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I have a board with some matches on it, in the shape of a number.

By adding one match, you make a larger number appear than before.

By adding a second match, you make it smaller.

By adding a third match, you make it bigger again.

And with a fourth match, you get something the same as I started with. What is the largest amount of matches I could have started with? Prove your concept and how to add the extra matches.

Additional rules:

Once a match is placed, it is stuck on the board in that position. The starting configuration may be determined by you as necessary to complete the puzzle.

Matches must be in identifiably different positions. Overlaid matches will not count.

"Smaller" means magnitude. -1 is the same size as +1. -3 is larger than +2.

The board does not allow matches to pass through it.

All matches must be undamaged.

Numerals must be Arabic, seven-segment, or Roman. Votes will serve as judgement of what is close enough to recognise as a number.

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  • $\begingroup$ Do you at least have one solution in mind here? $\endgroup$ – Deusovi Jun 28 '16 at 2:42
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    $\begingroup$ I had a couple, but they are (apparently very) suboptimal. Wanted to see what else others could come up with. $\endgroup$ – Nij Jun 28 '16 at 2:44
  • $\begingroup$ Why the hold-putting? There's clearly one correct answer. The fact it can be and has been shown in several ways doesn't give it multiplicity. Nor are good answers to the puzzle long at all: the best is only a few lines. There are puzzles requiring far more explanation and which haven't been definitively answered, that remain open. Others have explicitly open-ended possibilities and likewise. Why is this obviously precise and concisely-solved puzzle considered broader or necessarily longer to respond to? $\endgroup$ – Nij Jun 29 '16 at 13:19
  • $\begingroup$ @f'' JonMarkPerry, CodeNewbie, Deusovi, EngineerToast: could you respond to these concerns? $\endgroup$ – Nij Jun 30 '16 at 1:28
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A solution that fits the stricter criteria that numbers must be smaller or larger than all preceding numbers (in a similar fashion to my original, below):

$(6-5)\space\space\space\space\space(6-5)=1$
$(8-5)\space\space\space\space\space(6-5)=3$
$(8-5)\space\space\space\space\space(6-6)=0$
$(8-5)\space\space\space\space\space(8-6)=6$
$(8-5)-(8-6)=1$

Again we can

Prepend our numbers with $8888\cdots8$
$(8888\cdots86-8888\cdots85)\space\space\space\space\space(8888\cdots86-8888\cdots85)=1$
$(8888\cdots88-8888\cdots85)\space\space\space\space\space(8888\cdots86-8888\cdots85) = 3$
$(8888\cdots88-8888\cdots85)\space\space\space\space\space(8888\cdots86-8888\cdots86) = 0$
$(8888\cdots88-8888\cdots85)\space\space\space\space\space(8888\cdots88-8888\cdots86) = 6$
$(8888\cdots88-8888\cdots85)-(8888\cdots88-8888\cdots86) = 1$

For example:

With two leading $8$s on each number, yellow matches are those added:

enter image description here


My first solution...

This uses

$11$ initial matches:

 _     _
|_  _ |_
 _|    _|

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

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

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

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

To increase the number simply

prepend both sides of the difference with $8888\cdots8$

$8888\cdots85-8888\cdots85=0$
$8888\cdots86-8888\cdots85=1$
$8888\cdots86-8888\cdots86=0$
$8888\cdots88-8888\cdots86=2$
$8888\cdots88-8888\cdots88=0$

Alternatively

start with $n$ matches side-by-side for each segment of the three seven segment displays originally shown

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  • $\begingroup$ Very good, it gives me ideas for other solutions. $\endgroup$ – Nij Jun 28 '16 at 3:35
  • $\begingroup$ oops, missed last case, added it. $\endgroup$ – Jonathan Allan Jun 28 '16 at 3:37
  • $\begingroup$ You can easily use large number of matchsticks; simply start with 111...15-111...15. $\endgroup$ – Ankoganit Jun 28 '16 at 3:38
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    $\begingroup$ @Ankoganit thought that was obvious, added text to that effect $\endgroup$ – Jonathan Allan Jun 28 '16 at 3:40
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    $\begingroup$ @Ankoganit - replace each match in my original example with $n$ matches side-by-side, the digits become more and more "squished" and the inserted matches are much thinner segments than the originals, but the result represent the same things. $\endgroup$ – Jonathan Allan Jun 28 '16 at 3:55
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We can begin with

$2n+1$ matchsticks for any positive integer $n$.

