What is the largest number you can create with 0000 by moving only two sticks?

Question

Here's what I'm saying

Moving 2 sticks, what is the largest number that you can create?

Some answers use the argument that 1/0=infinity. Unless you use the fact that 1/x --> INF when x --> 0, that's not true. So... Do you consider that a valid answer? — Barranka, Sep 02 '16 at 13:39
and if you accept $1/0$ as ∞, then you have to accept it to be -∞ as well, which is, well, a very small number — Radovan Garabík, Sep 02 '16 at 15:39
Hello @bleh, did you make this puzzle yourself? Can I add it to my website? Of course, I will post it under your name - bleh (StackExchange). — Puzzle Prime, Sep 03 '16 at 13:45
@ArturKirkoryan Sure! I made it myself, inspired by other problems. You can use it freely. — bleh, Sep 03 '16 at 14:02
At first make 11000, when stand two sticks, 11000・・. This is largest number. — Takahiro Waki, Sep 03 '16 at 14:05
Move the top and bottom sticks from the first 1 to make 111000, then interpret it in base Graham's number. — user253751, Sep 04 '16 at 12:23
Sad that googol is not written in English as it is in French (gogol)... — Puck, Sep 05 '16 at 12:53

greenturtle3141 · Accepted Answer · 2016-09-02T02:09:14.567

97

How about...

Take the top and bottom sticks of the first zero, and place them upright on the bottom left, making $11^{11000}$?

If that's unrealistic...

... because the base is usually in bigger font, then $11000^{11}$ could work as well.

edited Sep 02 '16 at 02:09

answered Sep 02 '16 at 02:00

greenturtle3141

9,937
1
35
70

2

I think that's moving four sticks, not two. – EKons Sep 02 '16 at 06:46
35

@ΈρικΚωνσταντόπουλος - no, it's moving two sticks; losing the top and bottom stick of the first 0 makes it an 11, and those two sticks are what gets moved to make the second, smaller 11 (of one stick each). – Megha Sep 02 '16 at 06:50
That means one stick makes 1 and two sticks also make 1. I think to make 1 number, it should be made by 2 sticks – Neo Sep 03 '16 at 04:00
7

I accounted for that. Exponents are in smaller font, so it's more aesthetically pleasing if you end up putting the tiny "11" on the top right instead. – greenturtle3141 Sep 03 '16 at 04:07
@Megha thanks you made it crystal clear! ;) – Aneek Sep 03 '16 at 13:33

score 95 · Answer 2 · answered Sep 02 '16 at 06:20

95

How about:

As in,

g₉₀₀, the 900th term of the series used to generate Graham's number.

answered Sep 02 '16 at 06:20

MooseBoys

984
5
11

11

Not gonna get much bigger than that – tupto Sep 02 '16 at 12:41
27

This one is unimaginably larger than the top voted and accepted answer of 11^11000. Even just trying to write g900 using only digits and exponents would not be possible to fit on a piece of paper that filled the visible universe. – Shufflepants Sep 02 '16 at 15:32
2

And your answer inspired me to attempt to outdo you using the Busy Beaver function. See my answer below :) – Shufflepants Sep 02 '16 at 16:04
The reason I didn't accept is because it was more of a function, not a value – bleh Sep 03 '16 at 00:21
2

You could even have made $g_{17130}$. – Jonathan Allan Sep 03 '16 at 11:29
3

@bleh g900 is definitely a value, in the same way Pi or 11^11000 are values, it's just so large that it can only be expressed through recursion. – MooseBoys Sep 03 '16 at 21:36
No, I dont think so mainly because you have to use a function to get to it. $11^11000$ is not a function-user because it only uses an operation instead of a function like G – bleh Sep 03 '16 at 21:59
Excellent idea, but mathematical notation is very case-sensitive: one wouldn’t ever write “G900” to mean g(900). 8bittree’s answer below fixes this quibble. – Peter LeFanu Lumsdaine Sep 03 '16 at 22:22
@bleh You seem to differentiate between operations and functions. What is the difference? – JiK Sep 05 '16 at 12:09
It was an inferred limitation. :P – bleh Sep 05 '16 at 13:17
1

