-1
$\begingroup$

The Math Game

Tilting the Image is not allowed. You can cannot change order of digits. You cannot change the Size if Digits unless you are harry potter. You can increase or decrease space between digits though.

Hint:Think different

$\endgroup$
  • 3
    $\begingroup$ youtube.com/watch?v=9m6S0x-AKNU $\endgroup$ – ielyamani Apr 23 at 19:54
  • 5
    $\begingroup$ Waiting for Harry Potter to get a PSE account in order to answer this puzzle with an advantage :) $\endgroup$ – Brandon_J Apr 23 at 21:49
  • $\begingroup$ @Brandon_J correct He knows magic spells but is missing math spells $\endgroup$ – AmanSharma Apr 23 at 23:45
  • $\begingroup$ Are the typos intended? "can cannot", and "the Size if Digits"? $\endgroup$ – Rafalon Apr 24 at 10:49
  • $\begingroup$ Can you break any of the sticks into pieces? If you can, what is the maximum number of pieces you can break a stick into? $\endgroup$ – asgallant Apr 24 at 16:09

15 Answers 15

30
$\begingroup$

Wisest answer:

a. $5118^{11}$ By removing the two sticks of the zero and placing them on exponent :p

Debatable:

b. $5$^$118$ $= 5^{118}$ by using the caret symbol

Still debatable, allowing different sizes:

c. $11^{5118}$ Allowing that digits can have different sizes

Extremely debatable solution:

d. $5118! \ge 5.10^{16762}$ by cutting sticks

This solution could be acceptable for mathematicians' haters...

e. $56/0 \longrightarrow +\infty$, yes, the divide symbol is smaller than digits...

All of these solutions still remain if you can:

flip by 180° the sheet of paper (or your computer (or your head)), and that is also a debatable action !

Which gives:

a. $8115^{11}$
b. $8$^$115$ $= 8^{115}$
c. $11^{8115}$
d. $8115!$

$\endgroup$
  • $\begingroup$ @EricDuminil I don't get your idea, $5^{118} \simeq 3E82 \ge 51^{18} \simeq 5E30$, am I missunderstanding something ? $\endgroup$ – J.Khamphousone Apr 23 at 19:53
  • 2
    $\begingroup$ rot13(svsgl-avar bire mreb vf n ovttre vasvavgl guna svsgl-fvk bire mreb) $\endgroup$ – ielyamani Apr 23 at 19:58
  • 1
    $\begingroup$ +1, I think that your "a" is the correct solution. $\endgroup$ – Brandon_J Apr 23 at 21:39
  • 1
    $\begingroup$ @J.Khamphousone Your b is correct solution . Hats off to your genuises $\endgroup$ – AmanSharma Apr 23 at 23:44
  • 2
    $\begingroup$ @EricDuminil original poster wrote "You can increase or decrease space between digits though". $\endgroup$ – J.Khamphousone Apr 24 at 10:45
20
$\begingroup$

Without going too far out of the box:

15118 by removing the top and bottom matches from the 0 (creating 2 ones) and using them to create a one at the front

Actually, that should be:

51181 using the same method but putting the new digit at the end.

$\endgroup$
  • 3
    $\begingroup$ Great for a pure number solution without any "tricks" with operators or archaic symbols. $\endgroup$ – JPhi1618 Apr 24 at 17:40
20
$\begingroup$

There are a few

notations for insanely large numbers. The Knuth up arrow operator is one:

Move the top and bottom lines from the zero, to surround the left two lines of the zero:

5 ↑ 18

Trouble is

you need (at least) two of those arrows, or a superscripted exponent, to get really huge numbers. Not easy by moving two lines.

So we try escalating our approach a "few" quadrillion times ...

Alternative solution:

move the left two lines of the zero inward at a diagonal to get this:

5 Σ 18

I defy anyone to compute Σ 18, the value of the Busy Beaver function for an input of 18 ..... let alone 5 x that value.

Evaluation hint: Go directly to "inconceivably big". Do not pass Go. Do not collect £200.

Alternatively,

start at "inconceivably vast" and then scale up an inconceivably vast number of times, or something like that.

$\endgroup$
  • 3
    $\begingroup$ I would have tried something along these lines as well. Have my upvote instead. Still waiting for the "long long long long .... way behind the largest Number" comment by the OP :) $\endgroup$ – Arnaud Mortier Apr 23 at 20:48
  • 1
    $\begingroup$ Not in this case, I fear ;-) $\endgroup$ – Stilez Apr 23 at 20:51
  • 1
    $\begingroup$ Assuming 5 x BB(18) is a valid answer, I think it's probably smaller than g98 (as suggested by Sconibulus). Extrapolating blindly from the known bounds on BB(5), BB(6), and BB(7), the exponents don't seem to grow fast enough to beat out Graham's Number before BB(18). (Interestingly, there must be some N for which g(N) < BB(N), since the BB function grows faster than any computable function, which surprisingly includes g(N).) But proving which is larger may literally be beyond the capabilities of mathematics and computation, so... I'll call it a draw. $\endgroup$ – Gilad M Apr 23 at 22:35
  • 1
    $\begingroup$ This seems to beat the intended and currently accepted answer. $\endgroup$ – Jay Apr 24 at 2:31
  • 5
    $\begingroup$ @AmanSharma - come on, it might or might not be calculable (in a practical sense or at all, we don't know which), but that wasn't a requirement. The requirement was a big number, and its no surprise if you ask for a big number you get a huuuge one. Its definitely the larger number: it definitely has a specific but unknown value, and a huuuge known lower bound. Even non-calculable would mean we cant evaluate it, not that it doesn't exist. Seems as if you already have a specific smaller number fixed in your mind, which isn't the spirit of such puzzles. Admit it, the expected answer got beaten :) $\endgroup$ – Stilez Apr 24 at 5:45
14
$\begingroup$

Slightly out of the box, but probably legal.

