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While Artur is willing to settle for a billion, I'd like a much bigger return on my two matchstick investment.

Moving only two matchsticks, what is the largest number that can be created from the following pattern?

enter image description here

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  • 6
    $\begingroup$ Dodgy drawings should be a requirement ;) $\endgroup$ – BreakingMyself Jul 17 '17 at 13:04
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    $\begingroup$ Move the bottom horizontal stick of the last zero to 6 making it B and we get Billion, Allow three sticks and we can make 'zillion' $\endgroup$ – user9174 Jul 17 '17 at 19:16
  • $\begingroup$ while we can only move two, can we burn the others? :P $\endgroup$ – Quintec Jul 18 '17 at 0:13
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    $\begingroup$ So... is there an answer? $\endgroup$ – Forklift Jul 18 '17 at 13:30
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    $\begingroup$ very similar to puzzling.stackexchange.com/questions/41873/… $\endgroup$ – bleh Jul 19 '17 at 1:33

18 Answers 18

36
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enter image description here

6^1110G

Where G is Graham's number

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  • $\begingroup$ This would be quite large... But relatively equal to Graham's number (I think? Any answers?) $\endgroup$ – Henry Jul 18 '17 at 17:01
  • $\begingroup$ I think this beats also the Tetration (and even the Pentation) answer of Aaron above. Congratz. $\endgroup$ – Hjan Jul 18 '17 at 19:08
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    $\begingroup$ @Henry: Once you start using numbers like Graham's number (or even mere Tetration) a single exponentation step doesn't seem like much. But it really is a lot bigger. $\endgroup$ – James Hollis Jul 19 '17 at 10:07
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    $\begingroup$ G could also be the Universal constant of gravitation (6.67408(31)×10−11) ;) $\endgroup$ – tuskiomi Jul 20 '17 at 20:06
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    $\begingroup$ But this answer moves more than TWO matchsticks right? $\endgroup$ – Denny Jul 22 '17 at 22:48
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Here's one way to make something quite big:

Move the top two matches of the 6 to turn it into a 9 lower down: $9^{111100}$.

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  • $\begingroup$ To preserve font size, the second match could be used to turn a 0 into an 8. Or perhaps the first 1 into a 7 (which messes up the kerning) $\endgroup$ – Hugh Meyers Jul 17 '17 at 15:45
  • $\begingroup$ "To preserve font size" meaning to increase the numbers while avoiding the typographical problem of Deusovi's solution? Yup, those would work, except that the 9 is then "wrong" (i.e., not what a 9 generally looks like on a 7-segment display). $\endgroup$ – Gareth McCaughan Jul 17 '17 at 16:53
  • $\begingroup$ Ah, of course. I forgot those displays have a lower crossbar on the 9. $\endgroup$ – Hugh Meyers Jul 17 '17 at 17:24
  • $\begingroup$ @HughMeyers: Real-world devices usually have a crossbar on the 6, and often have a crossbar on the 9, but some devices only have it on the 6, and some devices--especially older ones--omit both. Showing the crossbar on the 6 often requires a little more circuitry than omitting it--a non-issue when billions of transistors can fit on one chip, but a definite consideration on devices built out of discrete components like transistors, diodes, and resistors. $\endgroup$ – supercat Jul 17 '17 at 19:55
  • $\begingroup$ In scientific notation, this equals approximately 2.202 × 10^106016. (Calculated with Wolfram Alpha) $\endgroup$ – Stevoisiak Jul 17 '17 at 20:31
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using Tetration, we can get much larger than the simple exponentiation..

Move the top and bottom matchstick from the zero in the ten's place, and form a subscript "11" after the resulting "61111110" to form:

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

Or: (provided you can ignore the incorrect relative font size between the intended superscript vs regular size)

6111111011

I'm not sure it would be reasonable to even try to determine how long this would be in digits, as it is equivalent to:

1111...11 (61,111,110 11's)

*edit: thanks to @Phlarx for catching the 7->11 I probably should have thought of that.. just realized I could swap the position of the sub / superscript to get many more exponentiation iterations vs. larger starting number (which quickly pays off)

