Here's what I'm saying
Moving 2 sticks, what is the largest number that you can create?
$\begingroup$
$\endgroup$
9
24 Answers
$\begingroup$
$\endgroup$
5
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.
answered Sep 2, 2016 at 2:00
-
2
-
35$\begingroup$ @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). $\endgroup$– MeghaSep 2, 2016 at 6:50
-
$\begingroup$ 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 $\endgroup$– NeoSep 3, 2016 at 4:00
-
7$\begingroup$ 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. $\endgroup$ Sep 3, 2016 at 4:07
-
$\begingroup$
$\endgroup$
14
-
11
-
27$\begingroup$ 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. $\endgroup$ Sep 2, 2016 at 15:32
-
2$\begingroup$ And your answer inspired me to attempt to outdo you using the Busy Beaver function. See my answer below :) $\endgroup$ Sep 2, 2016 at 16:04
-
2$\begingroup$ You could even have made $g_{17130}$. $\endgroup$ Sep 3, 2016 at 11:29
-
3$\begingroup$ @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. $\endgroup$ Sep 3, 2016 at 21:36
$\begingroup$
$\endgroup$
13
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.
-
1$\begingroup$ Radovan Gabarík had the same idea earlier, but the funny thing is, his answer is ridiculously small compared to this one. $\endgroup$– JiKSep 2, 2016 at 16:46
-
-
4
-
2$\begingroup$ @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. $\endgroup$ Sep 6, 2016 at 14:31
-
1$\begingroup$ @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. scottaaronson.com/writings/bignumbers.html $\endgroup$ Oct 19, 2016 at 13:47
$\begingroup$
$\endgroup$
4
Move the top and bottom stick of the first zero to create another 1, thus making 111000
-
-
7
-
6
-
$\begingroup$
$\endgroup$
10
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)
$7717130$
_ _ _ _ _ _ _ _ _
| || || || | --> | || ||_|| |
|_||_||_||_| --> | || ||_||_|
answered Sep 2, 2016 at 2:10
-
$\begingroup$ I'm not sure that middle bit is true. (though it is still very large) $\endgroup$– WillSep 2, 2016 at 2:21
-
-
1$\begingroup$ Doesn't the 778817 solution have an extra stick? $\endgroup$ Sep 2, 2016 at 2:46
-
2
-
1$\begingroup$ 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 $11^{11000} = 2.087\times10^{11455}$ $\endgroup$ Sep 2, 2016 at 12:24
$\begingroup$
$\endgroup$
8
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.
-
1$\begingroup$ this is very clear, the largest one, it should have been accepted, i think it's not even computable? $\endgroup$ Sep 9, 2016 at 13:39
-
$\begingroup$ @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.... $\endgroup$ Oct 1, 2016 at 22:29
-
$\begingroup$ @RudolfL.Jelínek It can't be computed even with supercomputer. $\endgroup$ Oct 2, 2016 at 2:19
-
$\begingroup$ @garg10may Maybe I could estimate the number of digits ofthe number of digits ofthe number of digits ofthe number of digits of the number :) $\endgroup$ Oct 2, 2016 at 6:47
-
$\begingroup$ @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 :) $\endgroup$ Oct 2, 2016 at 7:45
$\begingroup$
$\endgroup$
5
9900
Here are the sticks on their initial positions:+--------+ +--------+ +--------+ +--------+ | | | | | | | | | | | | | | | | | | | | | | | | + + + + + + + + | | | | | | | | | | | | | | | | | | | | | | | | +--------+ +--------+ +--------+ +--------+Just flip two, and...+--------+ +--------+ +--------+ +--------+ | | | | | | | | | | | | | | | | | | | | | | | | +--------+ +--------+ + + + + | | | | | | | | | | | | | | | | | | +--------+ +--------+ +--------+ +--------+
-
2$\begingroup$ 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. $\endgroup$ Sep 2, 2016 at 15:04
-
1$\begingroup$ This was the already the first answe (Jon Mark Perry's below). $\endgroup$– z100Sep 2, 2016 at 15:18
-
1$\begingroup$ @z100 Nope. The answer you specify is invalid, because infinity is a mathematical concept, not a number. $\endgroup$– EKonsSep 2, 2016 at 15:20
-
-
$\begingroup$
$\endgroup$
22
Take the top and bottom of the first 0 and place around the remaining matches to form the absolute value function. Then we have $|\dfrac{1}{000}|=\infty$.
