# Move matches and make largest possible number

a) Move three matches to make the largest possible number.

b) Do so moving four matches. • Does a 1 require 2 matches to form, or is 1 sufficient? – rtaft Aug 3 '20 at 13:10
• puzzling.stackexchange.com/questions/100767/… – Bernardo Recamán Santos Aug 3 '20 at 15:29
• @rtaft 2 matches – María Lucía Uribe Aug 3 '20 at 16:42
• Math Nerd points out there is no "largest possible number" Pick your infinity and then generate $2^{yourInfinity}$ . :-) //yes I know that's not the point here – Carl Witthoft Aug 5 '20 at 12:00

I can't beat @Excited Raichu's a) but for b):

• If g is allowed, then 3 matches gets one $g1173$, the 1173rd value in the sequence that includes Graham's Number at $g64$. It's slightly bigger than that result. (-: – JdeBP Aug 4 '20 at 0:57
• @JdeBP I don't know, anything that needs half a pargaraph and a wikipedia link for explanation doesn't really work IMO, :-P – Paul Panzer Aug 4 '20 at 8:05

a) Not certain this is the biggest, but I can't find anything bigger than

b) Moving 4 matches allows

which is pretty big. Not sure if it's the absolute biggest, but it's up there.

This is, of course, assuming that the digit 1 must be two sticks high. There's definitely a higher ceiling if it can only be one.

• For b) you could do $7^{43^9}$. – Paul Panzer Aug 2 '20 at 23:55
• @PaulPanzer how is that? Making a 7 and a 9, both on other lines besides the main one, would require 9 movements. – Excited Raichu Aug 3 '20 at 0:02
• You only offset half a line which is more within typographic conventions anyway. So for the $7$ you keep two sticks of the first $4$ exactly where they are and for the $9$ three sticks of the second $3$. – Paul Panzer Aug 3 '20 at 0:04
• Ok, ok. If you absolutely don't want it I will make my own answer ;-) – Paul Panzer Aug 3 '20 at 0:36
• @Chronocidal You can look at the first revision of my answer for that. Note that one of the new sticks in the $9$ does not come from the $3$ but is left over from the $4$ that becomes a $7$. – Paul Panzer Aug 3 '20 at 8:21

a)
Is it:

4431111

b)
Is it:

44771111
I added 4 extra digits which multiply the digit by 10 and add 1 each time it does so. I added them on the back since 4 is a bigger number than 1. (Obviously, but for explanation's sake it's here). This logic goes for a) as well.

• For b) you could do $7^{43^9}$ - sorry, I meant to comment on the other answer. – Paul Panzer Aug 2 '20 at 23:51
• yes i could, but why do that when you could post an answer yourself? I don't really want to leach off of other people. Im sure you understand ;) – TruVortex_07 Aug 2 '20 at 23:55
• As I said I meant to comment on the other answer and since my contribution is just a minor optimization on their major idea I thought it wouldn't warrant an answer of my own. – Paul Panzer Aug 2 '20 at 23:58

For a)

For b)

Turn the first 4 into two 1s to make 1,111,731

Hard to go wrong with

4477

As obviously

there is no such thing as a "largest possible number", but infinity is a common misconception ;)

Moving 4 matches to form

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


and viewing it upside down gives S S 1111 where S 1111 is the maximum shifts function of the busy beaver game with 1111 states http://en.wikipedia.org/wiki/Busy_beaver#Maximum_shifts_function_S . S 6 is 3.515E+18267 and S 7 is already 10^10^10^10^18705353.

S S 1111 is the maximum shifts function of the busy beaver game with S 1111 states, a number uncomputably large, but still finite.

(b) Move 4 matches to form

            _  _
|_| /|\ /|\  | _|
|          | _|


or 4 ↑↑ 73, where ↑ is Knuth's up-arrow notation http://en.wikipedia.org/wiki/Knuth%27s_up-arrow_notation

4^4^4^4^4^4^4^4^4^4^4^4^4^4^4^4^4^4^4^4^4^4^4^4^4^4^4^4^4^4^4^4^4^4^4^4^4^4^4^4^4^4^4^4^4^4^4^4^4^4^4^4^4^4^4^4^4^4^4^4^4^4^4^4^4^4^4^4^4^4^4^4^4 (73 4's) is too large to compute in Wolframalpha.

            _  _
|_| /|\ |||  | _|
|          | _|


A convention allows multiple ↑'s to be specified with a superscript, so 4 ↑¹¹¹ 73 = 4 ↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑ 73 (111 ↑'s), a number too large to represent even with power towers.

111153

Moving 2 Match sticks, the 4s can be made into 11s. And the third match stick would be moved to get 5.

Moving 3 matches, I'd

take them all from the last 3 to make a 1

giving

74431

My best guess for 4 moves is to

borrow 2 from each 3, and use them to make 1s

to get

447711

I think that's the highest so far, if each one needs two matchsticks.

• Hi Cristobol. These are certainly not higher than those given by Paul Panzer and Excited Raichu above. – user69943 Aug 3 '20 at 19:03
• I keyed off the OP asking for a number, not an equation. – Cristobol Polychronopolis Aug 3 '20 at 19:05
• Those are expressions, not equations. – Ross Millikan Aug 3 '20 at 19:58
• Okay, got me there...still not numbers, even if the operators aren't visible on their own. – Cristobol Polychronopolis Aug 3 '20 at 20:03
• Cristobol, what about mine? – TruVortex_07 Aug 3 '20 at 21:17

(b) Move 4 matches to form

      _      _
| |_| _|  / |_
|   | _| /  |_


The slash has to be squeezed in between the 3's to make 143/ε. The last character represents epsilon, an arbritarily small positive quantity used in the definition of limit in calculus, which is greater than zero to avoid division by zero.

(a) Move 3 matches to form

     _  _     _
|_| | | _| / |_
|   | _|    _


or 473/ε. The vertical match of ε is centered over the middle bar.

(b) Move 4 matches to form

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


which viewed upside down is Γ Γ 661, where Γ is the gamma function http://en.wikipedia.org/wiki/Gamma_function

Γ 661 ~ 10^1576 and Γ Γ 661 ~ 10^10^1579

which is slightly larger than 7^7^473 ~ 10^10^400.

Interpreting it as Γ Γ bb1 where bb1 in hexadecimal is 2993 in decimal,

Γ 2993 ~ 10^9103 and Γ Γ 2993 is too large to compute in Wolframalpha but is larger than Γ Γ 661.

Even better are

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


Γ Γ 66¹¹ and Γ Γ bb¹¹, respectively.