7
$\begingroup$

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

b) Do so moving four matches.

enter image description here

$\endgroup$
4
  • 1
    $\begingroup$ Does a 1 require 2 matches to form, or is 1 sufficient? $\endgroup$
    – rtaft
    Aug 3 '20 at 13:10
  • $\begingroup$ puzzling.stackexchange.com/questions/100767/… $\endgroup$ Aug 3 '20 at 15:29
  • $\begingroup$ @rtaft 2 matches $\endgroup$ Aug 3 '20 at 16:42
  • $\begingroup$ 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 $\endgroup$ Aug 5 '20 at 12:00

11 Answers 11

19
$\begingroup$

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

enter image description here

$\endgroup$
2
  • 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 '20 at 0:57
  • 2
    $\begingroup$ @JdeBP I don't know, anything that needs half a pargaraph and a wikipedia link for explanation doesn't really work IMO, :-P $\endgroup$ Aug 4 '20 at 8:05
14
$\begingroup$

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.

$\endgroup$
7
  • 1
    $\begingroup$ For b) you could do $7^{43^9}$. $\endgroup$ Aug 2 '20 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. $\endgroup$ Aug 3 '20 at 0:02
  • 1
    $\begingroup$ 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 $3$. $\endgroup$ Aug 3 '20 at 0:04
  • 2
    $\begingroup$ Ok, ok. If you absolutely don't want it I will make my own answer ;-) $\endgroup$ Aug 3 '20 at 0:36
  • 2
    $\begingroup$ @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 $7$. $\endgroup$ Aug 3 '20 at 8:21
7
$\begingroup$

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.

$\endgroup$
3
  • 2
    $\begingroup$ For b) you could do $7^{43^9}$ - sorry, I meant to comment on the other answer. $\endgroup$ Aug 2 '20 at 23:51
  • 2
    $\begingroup$ 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 ;) $\endgroup$ Aug 2 '20 at 23:55
  • 1
    $\begingroup$ 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. $\endgroup$ Aug 2 '20 at 23:58
6
$\begingroup$

For a)

enter image description here

For b)

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

$\endgroup$
2
$\begingroup$

Hard to go wrong with

4477

As obviously

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

$\endgroup$
2
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

(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.

$\endgroup$
1
$\begingroup$

111153

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

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
6
  • $\begingroup$ Hi Cristobol. These are certainly not higher than those given by Paul Panzer and Excited Raichu above. $\endgroup$
    – user69943
    Aug 3 '20 at 19:03
  • $\begingroup$ I keyed off the OP asking for a number, not an equation. $\endgroup$ Aug 3 '20 at 19:05
  • $\begingroup$ Those are expressions, not equations. $\endgroup$ Aug 3 '20 at 19:58
  • $\begingroup$ Okay, got me there...still not numbers, even if the operators aren't visible on their own. $\endgroup$ Aug 3 '20 at 20:03
  • $\begingroup$ Cristobol, what about mine? $\endgroup$ Aug 3 '20 at 21:17
1
$\begingroup$

(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.

$\endgroup$
1
$\begingroup$

(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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.