-
25$\begingroup$ I'll be "that guy" ... not technically true ... 1/0 is not infinity, it's just undefined $\endgroup$ Sep 2, 2016 at 8:56
-
2$\begingroup$ mathmo's write $\lim\limits_{x\to0^+}\dfrac1x=\infty$, so it is. $\lim\limits_{x\to0^-}\dfrac1x=-\infty$ on the other hand $\endgroup$– JMPSep 2, 2016 at 9:45
-
11$\begingroup$ infinity is not a number, its a mathematical concept $\endgroup$– crowieSep 2, 2016 at 12:34
-
3$\begingroup$ @JonMarkPerry: "so it is" no because you just showed a different thing. $\endgroup$ Sep 4, 2016 at 1:19
-
2$\begingroup$ @JonMarkPerry: No, infinity is not a quantity. $\endgroup$ Sep 4, 2016 at 13:38
$\begingroup$
$\endgroup$
3
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
-
3
-
4$\begingroup$ 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). $\endgroup$– 8bittreeSep 2, 2016 at 18:44
-
$\begingroup$
$\endgroup$
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$
$\begingroup$
$\endgroup$
Assuming calculator font only with no superscript or letters or other formatting changes, but still allowing the addition of numbers:
_ _ _ _ _ _ _ _ | || || || | -> || || || || | |_||_||_||_| | ||_||_||_|Thus 17000
$\begingroup$
$\endgroup$
Answer:
gogol, which is French for Googol ($10^{100}$)
Why:
_ _ _ _ _ _ _ _ | || || || | --> | | || | | | |_||_||_||_| |_||_||_||_| |
$\begingroup$
$\endgroup$
6
Move the two horizontal bars of the first 0 to make a slash:
11/000
which evaluates to infinity!
-
-
7$\begingroup$ As written, I think it's indeterminate, not infinity. $\endgroup$ Sep 2, 2016 at 12:20
-
2
-
1$\begingroup$ sigh, how can you evaluate infinity considering it has no value... $\endgroup$– crowieSep 2, 2016 at 12:35
-
3$\begingroup$ @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. $\endgroup$ Sep 2, 2016 at 12:46
$\begingroup$
$\endgroup$
1
_ _ _ _ _ _ _ _
| || || || | --> |_||_|| || |
|_||_||_||_| --> |_||_| ||_|
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 2, 2016 at 15:53
-
8$\begingroup$ Except the answer is indistinguishable from 8870. $\endgroup$– ablighSep 2, 2016 at 17:18
$\begingroup$
$\endgroup$
1
_ _ _ _ _ _ _ _
| || || || | --> |_||_|| || |
|_||_||_||_| --> | | _||_||_|
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.
-
$\begingroup$ 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 :) $\endgroup$– Doktor JSep 3, 2016 at 18:51
$\begingroup$
$\endgroup$
1
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?
-
3$\begingroup$ nice try, +1 for making hearts out of ms paint strokes $\endgroup$– blehJul 27, 2017 at 3:00
$\begingroup$
$\endgroup$
1
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
-
3
$\begingroup$
$\endgroup$
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.
$\begingroup$
$\endgroup$
3
How about:
Which is read as:
$$9^{{1717}^6}$$
That is rougly:
-
$\begingroup$ They should be the same size at least if you're going to do that $\endgroup$– blehSep 4, 2016 at 18:44
-
$\begingroup$ @bleh doesn't say that anywhere in the question, seems readable enough to me. Though h34's answer beats it anyways, so somewhat irrelevant. $\endgroup$ Sep 4, 2016 at 18:49
-
$\begingroup$ But if the size of the 1 is an issue, then not a problem. Try the new version out for size, much larger value. $\endgroup$ Sep 4, 2016 at 18:56
$\begingroup$
$\endgroup$
88110
Here are the sticks on their initial positions:
+--------+ +--------+ +--------+ +--------+
| | | | | | | |
| | | | | | | |
| | | | | | | |
+ + + + + + + +
| | | | | | | |
| | | | | | | |
| | | | | | | |
+--------+ +--------+ +--------+ +--------+
Just flip two, and...
+--------+ +--------+ + + +--------+
| | | | | | | |
| | | | | | | |
| | | | | | | |
+--------+ +--------+ + + + +
| | | | | | | |
| | | | | | | |
| | | | | | | |
+--------+ +--------+ + + +--------+
$\begingroup$
$\endgroup$
4
Move two sticks to make 1/0 (can be infinite). I change the 0 into a C in the process.
1
_
000C
-
1
-
1$\begingroup$ What do you mean by
C? The speed of light? An hexadecimal 12? Either case, ifCis dividing the number, then you have a small number, not a big one $\endgroup$– BarrankaSep 2, 2016 at 16:30 -
$\begingroup$ @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. $\endgroup$ Sep 2, 2016 at 21:52
-
$\begingroup$ @TonyRuth Not necessarily... if it's
000Chexadecimal, it's 12, not 0. $\endgroup$ Sep 2, 2016 at 23:12
$\begingroup$
$\endgroup$
1
110,001
can be made as follows:
_ _ _
| | | | | | | | |
| | |_| |_| |_| |
-
$\begingroup$ Another case of an edit after an auto flag yielding a down vote. $\endgroup$ Sep 3, 2016 at 11:47
$\begingroup$
$\endgroup$
2
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
-
2
-
3$\begingroup$ You're moving 4 matches, and you're only allowed to move 2. $\endgroup$ Sep 2, 2016 at 14:44
1/0=infinity. Unless you use the fact that1/x --> INFwhenx --> 0, that's not true. So... Do you consider that a valid answer? $\endgroup$