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?
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10
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6$\begingroup$ Dodgy drawings should be a requirement ;) $\endgroup$– BreakingMyselfCommented Jul 17, 2017 at 13:04
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3$\begingroup$ 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' $\endgroup$– user9174Commented Jul 17, 2017 at 19:16
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$\begingroup$ while we can only move two, can we burn the others? :P $\endgroup$– QuintecCommented Jul 18, 2017 at 0:13
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4$\begingroup$ So... is there an answer? $\endgroup$– ForkliftCommented Jul 18, 2017 at 13:30
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3$\begingroup$ very similar to puzzling.stackexchange.com/questions/41873/… $\endgroup$– blehCommented Jul 19, 2017 at 1:33
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18 Answers
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$\begingroup$ This would be quite large... But relatively equal to Graham's number (I think? Any answers?) $\endgroup$– HenryCommented Jul 18, 2017 at 17:01
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$\begingroup$ I think this beats also the Tetration (and even the Pentation) answer of Aaron above. Congratz. $\endgroup$– HjanCommented Jul 18, 2017 at 19:08
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4$\begingroup$ @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. $\endgroup$ Commented Jul 19, 2017 at 10:07
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3$\begingroup$ G could also be the Universal constant of gravitation (6.67408(31)×10−11) ;) $\endgroup$– tuskiomiCommented Jul 20, 2017 at 20:06
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2$\begingroup$ But this answer moves more than TWO matchsticks right? $\endgroup$– DennyCommented Jul 22, 2017 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: $9^{111100}$.
answered Jul 17, 2017 at 12:44
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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) $\endgroup$ Commented Jul 17, 2017 at 15:45
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$\begingroup$ "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 ♦Commented Jul 17, 2017 at 16:53
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$\begingroup$ Ah, of course. I forgot those displays have a lower crossbar on the 9. $\endgroup$ Commented Jul 17, 2017 at 17:24
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$\begingroup$ @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. $\endgroup$– supercatCommented Jul 17, 2017 at 19:55
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$\begingroup$ In scientific notation, this equals approximately
2.202 × 10^106016. (Calculated with Wolfram Alpha) $\endgroup$ Commented Jul 17, 2017 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:
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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. $\endgroup$– PhlarxCommented Jul 17, 2017 at 21:24
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5$\begingroup$ That looks like a subscript 11, which would be 61,111,110 in base 11, which is 118871742 $\endgroup$– isaacgCommented Jul 18, 2017 at 6:25
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$\begingroup$ 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! $\endgroup$– HjanCommented Jul 18, 2017 at 9:09
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$\begingroup$ @isaacg I did state the intended interpretation, and wasn't dodgy drawing a proposed requirement? ;) $\endgroup$– AaronCommented Jul 18, 2017 at 13:33
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$\begingroup$ @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. $\endgroup$– AaronCommented Jul 18, 2017 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
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or 1E111190, 1 followed by 111,190 Zeroes.
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Using knuth's arrow notation
${\displaystyle 6\uparrow \uparrow 100}$
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.. $\endgroup$– AaronCommented Jul 17, 2017 at 17:10
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5$\begingroup$ take the arrow halves from the zero in the tens place to get 6^^11110. definitely bigger than mine.. $\endgroup$– AaronCommented Jul 17, 2017 at 17:14
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$\begingroup$ @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. $\endgroup$– HjanCommented Jul 17, 2017 at 17:25
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3$\begingroup$ 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. $\endgroup$– HjanCommented Jul 18, 2017 at 8:58
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2$\begingroup$ :-) Hope you enjoy my link to the Frivolous Theorem of Arithmetic. $\endgroup$ Commented Jul 21, 2017 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 ♦Commented Jul 17, 2017 at 12:47
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22$\begingroup$ And have typesetters laugh at you, because the base's font is half as big as the exponent's?
;)$\endgroup$– M OehmCommented Jul 17, 2017 at 12:47 -
$\begingroup$ In scientific notation, this equals approximately
8.628 × 10^63640662. (Calculated with Wolfram Alpha) $\endgroup$ Commented Jul 17, 2017 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 $\endgroup$ Commented Jul 19, 2017 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. $\endgroup$ Commented Jul 17, 2017 at 15:28
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5$\begingroup$ pretend it is an exclamation mark Clever. $\endgroup$– cyberbitCommented Jul 17, 2017 at 17:18
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Why stop at two up-arrows, when you can have 111
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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 $\endgroup$ Commented Jul 22, 2017 at 15:55
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6^1111110?
At least that`s the best I can do
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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.
answered Jul 19, 2017 at 8:26
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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. $\endgroup$ Commented Jul 20, 2017 at 19:59
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$\begingroup$ 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. $\endgroup$ Commented Jul 20, 2017 at 21:53
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$\begingroup$ @FlorianF Breaking is not that important, my first idea was without it - but I went for this, just because :-) $\endgroup$ Commented Jul 21, 2017 at 6:34
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$\begingroup$ @SimplyBeautifulArt Your solution is a nice complement, since it produces an ordinal, while mine gives a cardinal... $\endgroup$ Commented Jul 21, 2017 at 6:35
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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. $\endgroup$– ForkliftCommented Jul 17, 2017 at 17:14
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1$\begingroup$ 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 $\endgroup$– HawkeyeCommented Jul 18, 2017 at 17:18
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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 $6111\infty$.
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"
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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 $\endgroup$– user9174Commented Jul 17, 2017 at 17:26
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$\begingroup$ In a similar spirit: one can divide by 0. $\endgroup$– WintherCommented Jul 17, 2017 at 17:36
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$\begingroup$ @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. $\endgroup$– user9174Commented Jul 17, 2017 at 17:50
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$\begingroup$ Neither is $\infty$. However integer division by 0 (mainly in programming languages) is often defined as "inf". $\endgroup$– WintherCommented Jul 17, 2017 at 17:54
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$\begingroup$ Division by zero is defined on the projectively extended real line. $\endgroup$ Commented Jul 18, 2017 at 12:14
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Try this:
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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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3$\begingroup$ Aaron discovered this already in the comments to Hjan's answer. That's what I get for not refreshing the page before answering. $\endgroup$– PhlarxCommented Jul 17, 2017 at 19:19
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$\begingroup$ 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. $\endgroup$– RobertCommented Jul 22, 2017 at 18:25
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$\begingroup$ @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. $\endgroup$– PhlarxCommented Jul 25, 2017 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 $\endgroup$– KarunCommented Jul 19, 2017 at 20:50
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Move the top left and middle sticks to the bottom left and into the left zero to get:
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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,
Where ω1 is the first uncountable ordinal and the end result is
ω111100
answered Jul 20, 2017 at 21:14
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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. $\endgroup$ Commented Jul 20, 2017 at 21:51
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$\begingroup$ 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... $\endgroup$ Commented Jul 21, 2017 at 6:38
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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 !!
answered Jul 17, 2017 at 12:48
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4$\begingroup$ 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 ♦Commented Jul 17, 2017 at 12:56
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$\begingroup$ 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 ;-) $\endgroup$ Commented Jul 17, 2017 at 13:01
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$\begingroup$ Well, @GarethMcCaughan ... | does not mean divide by...but means modulo (in Number Theory !) and again we boil down to division ...I think... $\endgroup$ Commented Jul 17, 2017 at 13:36
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2$\begingroup$ 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 ♦Commented Jul 17, 2017 at 13:48
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2$\begingroup$ No, it's merely false because 11100 is not 0 times any integer :-). $\endgroup$– Gareth McCaughan ♦Commented Jul 17, 2017 at 15:19
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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.