I will describe the case $n=3.$

Begin with

$|||-|||$ which is $111-111=0$.

Add a matchstick to get

$||||-|||$ which is $1111-111=1000$.

Add another matchstick to get

$-||||-|||$ which is $-1111-111=-1222.$

Another matchstick may be suitably placed to obtain

$+||||-|||$ which is $+1111-111=1000$.

Now comes the fourth matchstick, which yields

$+||||-||||$ which is $+1111-1111=0$.

So we get back to where we started.

If $+$ counts as overlay, then we can do the following, as suggested by Jonathan Allan:

$|||-|||=0\to ||||-|||=1000\to ||||-||||=0\to |||||-||||=10000\to |||||-|||||=0.$

Obviously, for general $n$, we can start with

$\underbrace{|||\cdots | |}_{n\text{ sticks}}-\underbrace{|||\cdots ||}_{n\text{ sticks}}$.

Then we procced as above.

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  • $\begingroup$ "Overlaid matches will not count." $\endgroup$ – Jonathan Allan Jun 28 '16 at 3:30
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    $\begingroup$ @JonathanAllan I interpreted it as matches which are exactly superimposed, since the preceding sentence says that the matches just have to be in identifiably different position. $\endgroup$ – Ankoganit Jun 28 '16 at 3:33
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    $\begingroup$ To get rid of the overlaid problem with $+$, you could do $111-111=0$, $1111-111=1000$, $1111-1111=0$, $11111-1111=10000$, $11111-11111=0$ $\endgroup$ – Jonathan Allan Jun 28 '16 at 3:50
  • $\begingroup$ @JonathanAllan that's what I had in mind at first; but I was trying to satisfy the stricter condition which the OP mentioned in the comments to gtwebb's answer. Anyway, I'll add this. Thanks for the suggestion. $\endgroup$ – Ankoganit Jun 28 '16 at 3:53
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    $\begingroup$ He mentioned in the OP that it is magnitude so -1222 is actually bigger then +1000 which doesn't match the conditions $\endgroup$ – gtwebb Jun 28 '16 at 4:21
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My solution assumes

Operators are allowed since they aren't explicitely forbidden

Solution

matches ads high as roman numerals go, Will get an actual number tomorrow when I have more time if this method is ok.
DCC..XXXI- DCC..XXXI = 0
DCC..XXXII- DCC..XXXI = 1 larger
DCC..XXXII- DCC..XXXII = 0 smaller
DCC..XXXIII- DCC..XXXII = 1 larger
DCC..XXXIII- DCC..XXXIII = 0 same as original

Or

COULD also us arabic transforming 3s to 8s instead of adding 1s which would make it basically unlimited matchstick.

To make them larger and smaller then the original

8333 - 3333 = 5000
8383 - 3333 = 5050 larger
8383 - 3833 = 4550 smaller then original and preceding
8883 - 3833 = 5050 larger then original and preceding
8883 - 3883 = 5000 back to original
proceed with as many match numbers as you want

Next logical tightening is

larger or smaller then all proceeding numbers, don't have a solution for that yet

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  • $\begingroup$ Good effort. Though it might be argued that 0 is the same as 0 still (larger/smaller than the original, not just the preceding number, would be a tighter condition). $\endgroup$ – Nij Jun 28 '16 at 2:47
  • $\begingroup$ It just says smaller which I read as smaller then the previous $\endgroup$ – gtwebb Jun 28 '16 at 2:51
  • $\begingroup$ Of course, and fair enough to do so. Can you get something similarly large with the stricter condition? $\endgroup$ – Nij Jun 28 '16 at 2:52

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