# 5008 out of the box

So I came across this puzzle somewhere:

And the intended answer to it was:

But that's pretty much straight forward. Thinking out of the box you can get a lot bigger "numbers" (values), for example:

801115 only by flipping the intended answer 180 degrees.

Or

800^2 which is 640000

Or

90E8 which is 9000000000

Or even

51108^11

So, my question is, if anything is allowed, for instance, breaking the matchsticks, what would be the maximum value we can achieve?

• So, you can break the matchstick to any number of pieces?
– Sid
Oct 16 '16 at 14:39
• @Gintas "breaking" the matchsticks is not very well defined, I think you should narrow it there, or "all" becomes possible (see answers below). Also there is the question of how the shown numbers are INTERPRETED. If I would use the number 100 to the base 10, well, it's hundert. But what if I use to a (near)-infinite base (X)? then it is X**3 at once. So I guess "all is allowed" is just "too broad" to be meaningful. Oct 16 '16 at 14:44
• @BmyGuest well basically I meant bending the matches, not breaking, for example making ^ symbol Oct 16 '16 at 14:54
• @BmyGuest the intention was to get some creative but at the same time realistic answers Oct 16 '16 at 15:03
• Using en.wikipedia.org/wiki/Knuth%27s_up-arrow_notation and sticking to the same rules as set in the question it is possible to make 5↑108 ~ 10^75 which is significantly larger than 51108^11 (~10^51). Moving six match sticks you can make 5↑↑↑8 which is truly vast. Oct 16 '16 at 17:11

$11^{81105}$
or if you want outside the box, light up a match, burn everything, then make the infinity symbol with the ashes!

• I was definitely wasting my time trying to spell GOOGOL, which would have been much smaller.
– humn
Oct 16 '16 at 18:37

Split the matches into their atoms, then build a row of atoms and just count...

• Why stop there when you can get the subatomic particles? Anyway, I wouldn't be surprised if this wasn't the biggest answer. Oct 17 '16 at 0:03
• @greenturtle3141 The OP's answer $51108^{11}\approx 10^{51.8}$ is already larger; $10^{51}$ protons are as massive as the Earth. Oct 17 '16 at 0:25
• @greenturtle3141. Don't split the atom. You want to...well -> s-media-cache-ak0.pinimg.com/564x/a2/0f/ea/… ? Oct 17 '16 at 14:04
• Considering other posted solutions this one gives surprisingly low number. Oct 19 '16 at 11:08

This is quite similar to a previous question. One of the answers used the Busy Beaver function. Here, we can move two sticks to create 61188, turn it upside down to get 88119, then interpret 88 as BB to get BB119 which could be the 119th term of the Busy Beaver function.

EDIT: I have an even better answer!

You can make any number you want. Simply take any match, break it up, then use Banach - Tarski to make more matches!

• I was about to write up a similar answer! I was trying to get at least $BB(7,918)$, whose value is so large it cannot be proven in ZFC, even given unlimited computational power. Oct 17 '16 at 0:28
• Actually the symbol to denote Busy Beaver function is not BB but Σ. Oct 19 '16 at 13:15
• Is that not summation? In a previous puzzle I'm pretty sure we reached the consensus that BB was totally ok... Oct 19 '16 at 14:44
• @oleslaw Actually, it is well accepted that BB and Σ are both common notations for the busy beaver function. Jul 21 '17 at 0:20

Consider:

1. Take 2 matches from the first zero (51108)
2. Turn everything upside down (80115)
3. Bend 1 match to make an arrow (80↑15) (Knuth up-arrow notation)
4. Break the other match to make 11 over the arrow (80↑^11 15)
5. This is equal to 80↑↑↑↑↑↑↑↑↑↑↑15 which is pretty darn large and persists the numbers' sizes.

Maybe this is a little too "outside the box", but you could have

E8105 , turning upside down and moving the 1 stick from the 8 into the 0 and the other after the 0.

Or, even better,

E805 ! , if a matchstick on its own can be an exclamation mark, I think (10^805)! would be one of the largest possible.

Edit: no breaking or burning required!

Edit2: or, even EVEN better

81105_11, which would look like 11 with 81105 in the top left (tetration), i.e. 11^11^11^11^11^11..... 81105 times. Given the number here http://www.had2know.com/academics/2tetrate5.pdf , I think this may win!

An idea similar to @greenturtle3141's:

Take two matchsticks from the last digit to make it a five. Place one in the center of the second digit, turning it into an 8, then one in the bottom left of the first digit, making it a G. This gives us the G805, or the 805th number in the Graham series. I am unsure if this beats the busy beaver answer.

• Actually G (Graham's number) is a number, not a series. It is a 64th element of the g_n series. Oct 19 '16 at 12:51
• @oleslaw Actually, g_n is not a series, its a sequence. Jul 21 '17 at 0:21

Without breaking any matchsticks or resorting to arrow notation, I believe this answer gives a value ridiculously larger than other answers given so far -

Remove two matchsticks from the final 8, leaving 2. take all matchsticks from all other digits. This provides 19 sticks.
now arrange them thus -

111
11
11
11
11
11
11
11
11
2

• You are supposed to move only 2 sticks.. Oct 19 '16 at 11:30
• @oleslaw most other people didn't do that...
– matt
Jun 9 '20 at 23:11

Respecting the 'move only 2 matches rule and ignoring HEX (dec value only)'I got

9909. 1st match you move from second digit to 1st to get a 9. 2nd match you move from last digit to the second.