = <span class="math">\sqrt{(\sqrt{(2+2)}+2)}-2[/spoiler]
1 = <span class="math">(\sqrt{(\sqrt{(2+2)}+2)}-2)![/spoiler]
2 = <span class="math">\sqrt{(\sqrt{(\sqrt{(2+2)}+2)}+2)}[/spoiler]
3 = <span class="math">(\sqrt{(2+2)}-2)!+2[/spoiler]
4 = <span class="math">\sqrt{(\sqrt{(2+2)}+2)}+2[/spoiler]
5 = ((2-2)!+2)+2
6 = <span class="math">(\sqrt{(2+2)}+2)+2[/spoiler]
8 = <span class="math">(\sqrt{(2+2)}+2)*2[/spoiler]
9 = <span class="math">((2-2)!+2)^{2}[/spoiler]
10 = ((2+2)!/2)-2
11 = ((2+2)!-2)/2
12 = <span class="math">(\sqrt{(2+2)}+2)!/2[/spoiler]
13 = ((2+2)!+2)/2
14 = ((2+2)!/2)+2
16 = <span class="math">(\sqrt{(2+2)}+2)^{2}[/spoiler]
18 = <span class="math">((2+2)^{2})+2[/spoiler]
20 = ((2+2)!-2)-2
21 = 22-(2-2)!
22 = <span class="math">(\sqrt{(2+2)}+2)!-2[/spoiler]
23 = (2+2)!-(2-2)!
24 = <span class="math">(\sqrt{(\sqrt{(2+2)}+2)}+2)![/spoiler]
25 = (2-2)!+(2+2)!
26 = <span class="math">(\sqrt{(2+2)}+2)!+2[/spoiler]
28 = ((2+2)!+2)+2
32 = <span class="math">((2+2)^{2})*2[/spoiler]
36 = <span class="math">((2+2)+2)^{2}[/spoiler]
40 = (22-2)*2
42 = (22*2)-2
44 = ((2+2)!-2)*2
46 = ((2+2)!*2)-2
48 = <span class="math">(\sqrt{(2+2)}+2)!*2[/spoiler]
50 = ((2+2)!*2)+2
52 = ((2+2)!+2)*2
62 = <span class="math">\sqrt{\sqrt{(2^{(2+2)!})}}-2[/spoiler]
64 = <span class="math">((2+2)*2)^{2}[/spoiler]
66 = <span class="math">\sqrt{\sqrt{(2^{(2+2)!})}}+2[/spoiler]
88 = (22*2)*2
96 = ((2+2)!*2)*2

[Anonymous No.3326663]

okay, here some Mathematica code:

ops = {?????????};

Permutations@%;
Table[Take[#, n], {n, Length@ops}] & /@ %;
Flatten[%, 1];
Join[#, {"2", "2", "2", "2"}] & /@ %;
Permutations /@ %;
strings = Flatten[%, 1];

ToExpression[RowBox[#]] & /@ %;
sols = DeleteDuplicates@%;
depending on your computing power, you can find all possible solutions...

[Anonymous No.3326675]

>>3326663
here I use four 2's and ops = {"+", "+", "-", "/"};

max level would probably be
ops = {"+", "+","+", "+", "-","-","-","-", "/", "/", "/", "/","brackets"};
where "brackets" stands for lots and lots of brakets ( and ). However, generating the permutations of a string with 30 characters and acting on them might take a while

[Anonymous No.3326722]

>>3326302

k = 2 - 2 + 2/2
1 = k
2 = k + k
3 = k + k + k
etc

[Not Izzy No.3326737]

2/2/2/2
2/2/2*2
2/2/2-2
2/2/2+2
2/2*2/2
2/2*2*2
2/2*2-2
2/2*2+2
2/2-2/2
2/2-2*2
2/2-2-2
2/2-2+2
2/2+2/2
2/2+2*2
2/2+2-2
2/2+2+2
2*2/2/2
2*2/2*2
2*2/2-2
2*2/2+2
2*2/2-2
2*2*2*2
2*2*2-2
2*2*2+2
2*2-2/2
2*2-2*2
2*2-2-2
2*2-2+2
2*2+2/2
2*2+2*2
2*2+2-2
2*2+2+2
2-2/2/2
2-2/2*2
2-2/2-2
2-2/2+2
2-2*2/2
2-2*2*2
2-2*2-2
2-2*2+2
2-2-2/2
2-2-2*2
2-2-2-2
2-2+2/2
2+2-2*2
2-2+2-2
2-2+2+2
2+2/2/2
2+2/2*2
2+2/2-2
2+2/2+2
2+2*2/2
2+2*2*2
2+2*2-2
2+2*2+2
2+2-2/2
2+2-2*2
2+2-2-2
2+2-2+2
2+2+2/2
2+2+2*2
2+2+2-2
2+2+2+2

Happy op?

[Anonymous No.3326860]

15= (2+2+2)choose2