While I admit it could have been clearer that functions are not allowed, I think it's silly to not have understood that to begin with. Otherwise I'm going to declare a function called the A&%-hole function (A) that is simply the largest number on this page +1. Then all I have to do is move one stick on the first 0 to make it A000 (oh yeah, the function ignores any trailing numbers) and I will always win forever. – thanby Sep 06 '16 at 14:10
If this is a function, so is anything involving an exponent. It evaluates to a number, it's just a number so big you'd have to write it in this form because there isn't enough space in the visible universe to write down all its numerals. – Shufflepants Sep 06 '16 at 14:27
2

@bleh How is $g$ a function? $g_{900}$ is a term in an existing series, just like how in some existing arbitrary series $a={2, 4, 6, 8, 10}$, $a_2$ is $4$. You don't necessarily need to use a function to determine $a_2$'s value, nor do you need one to determine that of $g_{900}$ (though computing that is an entirely different matter). – TNT Sep 07 '16 at 03:06
@bleh as TNT mentions, there is no g function. What there is, though, is a sequence, where g64 is Graham's number. – Simply Beautiful Art Jul 20 '17 at 22:10

score 35 · Answer 3 · answered Sep 02 '16 at 16:03

35

Move the top and bottom matchsticks from the 3rd zero to make:

Where the first two characters are B's thus: BB110 And BB refers to the Busy Beaver Function The Busy Beaver function is non-computable and therefor grows faster than any computable function such as exponentiation or the series to create Grahams Number.

The first few entries:

BB2 = 6 
BB3 = 21
BB4 = 107   
BB5 ≥ 47,176,870
BB6 > 7.4 × 10^36534 which is already greater than greenturtle3141's answer
BB12 > g1

which is already close enough to catching up to Graham's Sequence to pretty safely say that BB110 > g900 considering BB continues to ramp up faster than any computable function.

answered Sep 02 '16 at 16:03

Shufflepants

691
4
8

1

Radovan Gabarík had the same idea earlier, but the funny thing is, his answer is ridiculously small compared to this one. – JiK Sep 02 '16 at 16:46
Still a function.... – bleh Sep 03 '16 at 04:09
You could be pedantic and define the Busy Beaver function to be B. In that case, your answer becomes B(B(110)), which is just silly. – Joel Harmon Sep 03 '16 at 04:41
This gets my vote (you could up it to $BB1717$ though). @JoelHarmon you have no sticks with which to do so, whereas $BB$ is already defined. – Jonathan Allan Sep 03 '16 at 11:36
@JonathanAllan Could you clarify what you mean when you say I'm out of sticks? I've seen Busy Beaver abbreviated as both $BB$ and $B$. And while I agree that $BB$ is already defined, I'm much less certain it's uniquely defined. If it was, then the respondent wouldn't have needed to clarify that Busy Beaver was intended. – Joel Harmon Sep 03 '16 at 12:38
@JoelHarmon I think BB is common, but B would need to be defined, so you'd need sticks to define B (and that the answer only needs to say what BB is because many may never have heard of it before). – Jonathan Allan Sep 03 '16 at 13:17
@bleh You asked for the largest number; not the largest numeral. This is a perfectly valid answer imho. – noisypixy Sep 04 '16 at 22:33
4

Also it's 88110, not BB110 :{ – bleh Sep 05 '16 at 13:17
2

@bleh If you want to be pedantic about how characters are represented, none of these answers have any numbers in them, just a bunch of squares and lines made by matchsticks. Clearly it's not 88110 either since clearly '8' is two circles, not two squares. – Shufflepants Sep 06 '16 at 14:31
But then, I guess there are no letters in this image http://www.romanblack.com/led_no/led_no09.jpg – Shufflepants Sep 06 '16 at 14:36
Isn't the Busy Beaver function symbol "Σ" rather than "BB" anyway? – oleslaw Oct 19 '16 at 13:18
1