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  • 3
    $\begingroup$ Instead of 7, you could make a little 11 with the two matchsticks. $\endgroup$ – Phlarx Jul 17 '17 at 21:24
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    $\begingroup$ That looks like a subscript 11, which would be 61,111,110 in base 11, which is 118871742 $\endgroup$ – isaacg Jul 18 '17 at 6:25
  • $\begingroup$ Hi Aaron, if you move the two matches that form 11 at the right bottom to the left bottom as pentation (also here )notation, which is the equivalent of 3 arrow notation, it will be even unimaginably more. maybe we can team up some times! $\endgroup$ – Hjan Jul 18 '17 at 9:09
  • $\begingroup$ @isaacg I did state the intended interpretation, and wasn't dodgy drawing a proposed requirement? ;) $\endgroup$ – Aaron Jul 18 '17 at 13:33
  • $\begingroup$ @Hjan I could only find a single paper that mentioned that form a single time, whereas the tetration form was used much more often (probably due to the relative frequency of use in general). It would seem for pentation and higher operations, the up arrow or chained arrow notation is preferred. The only real use of pentation I could find was in describing the Ackermann function given various levels of recursion. $\endgroup$ – Aaron Jul 18 '17 at 13:38
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You could move the lower right stick in the six and the lower left stick in the first zero to make

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

or 1E111190, 1 followed by 111,190 Zeroes.

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Using knuth's arrow notation

6↑↑100 in matches ${\displaystyle 6\uparrow \uparrow 100}$

which represents ${{{6}^6}^{...}}^6$ (recursively raised to the power 100 times) which is a number so huge that i can not express in in normal notation but of course almost all natural numbers are very very large. In fact, almost every natural number is larger than mine.

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  • 3
    $\begingroup$ ooh, how does that compare to mine I wonder... how would one calculate that? I get the feeling this might be larger.. $\endgroup$ – Aaron Jul 17 '17 at 17:10
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    $\begingroup$ take the arrow halves from the zero in the tens place to get 6^^11110. definitely bigger than mine.. $\endgroup$ – Aaron Jul 17 '17 at 17:14
  • $\begingroup$ @Aaron, ah yes i didnt think about splitting the zeros. That is a good idea. I would not know how to compare our original answers. interesting problem. $\endgroup$ – Hjan Jul 17 '17 at 17:25
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    $\begingroup$ Hi Tuskiomi, i am also not sure about how valid half arrows are. I thought about bending or breaking the matches to make a full arrow, but i was not sure. I am not claiming any prizes. I liked the question and the creative answers are people are giving. $\endgroup$ – Hjan Jul 18 '17 at 8:58
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    $\begingroup$ :-) Hope you enjoy my link to the Frivolous Theorem of Arithmetic. $\endgroup$ – Simply Beautiful Art Jul 21 '17 at 0:18
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You could

move the top and bottom of the first 0, down and left of the whole thing, to make "$11^{61111110}$".

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  • $\begingroup$ Ah yes, that's clearly better than my similar answer -- except that it produces kinda-misformed digits :-). $\endgroup$ – Gareth McCaughan Jul 17 '17 at 12:47
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    $\begingroup$ And have typesetters laugh at you, because the base's font is half as big as the exponent's? ;) $\endgroup$ – M Oehm Jul 17 '17 at 12:47
  • $\begingroup$ In scientific notation, this equals approximately 8.628 × 10^63640662. (Calculated with Wolfram Alpha) $\endgroup$ – Stevoisiak Jul 17 '17 at 20:35
  • $\begingroup$ Arguably, this isn't mere exponentation, but pentation, or 61111110↑↑↑11. Viewed like this, the subscript is correctly smaller and as a bonus the number is much larger. tetration.org/Tetration/index.html $\endgroup$ – James Hollis Jul 19 '17 at 12:24
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With some imagination,

6111101! -> remove 2 matches from last 0 to make 11 out of it, then put them in a cross below the last 1 and pretend it is an exclamation mark.

EDIT: @maxathousand gave me an idea that matches need not lie horizontally.

8111101! -> Similar to the idea above, remove 2 matches from the last 0, put one of them to make 8 out of 6 and put the other vertically below the last 1. Looking from the top, exclamation mark should be clearly visible.

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  • 3
    $\begingroup$ Better yet, position one of them as the "stem" of the exclamation mark, then balance the second on its end with the match head standing up so that, from above, you clearly see an exclamation mark. $\endgroup$ – maxathousand Jul 17 '17 at 15:28
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    $\begingroup$ pretend it is an exclamation mark Clever. $\endgroup$ – cyberbit Jul 17 '17 at 17:18
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Why stop at two up-arrows, when you can have 111

 _              _   _
|_  |\ | | | | | | | |
|_| |        | |_| |_|

This would be

$6\uparrow^{111}100$

A hyperoperation of rank 113, which is very large.

