7
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a) Move three matches to make the largest possible number.

b) Do so moving four matches.

enter image description here

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11 Answers 11

19
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I can't beat @Excited Raichu's a) but for b):

enter image description here

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  • 3
    $\begingroup$ 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. (-: $\endgroup$ – JdeBP Aug 4 at 0:57
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    $\begingroup$ @JdeBP I don't know, anything that needs half a pargaraph and a wikipedia link for explanation doesn't really work IMO, :-P $\endgroup$ – Paul Panzer Aug 4 at 8:05
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a) Not certain this is the biggest, but I can't find anything bigger than

enter image description here

b) Moving 4 matches allows

enter image description here

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.

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    $\begingroup$ For b) you could do $7^{43^9}$. $\endgroup$ – Paul Panzer Aug 2 at 23:55
  • $\begingroup$ @PaulPanzer how is that? Making a 7 and a 9, both on other lines besides the main one, would require 9 movements. $\endgroup$ – Excited Raichu Aug 3 at 0:02
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    $\begingroup$ 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 $3$. $\endgroup$ – Paul Panzer Aug 3 at 0:04
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    $\begingroup$ Ok, ok. If you absolutely don't want it I will make my own answer ;-) $\endgroup$ – Paul Panzer Aug 3 at 0:36
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    $\begingroup$ @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 $7$. $\endgroup$ – Paul Panzer Aug 3 at 8:21
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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.

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    $\begingroup$ For b) you could do $7^{43^9}$ - sorry, I meant to comment on the other answer. $\endgroup$ – Paul Panzer Aug 2 at 23:51
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    $\begingroup$ 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 ;) $\endgroup$ – TruVortex_07 Aug 2 at 23:55
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    $\begingroup$ 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. $\endgroup$ – Paul Panzer Aug 2 at 23:58
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For a)

enter image description here

For b)

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

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2
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Hard to go wrong with

4477

As obviously

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

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2
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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.

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2
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(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.

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1
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111153

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

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1
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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.

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  • $\begingroup$ Hi Cristobol. These are certainly not higher than those given by Paul Panzer and Excited Raichu above. $\endgroup$ – Daniel C Aug 3 at 19:03
  • $\begingroup$ I keyed off the OP asking for a number, not an equation. $\endgroup$ – Cristobol Polychronopolis Aug 3 at 19:05
  • $\begingroup$ Those are expressions, not equations. $\endgroup$ – Ross Millikan Aug 3 at 19:58
  • $\begingroup$ Okay, got me there...still not numbers, even if the operators aren't visible on their own. $\endgroup$ – Cristobol Polychronopolis Aug 3 at 20:03
  • $\begingroup$ Cristobol, what about mine? $\endgroup$ – TruVortex_07 Aug 3 at 21:17
1
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(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.

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1
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(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.

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