@oleslaw Many things have many representations. Sometimes multiplication is *, or a dot, or x, or sometimes entirely implicit. Here's an article written by a top quantum computer complexity theorist where he uses BB for the busy beaver function. http://www.scottaaronson.com/writings/bignumbers.html – Shufflepants Oct 19 '16 at 13:47
A function related to Busy Beaver is the maximum shifts function S http://en.wikipedia.org/wiki/Busy_beaver#Maximum_shifts_function_S . S800 is larger than BB110 and avoids the B's looking like 8's (slant the bar down slightly to avoid it looking like a 5). – Gnubie Aug 04 '20 at 13:14

score 34 · Answer 4 · edited Sep 02 '16 at 02:38

34

Move the top and bottom stick of the first zero to create another 1, thus making 111000

edited Sep 02 '16 at 02:38

JMP

35,612
7
78
151

answered Sep 02 '16 at 01:56

BlueFire

1,073
10
13

Ugh.... too quick – bleh Sep 02 '16 at 01:58
7

And I was happy with my 9800... – Chaotic Sep 02 '16 at 02:29
6

@Chaotic I did 9900. – EKons Sep 02 '16 at 06:52
And I thought I was clever with my E800. – Amit Naidu Sep 03 '16 at 03:24

Jonathan Allan · Answer 5 · 2016-09-02T03:57:11.537

22

How about

take the two bottom sticks from the leftmost zeros, break the end off of one and place it at the right hand side and place the other at the left hand side to make: $1171700!$ (or if the small 1 on the left is invalid make $771700!$

That's $factorial(1171700)$ (or $factorial(771700)$)

...and given that $100000!=2.824229408×10^{456573}$ both of the above examples are pretty big.

For a less lateral-thinking solution

- without breaking sticks, that is - one can make $7713170$ ($778170$ edit, credit @PaulGriffiths):

 _  _  _  _        _ _  _  _  _ 
| || || || |  -->   | ||_|| || |
|_||_||_||_|  -->   | ||_|| ||_|

Or for a smidgen more (due to @PaulGrffiths' observation on my previous attempt)

$7717130$

 _  _  _  _        _ _  _  _  _ 
| || || || |  -->   | || ||_|| |
|_||_||_||_|  -->   | || ||_||_|

edited Sep 02 '16 at 03:57

answered Sep 02 '16 at 02:10

Jonathan Allan

21,150
2
58
109

I'm not sure that middle bit is true. (though it is still very large) – Will Sep 02 '16 at 02:21
@Will fine, I'll change it :) – Jonathan Allan Sep 02 '16 at 02:23
1

Doesn't the 778817 solution have an extra stick? – greenturtle3141 Sep 02 '16 at 02:46
@greenturtle3141 Oh my, oops! Changed it to a slightly smaller number - thanks! – Jonathan Allan Sep 02 '16 at 03:01
2

Surely this is much bigger, at 7,713,170? – Crowman Sep 02 '16 at 03:16
@PaulGriffiths - and a little more - thanks! – Jonathan Allan Sep 02 '16 at 03:57
1

You don't have to break a matchstick; just stand it up on its end. Now you have $11000! = 3.1624624\times10^{39680}$ which is well over $11^{11000} = 2.087\times10^{11455}$ – Engineer Toast Sep 02 '16 at 12:24
@EngineerToast If they were matchsticks – Jonathan Allan Sep 02 '16 at 12:38
Argh, I mistyped. You can still stand a sick on end, though. The high def drawing by OP seems to show them as fairly straight. Actually, sticks would be easier because they would be stiff enough to push into the dirt and stand on their own. – Engineer Toast Sep 02 '16 at 12:54
1,171,700 seems larger than 771,780. – Dewi Morgan Sep 03 '16 at 04:52

score 14 · Answer 6 · answered Sep 03 '16 at 09:14

14

Expanding on greenturtle3141's answer:

Using tetration notation we can express $$^{11000}11$$ which is $\underbrace{11^{11^{...^{11}}}}_{11000\mbox{~times}}$