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  • $\begingroup$ The largest and best finite one... GG $\endgroup$ – Mega Man Jul 22 '17 at 15:55
11
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6^1111110? enter image description here

At least that`s the best I can do

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10
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Think big.

,\′ _ _ |_ | | | | | | | | |_| | | | | | | |_| The , and ' are two halves of a broken match (they should connect with the diagonal one, pardon my sloppy ASCII art). The result is $\aleph_{61111110}$, quite a large transfinite number.

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  • 3
    $\begingroup$ Indeed, nothing was said that "moving" a match doesn't allow breaking it. You could go further, I guess, and reduce it into a fine powder and use it to write a long essay on compared infinities. $\endgroup$ – Florian F Jul 20 '17 at 19:59
  • $\begingroup$ Ah shoot, appears someone got to the infinities before me. See my solution for another one that doesn't require breaking things or anything like that. $\endgroup$ – Simply Beautiful Art Jul 20 '17 at 21:53
  • $\begingroup$ @FlorianF Breaking is not that important, my first idea was without it - but I went for this, just because :-) $\endgroup$ – Radovan Garabík Jul 21 '17 at 6:34
  • $\begingroup$ @SimplyBeautifulArt Your solution is a nice complement, since it produces an ordinal, while mine gives a cardinal... $\endgroup$ – Radovan Garabík Jul 21 '17 at 6:35
9
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current provided number is 7-digit number

6111100

remove top and bottom horizontal matchsticks from 6th '0' digit and placing them in front of the whole number makes it a 9-digit number

161111110

or even bigger putting that '1' after '6'

611111110

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  • 1
    $\begingroup$ Could make it about 6 times bigger by moving the new 1 to the right of the 6. $\endgroup$ – Forklift Jul 17 '17 at 17:14
  • $\begingroup$ haha .. yes let me edit that $\endgroup$ – Karun Kamal Jul 17 '17 at 18:47
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    $\begingroup$ I honestly think this is the best answer, specifically "161111110" because it seems to me to be in the best spirit of the challenge and it preserves font size. +1 $\endgroup$ – Hawkeye Jul 18 '17 at 17:18
9
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This is not valid answer since infinity is not a number, but still just for fun, take two sticks from any "1", make a cross, place the cross between two zeros to make infinity sign "$\infty$", that makes $6111\infty$.

enter image description here

EDIT: A valid and "thinking out of the box" answer no one has posted:

Remove the two left side vertical sticks of the second zero from right, place one stick horizontally in it in the middle making it reverse "E", place second stick horizontally in the middle of first zero from right making it "8", $\underline{\text{look at this number from the opposite side of the table}}$, it is "8E11119"

enter image description here

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  • $\begingroup$ OR do not disturb 61111,(for first zero from left) rotate lower right side vertical stick to the right by 25 degrees and (for second zero from left) upper left side of the vertical stick to the left by 25 degrees to make infinity sign $\endgroup$ – user9174 Jul 17 '17 at 17:26
  • $\begingroup$ In a similar spirit: one can divide by 0. $\endgroup$ – Winther Jul 17 '17 at 17:36
  • $\begingroup$ @Winther, sorry,I did not get you, are you suggesting 611/00 ?, that is not possible because division by zero is not defined in maths. $\endgroup$ – user9174 Jul 17 '17 at 17:50
  • $\begingroup$ Neither is $\infty$. However integer division by 0 (mainly in programming languages) is often defined as "inf". $\endgroup$ – Winther Jul 17 '17 at 17:54
  • $\begingroup$ Division by zero is defined on the projectively extended real line. $\endgroup$ – Tanner Swett Jul 18 '17 at 12:14
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Try this:

_ _ |_ |\ |\ | | | | | | |_| | | | | | | |_|

Which is (with a little imagination):

$6\uparrow\uparrow11110$ (see up-arrow notation)

Which is equal to:

$$\left.6^{6^{6^{\cdots^{6}}}}\right\}\text{11110 copies of 6}$$

...which is big.

