6
$\begingroup$

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.

$\endgroup$
4
  • $\begingroup$ Do you at least have one solution in mind here? $\endgroup$
    – Deusovi
    Commented Jun 28, 2016 at 2:42
  • 1
    $\begingroup$ I had a couple, but they are (apparently very) suboptimal. Wanted to see what else others could come up with. $\endgroup$
    – Nij
    Commented Jun 28, 2016 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
    Commented Jun 29, 2016 at 13:19
  • $\begingroup$ @f'' JonMarkPerry, CodeNewbie, Deusovi, EngineerToast: could you respond to these concerns? $\endgroup$
    – Nij
    Commented Jun 30, 2016 at 1:28

3 Answers 3

10
$\begingroup$

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

$\endgroup$
9
  • $\begingroup$ Very good, it gives me ideas for other solutions. $\endgroup$
    – Nij
    Commented Jun 28, 2016 at 3:35
  • $\begingroup$ oops, missed last case, added it. $\endgroup$ Commented Jun 28, 2016 at 3:37
  • $\begingroup$ You can easily use large number of matchsticks; simply start with 111...15-111...15. $\endgroup$
    – Ankoganit
    Commented Jun 28, 2016 at 3:38
  • 1
    $\begingroup$ @Ankoganit thought that was obvious, added text to that effect $\endgroup$ Commented Jun 28, 2016 at 3:40
  • 1
    $\begingroup$ @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. $\endgroup$ Commented Jun 28, 2016 at 3:55
4
$\begingroup$

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.

$\endgroup$
5
  • $\begingroup$ "Overlaid matches will not count." $\endgroup$ Commented Jun 28, 2016 at 3:30
  • 1
    $\begingroup$ @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
    Commented Jun 28, 2016 at 3:33
  • 1
    $\begingroup$ To get rid of the overlaid problem with $+$, you could do $111-111=0$, $1111-111=1000$, $1111-1111=0$, $11111-1111=10000$, $11111-11111=0$ $\endgroup$ Commented Jun 28, 2016 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
    Commented Jun 28, 2016 at 3:53
  • 1
    $\begingroup$ He mentioned in the OP that it is magnitude so -1222 is actually bigger then +1000 which doesn't match the conditions $\endgroup$
    – gtwebb
    Commented Jun 28, 2016 at 4:21
4
$\begingroup$

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

$\endgroup$
3
  • $\begingroup$ 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
    Commented Jun 28, 2016 at 2:47
  • $\begingroup$ It just says smaller which I read as smaller then the previous $\endgroup$
    – gtwebb
    Commented Jun 28, 2016 at 2:51
  • $\begingroup$ Of course, and fair enough to do so. Can you get something similarly large with the stricter condition? $\endgroup$
    – Nij
    Commented Jun 28, 2016 at 2:52

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