Looking at the examples, even the number of digits of this number will be insanely high.

answered Sep 03 '16 at 09:14

Petr

241
1
4

1

this is very clear, the largest one, it should have been accepted, i think it's not even computable? – garg10may Sep 09 '16 at 13:39
@garg10may , Petr Pudlák, this is NOT computable with a normal computer. My computer is over-average and to calculate $5_{11}$ took 3 seconds, $9_{11}$ took to long (Java...) - after a minute I terminated it. $5_{11} =$ I can't paste the number here, beacuse it has cca 14000 digits! I'm allowed 600.... – user21233 Oct 01 '16 at 22:29
@RudolfL.Jelínek It can't be computed even with supercomputer. – garg10may Oct 02 '16 at 02:19
@garg10may Maybe I could estimate the number of digits ofthe number of digits ofthe number of digits ofthe number of digits of the number :) – user21233 Oct 02 '16 at 06:47
@garg10may I got it! The number of digits of the number has 10500 (plus minus 200) digits! Well, didn't get farther than that with my Java Super-Skills :) – user21233 Oct 02 '16 at 07:45
@RudolfL.Jelínek say if instead of 11 it was 10, then 10^10 has 100 digits. 10^10^10 has 1000 digits. 10^10^10^10 has 10,000 digits, we have done just 4 iterations. Similarly with 11000 iterations it would be 1 after it 11000 zeros and whatever that number is called, that many digits. I don't know how you got 10500 (plus minus 200), these few digits would be just with 4 iterations. Therefore I say not computable. – garg10may Oct 02 '16 at 08:09
@garg10may That's the number of digits of the number of digits of the number. Double-digited. – user21233 Oct 02 '16 at 09:48
@garg10may technically, all natural numbers are computable. What's not computable are certain real numbers. Computable doesn't refer to whether or not physical computers can 'compute' it, but rather whether a Turing machine can 'theoretically compute' it. – Simply Beautiful Art Jul 20 '17 at 22:15

EKons · Answer 7 · 2016-09-02T15:28:36.217

13

9900

Here are the sticks on their initial positions:

+--------+ +--------+ +--------+ +--------+
|        | |        | |        | |        |
|        | |        | |        | |        |
|        | |        | |        | |        |
+        + +        + +        + +        +
|        | |        | |        | |        |
|        | |        | |        | |        |
|        | |        | |        | |        |
+--------+ +--------+ +--------+ +--------+

Just flip two, and...

+--------+ +--------+ +--------+ +--------+
|        | |        | |        | |        |
|        | |        | |        | |        |
|        | |        | |        | |        |
+--------+ +--------+ +        + +        +
         |          | |        | |        |
         |          | |        | |        |
         |          | |        | |        |
+--------+ +--------+ +--------+ +--------+

edited Sep 02 '16 at 15:28

answered Sep 02 '16 at 06:51

EKons

1,253
12
26

2

what about 111000 ? Remove top and bottom from first 0, you get 11000, then use the 2 sticks to create another 1 to put in front. – CoffeDeveloper Sep 02 '16 at 15:04
1

This was the already the first answe (Jon Mark Perry's below). – z100 Sep 02 '16 at 15:18
1

@z100 Nope. The answer you specify is invalid, because infinity is a mathematical concept, not a number. – EKons Sep 02 '16 at 15:20
Sorry, BlueFire's (Jon just edited). – z100 Sep 02 '16 at 15:27
@z100 revering now....... – EKons Sep 02 '16 at 15:28

JMP · Answer 8 · 2016-09-02T09:50:32.667

11

Take the top and bottom of the first 0 and place around the remaining matches to form the absolute value function. Then we have $|\dfrac{1}{000}|=\infty$.