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    $\begingroup$ Aaron discovered this already in the comments to Hjan's answer. That's what I get for not refreshing the page before answering. $\endgroup$ – Phlarx Jul 17 '17 at 19:19
  • $\begingroup$ Oh, but Aaron uses a different notation? Also I think I spotted a typing error in your formula. Probably you meant to write G instead of 6. $\endgroup$ – Robert Jul 22 '17 at 18:25
  • $\begingroup$ @Robert In his answer, he did, but not in the comments of Hjan's answer. And yes, G^^11110 would be quite a bit larger than 6^^11110, but I think the crossbar would have to be slightly offset for that as in James Hollis' answer. $\endgroup$ – Phlarx Jul 25 '17 at 15:33
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9111190 ?

So you move a matchstick from the first and 5th digit.

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  • $\begingroup$ removing both and making a '1' and adds a digit , making it much much higher number, check out my answer $\endgroup$ – Karun Kamal Jul 19 '17 at 20:50
5
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Move the top left and middle sticks to the bottom left and into the left zero to get:

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

where the W refers to ω, the smallest transfinite ordinal, so the end result is

ω11180

Clearly larger than any finite numbers here.


If you want to make it even larger,

enter image description here

Where ω1 is the first uncountable ordinal and the end result is

ω111100

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3
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With a little bit of imagination, we can beat all of these answers!

  • Take the top and bottom matchstick out from the zero in the 10s spot to turn it into two 1s.
  • Place one of the matchsticks horizontally to the left of the 6 to create a minus symbol.
  • Place the other stick to the right of the last digit, and slightly above it, to raise our number to the power of 1
  • Slide our two moved matchsticks all the way out to the opposite edges of the universe.
  • Finally, exploit the fact that the original question never said which definition of largest to use, et voilà! The largest representation of the number -61111110$^1$ possible!

Unfortunately, I can't fit my dodgy drawing on your monitor. Get one a few billion lightyears across and we'll talk.

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  • $\begingroup$ Should still be smaller than my number, assuming a finite universe. $\endgroup$ – Simply Beautiful Art Jul 20 '17 at 21:51
  • $\begingroup$ And I have a finite (i.e. natural number) solution that is much, much larger than what you propose. But I've already used up my slot, so to speak... $\endgroup$ – Radovan Garabík Jul 21 '17 at 6:38
0
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611 | 00 ( the | is formed by 4 match sticks, two originally existing + two moved) which is quuite big, I guess !! By this I mean 611 is the numerator divided by (using |) 00 as the denominator !!

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    $\begingroup$ But (1) a long vertical bar doesn't mean "divided by" and (2) dividing by zero doesn't give you a number at all. $\endgroup$ – Gareth McCaughan Jul 17 '17 at 12:56
  • $\begingroup$ Well, @GarethMcCaughan...I did not settle for a billion as per the title of the riddle :-)... Anyways, here goes another, assuming operators operating on the numerics can be of different size, possible biggest number could be ((611100)!)!, the last two exclamatory signs are being obtained by inverting single match sticks...twice ;-) $\endgroup$ – Mea Culpa Nay Jul 17 '17 at 13:01
  • $\begingroup$ Well, @GarethMcCaughan ... | does not mean divide by...but means modulo (in Number Theory !) and again we boil down to division ...I think... $\endgroup$ – Mea Culpa Nay Jul 17 '17 at 13:36
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    $\begingroup$ Vertical bar doesn't mean modulo in number theory, it means "divides exactly into". 611|00, interpreted that way, would be true but still wouldn't be a number. $\endgroup$ – Gareth McCaughan Jul 17 '17 at 13:48
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    $\begingroup$ No, it's merely false because 11100 is not 0 times any integer :-). $\endgroup$ – Gareth McCaughan Jul 17 '17 at 15:19
0
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Some imagination is being used, but: 6111100 can be interpreted as 6iiii00. Then, changing the two 0's into 9's, we get 6 * i^4 * 99, which is undefined so it could be anything! And for these purposes, I interpret 6 * i^4 * 99 as infinity.

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  • $\begingroup$ I'm not sure what you mean. i^4 is equal to 1. $\endgroup$ – Phlarx Jul 25 '17 at 16:15
  • $\begingroup$ well then, we can change it to 6iii100, which isn't 1 (sorry, forgot my math for a moment there). $\endgroup$ – Hazard Jul 31 '17 at 8:06

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