edited Sep 02 '16 at 09:50

answered Sep 02 '16 at 06:39

JMP

35,612
7
78
151

25

I'll be "that guy" ... not technically true ... 1/0 is not infinity, it's just undefined – Chris Burt-Brown Sep 02 '16 at 08:56
2

mathmo's write $\lim\limits_{x\to0^+}\dfrac1x=\infty$, so it is. $\lim\limits_{x\to0^-}\dfrac1x=-\infty$ on the other hand – JMP Sep 02 '16 at 09:45
11

infinity is not a number, its a mathematical concept – crowie Sep 02 '16 at 12:34
Also, I think that the slash should look like a slash, not $|$. – EKons Sep 02 '16 at 15:23
3

@JonMarkPerry: "so it is" no because you just showed a different thing. – Lightness Races in Orbit Sep 04 '16 at 01:19
@LightnessRacesinOrbit; i guess, most people just accept the first example without questioning stuff too deeply i suppose – JMP Sep 04 '16 at 02:29
Prove that infinity is a number. – Numberknot Sep 04 '16 at 04:34
before i can, i need to know your definition of a number and your definition of infinity; @numberknot – JMP Sep 04 '16 at 04:40
Number:an arithmetical value, expressed by a word, symbol, or figure, representing a particular quantity and used in counting and making calculations. – Numberknot Sep 04 '16 at 04:48
like infinity?@numberknot – JMP Sep 04 '16 at 04:49
2

@JonMarkPerry: No, infinity is not a quantity. – Lightness Races in Orbit Sep 04 '16 at 13:38
1

@JonMarkPerry $\lim_{x \to c} f(x) = L \nRightarrow f(c) = L$ – Ovi Sep 04 '16 at 14:53
@ChrisBurt-Brown 0/0 is undefined. 1/0 is +infinity. – polkovnikov.ph Sep 05 '16 at 10:55
$1/0 = \infty$ if we are in the extended complex plane. Of course one could argue that the elements of the extend complex plane are not numbers, and/or that they are not ordered, and thus $\infty$ is not the largest of them (though it is in magnitude) – Frames Catherine White Sep 05 '16 at 10:56
1

I don't see how you can do this without moving more than two sticks. – LarsH Sep 05 '16 at 13:14
@polkovnikov If you could please provide the proof showing that the limit for x -> 0^- for 1/x is infinity (so how do I get latex to work here?) and not negative infinity I'd be much obliged. I'll even let you publish that break through paper first if you want to. – Voo Sep 05 '16 at 15:35
@polkovnikov.ph You have corrected me wrongly :) 1/0 is not +infinity. – Chris Burt-Brown Sep 05 '16 at 16:01
@Oxinabox: I'd argue it's acceptable to use the extended complex plane as long as you can convey that while moving only two sticks – Chris Burt-Brown Sep 05 '16 at 16:16
@ChrisBurt-Brown I totally forgot that most of puzzling readers are kids. Or did you mean something else by linking a question "if it's okay to tell kids $1/0=\infty$"? – polkovnikov.ph Sep 05 '16 at 21:33
@LightnessRacesinOrbit: Yes, $\infty$ of the projective numbers is a quantity. And so is the $\pm \infty$ of the extended reals. (and cardinal numbers like $\aleph_0$, although that's a mostly unrelated topic) – Sep 06 '16 at 03:58
I will grant, however, that whatever vague, formless notion of infinity the people who say "infinity is just a concept" have in mind is probably not a number. – Sep 06 '16 at 04:04
@polkovnikov.ph I am just providing a citation for 1/0 ≠ ∞ and that was the first one to come up, I am sure a little hunting around math.SE can provide a better link – Chris Burt-Brown Sep 06 '16 at 08:56

score 10 · Answer 9 · edited Apr 13 '17 at 12:50

10

Inspired by MooseBoys:

_ _ _ _ _ _ _ | || || || | --> | || || || | |_||_||_||_| |_|| ||_||_| _|

g1100 is quite a bit greater than g900. Not sure that it's greater than BB110, though.

edited Apr 13 '17 at 12:50

Community

1

answered Sep 02 '16 at 19:14

8bittree

211
1
4

It's way smaller than BB110, trust me... – Simply Beautiful Art Jul 20 '17 at 22:15

ElPedro · Answer 10 · 2016-09-02T10:46:39.867

9

I'm new here but couldn't resist this one. How about

Move the bottom sticks of the first two numbers up to the middle row which gives AA00 which is hexadecimal for 43520

EDIT

Can even go a little further with this one and

Move the two right sticks from the first 0 to the middle row of the middle two 0s giving C880 which is hex for 51328

edited Sep 02 '16 at 10:46

answered Sep 02 '16 at 08:27

ElPedro

373
2
11

3

C0001 is larger. – z100 Sep 02 '16 at 10:44
4

If you're going to change the numerical base, why stop at hexadecimal? C0001 in base 17 is bigger. C0001 in base G900 is even bigger. You can trivially beat any other answer by saying your >10 number is in base (their number). – 8bittree Sep 02 '16 at 18:44
I like that even more! – ElPedro Sep 03 '16 at 12:04

score 8 · Answer 11 · answered Sep 04 '16 at 09:45

If you count this as $900!$ (900 factorial) then it is

$6752680220964584158387906136180081422426942786958938431219826870368509164318041$$696913244695269830379422601037057867290859319834769988692859190650103158765184$$697675968111260952478709384800442863618689339527278445063035408024321764665802$$469665906595179375722352022923557754865383368110217097389374605464912641590914$$315017286072115668581065575923001145013299217645498322753869634011261044702900$$233700488787726638770458607729358543315161251880014776446118268082286709278669$$498283183864180099749981933920657941532564974848626523391891108711459244089659$$406267591429492581671986217837467927209263752478693903629003592427178225373805$$988693392344787776958300301670536333903141306915583751852476107834205263547563$$211316961877454927570148010693336299000373258937059355732529943473445929586672$$898874079417465439147992600084884668670872973671320728520371273220127241083083$$691305263536508288872517163608158715160346829110675464039823214667362737089593$$409077782882754955423243619046482799868392717924602991944325102646445233793959$$919852829782859112268996062036123824831315807164339584840504726141268003987773$$376184987444732386791171263002317174596827846578055856806703501388527508029213$$736049187516494772446422169353375503530006535006513749083203952338296374702618$$565305033183238099184484256075092354377518858209648747695025441836519899967468$$441728626544278665159440478162294690187916638293071419690822746013302760581786$$487737771219314213762543035371844826939073261577664528319882860291768022404108$$899389261050680219591724783890010691069805703037919057105760584932311330863445$$200817988116561644976764835416122506696796129760969874273792338939161520744115$$231939284568767331189924708532770342186297287164449540957225998556321547148208$$332565323177711327132657997031075560497396970894947737425497448029465242702243$$670538018406400885345721451851527098556319541299314527405768863444881244944580$$061763116276824312560642484470937202214990846357225491265490776344575854398099$$914912299810437896562678189865522144326360140515207319970658508028873504020541$$737127725309624320000000000000000000000000000000000000000000000000000000000000$$000000000000000000000000000000000000000000000000000000000000000000000000000000$$000000000000000000000000000000000000000000000000000000000000000000000000000000$$0000000$

With number of trailing 0s $= 224$ and total number of digits $= 2270$

score 6 · Answer 12 · answered Sep 03 '16 at 01:05

Assuming calculator font only with no superscript or letters or other formatting changes, but still allowing the addition of numbers:

 _  _  _  _             _  _  _  _ 
| || || || |    ->    || || || || |
|_||_||_||_|          |  ||_||_||_|

Thus 17000

Puck · Answer 13 · 2016-09-05T12:41:16.413

6

Answer:

gogol, which is French for Googol ($10^{100}$)

Why:

_ _ _ _ _ _ _ _ | || || || | --> | | || | | | |_||_||_||_| |_||_||_||_| |

edited Sep 05 '16 at 12:41

answered Sep 05 '16 at 12:34

Puck

383
2
9

score 5 · Answer 14 · edited Sep 02 '16 at 10:12

5

Move the two horizontal bars of the first 0 to make a slash:

11/000

which evaluates to infinity!

edited Sep 02 '16 at 10:12

IAmInPLS

11,471
4
43
112

answered Sep 02 '16 at 09:59

Gnubie

511
2
10

or 011/00 or 0011/0. – Gnubie Sep 02 '16 at 10:00
7

As written, I think it's indeterminate, not infinity. – Engineer Toast Sep 02 '16 at 12:20
2

0/0 is indeterminate – Gnubie Sep 02 '16 at 12:31
1

sigh, how can you evaluate infinity considering it has no value... – crowie Sep 02 '16 at 12:35
3

@Gnubie 1/+0 goes to positive infinity. 1/-0 goes to negative infinity. 1/0 is indeterminate. All this applies only when it's part of a limit. Otherwise, it is undefined.$\frac{1}{0}=x\Rightarrow1=0x$ and there is no value for $x$ where this is true, not even "infinity" which is technically not even a number. – Engineer Toast Sep 02 '16 at 12:46
Unless this is utilizing Wheel Theory... – Nate Diamond Sep 02 '16 at 19:53

score 5 · Answer 15 · answered Sep 02 '16 at 15:53

5

 _  _  _  _          _  _  _  _   
| || || || |  -->   |_||_|| || |
|_||_||_||_|  -->   |_||_|  ||_|

i.e.

BB70, the 70th Busy Beaver number, which is uncomputable, most likely even independent from ZFC and thus bigger than any other natural number mentioned so far.

answered Sep 02 '16 at 15:53

Radovan Garabík

3,404
1
11
23

8

Except the answer is indistinguishable from 8870. – abligh Sep 02 '16 at 17:18

Joel Harmon · Answer 16 · 2016-09-03T18:55:34.463

4

 _  _  _  _          _  _  _  _   
| || || || |  -->   |_||_|| || |
|_||_||_||_|  -->   | | _||_||_|

That is, in the spirit of the Graham and Busy Beaver answers, A(900) where A is the one-argument version of the Ackermann Function.

I'm pretty sure this loses to the BB answer, but I'm less certain about Graham's. I just thought Ackermann could use some love, too. At the very least this is also an unimaginably large number.

edited Sep 03 '16 at 18:55

answered Sep 03 '16 at 04:39

Joel Harmon

226
1
5

I don't see how you can have accomplished this by only moving two sticks. It looks like you added a stick (the bottom of the leftmost zero moves up to make the A; one move ... the lower left of the second zero rotates 90˚ to make the 9; second move ... now where did the crosspiece come from to turn the third zero into an eight?). You can still have A(900) though :) – Doktor J Sep 03 '16 at 18:51

candied_orange · Answer 17 · 2017-07-27T02:51:50.847

3

Late answer but here's one for fans of out of the box thinking:

Pick up two sticks, take a walk down to the beach with someone you love, draw this in the sand

and define this symbol to represent a number larger than any ever conceived. That's right, it gets bigger the more you think about it. By definition x < ♡ is true for all x. Yes even that one. Now that's a big heart. Who could love you more than that?

edited Jul 27 '17 at 02:51

answered Jul 27 '17 at 02:07

candied_orange

276
1
9

3

nice try, +1 for making hearts out of ms paint strokes – bleh Jul 27 '17 at 03:00

score 2 · Answer 18 · answered Sep 02 '16 at 06:58

2

Theres a lot of good answers but I though this one is worth mentioning as it is the only one I can see that wouldn't result in badly justified or sized numbers.

Move the lower left of the first two digits up to be the middle horizontal bar giving 9900

answered Sep 02 '16 at 06:58

user3559247

211
1
2

3

"It's over 9000!" – IQAndreas Sep 04 '16 at 02:41

score 2 · Answer 19 · answered Sep 03 '16 at 12:22

Take the bottom stick from each of the first two digits, giving

_ _ _ _ _ _ _ _ | || || || | --> | || || || | |_||_||_||_| | || ||_||_|

i.e.

171700 .

Then

rotate them to make an 11, and raise this to the power 171700: _ _ _ _ | || || || | | || ||_||_| | |

i.e.

$11^{171700}$.

This is

$1.33 * 10^{178807}$

to 3 significant figures.

Tom Carpenter · Answer 20 · 2016-09-04T18:55:50.687

1

How about:

Which is read as:

$$9^{{1717}^6}$$

That is rougly:

A number with 24450093851501172475 decimal digits.

edited Sep 04 '16 at 18:55

answered Sep 04 '16 at 18:38

Tom Carpenter

456
2
10

They should be the same size at least if you're going to do that – bleh Sep 04 '16 at 18:44
@bleh doesn't say that anywhere in the question, seems readable enough to me. Though h34's answer beats it anyways, so somewhat irrelevant. – Tom Carpenter Sep 04 '16 at 18:49
But if the size of the 1 is an issue, then not a problem. Try the new version out for size, much larger value. – Tom Carpenter Sep 04 '16 at 18:56

score 1 · Answer 21 · answered Sep 05 '16 at 08:51

88110

Here are the sticks on their initial positions:

+--------+ +--------+ +--------+ +--------+
|        | |        | |        | |        |
|        | |        | |        | |        |
|        | |        | |        | |        |
+        + +        + +        + +        +
|        | |        | |        | |        |
|        | |        | |        | |        |
|        | |        | |        | |        |
+--------+ +--------+ +--------+ +--------+
Just flip two, and...
+--------+ +--------+ +        + +--------+
|        | |        | |        | |        |
|        | |        | |        | |        |
|        | |        | |        | |        |
+--------+ +--------+ +        + +        +
|        | |        | |        | |        |
|        | |        | |        | |        |
|        | |        | |        | |        |
+--------+ +--------+ +        + +--------+

Tony Ruth · Answer 22 · 2016-09-02T05:43:48.830

0

Move two sticks to make 1/0 (can be infinite). I change the 0 into a C in the process.

1
_
000C

edited Sep 02 '16 at 05:43

answered Sep 02 '16 at 05:36

Tony Ruth

1,867
11
15

1

The number is not as big as you expect. – EKons Sep 02 '16 at 10:24
1

What do you mean by C? The speed of light? An hexadecimal 12? Either case, if C is dividing the number, then you have a small number, not a big one – Barranka Sep 02 '16 at 16:30
@Barranka regardless of what C is, it's still 0C which is 0. I needed to find a way to make the denominator stay 0 when I removed some sticks so I made a C. – Tony Ruth Sep 02 '16 at 21:52
@TonyRuth Not necessarily... if it's 000C hexadecimal, it's 12, not 0. – ArtOfCode Sep 02 '16 at 23:12

score 0 · Answer 23 · edited Sep 03 '16 at 19:27

0

Take the right hand side 2 of the last two zero making them into two CC. Then create two 1 like this:

1100CC which is 1100*C squared where C is the speed of light

(as in E=MC squared)
1100 * 299792458 * 299792458 = 98,863,069,661,049,940,400
ie. 98 quintillion give or take a few quadrillion

edited Sep 03 '16 at 19:27

Beastly Gerbil

58,036
8
166
314

answered Sep 02 '16 at 14:14

Paul Hartness

1

2

It is not a number but a quantity. – z100 Sep 02 '16 at 14:22
3

You're moving 4 matches, and you're only allowed to move 2. – Jaap Scherphuis Sep 02 '16 at 14:44

score 0 · Answer 24 · edited Sep 03 '16 at 00:56

0

110,001

can be made as follows:

_ _ _
| | | | | | | | |
| | |_| |_| |_| |

edited Sep 03 '16 at 00:56

Rand al'Thor

116,845
28
322
627

answered Sep 03 '16 at 00:34

user29755

11

Another case of an edit after an auto flag yielding a down vote. – Jonathan Allan Sep 03 '16 at 11:47

What is the largest number you can create with 0000 by moving only two sticks?

24 Answers24

Linked

Related