6E8, moving the two right sticks of the 0. 9E8 might be possible, but I don't think that's the accepted way of making a digital 9.

So far out of the box it's probably illegal.

g98 in Graham's Notation, where Graham's Number is g64.

$\endgroup$
  • $\begingroup$ None of the comments here are pertinent. Telling someone their answer is not in the running isn't necessary when the evidence of that in other answers speaks for itself. Calling someone "bro" in such a comment doesn't need to be done. Discussing calling someone "bro" in subsequent comments is not what comments on an answer are for. Let's at least try to keep comments relevant and constructive, please. Various comments on this and other answers deleted. $\endgroup$ – Rubio Apr 25 at 2:22
7
$\begingroup$

Not a serious answer.

Take one stick, put it aside.

Take another stick, light it and set fire to everything remaining.

Use the stick you saved to shift the ashes around and form any number you wish!

I only moved two sticks...

and a bunch of ash. 😋

$\endgroup$
  • 2
    $\begingroup$ This also works if you turn the sticks to charcoal and write with them. $\endgroup$ – Mindwin Apr 24 at 13:10
6
$\begingroup$

Without adding extra digits and keeping with the digital-like format,

938 999 @Rafalon, great catch

You can do this by

Removing the two matches on the left side of the zero and turning on horizontal to make the 0 a 3, and the other match to the top right of the 5 to turn it into a 9.

Removing the bottom left stick from 0 to the middle, and moving the bottom-left stick from the 8 to the top-right of the 5

$\endgroup$
  • 1
    $\begingroup$ rot13(Lbh pbhyq ernpu 999 ol zbivat gur obggbz-yrsg fgvpx sebz 0 gb gur zvqqyr, naq zbivat gur obggbz-yrsg fgvpx sebz gur 8 gb gur gbc-evtug bs gur 5) $\endgroup$ – Rafalon Apr 24 at 10:54
  • $\begingroup$ @Rafalon Genius :] $\endgroup$ – Sensoray Apr 24 at 12:04
5
$\begingroup$

Without taking too many liberties with the possibilities when it comes to rules...

I would say that the largest number made by moving only two sticks and without invoking any sort of exponents is:

15118 created by moving the top sticks from the zero to make a one in front of the 5...

$\endgroup$
  • 1
    $\begingroup$ You can place the one at the end. But I agree with @AmanSharma remains a sub-optimal solution $\endgroup$ – J.Khamphousone Apr 23 at 19:39
2
$\begingroup$

I'm going to guess either

999 as that removes the possibility of any operators/exponents/etc.

OR

80E which converts to 100000001110 as binary

$\endgroup$
1
$\begingroup$

According to http://mathworld.wolfram.com/BusyBeaver.html, "some authors define a busy beaver as a Turing machine that performs a maximum number S(n) of steps when started on an initially blank tape before halting." and S "has the first few terms 1, 6, 21, 107, ... the next few terms of S(n) are not known, but explicit constructions give lower bounds of S(5)>=47176870 and S(6)>=3×101730."

Luckily, 5 and S are identical in matchstick font. So, the horizontal bars of 0 can be moved to make...

S11118 i.e. S(11118), or even S L118 i.e. S(50118) where L is the Roman number for 50.

Even better, the 0 can be turned into another S, leaving a matchstick to make a short 1: SS8ı i.e. S(S(81)).

These numbers are so big that they cannot be described in decimal form or even with power towers.

$\endgroup$
  • $\begingroup$ Spoiler tag is not working for me $\endgroup$ – Gnubie Apr 25 at 21:48
0
$\begingroup$

Without crazy rules, operations or extra digits:

980

How I did it:

Move the bottom-most match from the 5 to change the 5 to a 9. Then move the center match from the 8 to the 0, to form an 8.

$\endgroup$
0
$\begingroup$

2

How I got it:

I took two sticks since that's all I could move.

Subsequently,

I made a '1' out of them...

$\endgroup$
  • 2
    $\begingroup$ It appears that the question wording has confused you. The point of the puzzle is to combine the two moved matchsticks with the remaining (unmoved) matchsticks to create a number, not merely use the two used matchsticks. $\endgroup$ – Brandon_J Apr 23 at 22:24
0
$\begingroup$

Take bottom left stick from 0 and place it at Top Right of 5. Now take bottom left stick from 8 and place middle of previously 0 digit. Thus, getting 999.

$\endgroup$
0
$\begingroup$

407.7 (but how... ;)

150e, where e is Euler's constant; approximately 2.71828.

$\endgroup$
  • $\begingroup$ I have already marked the correct answer See if it helps YOu. $\endgroup$ – AmanSharma Apr 24 at 3:07
  • $\begingroup$ I just wanted to use that 'number', but I forgot how small it was. 'why' is even smaller, and 'eye' doesn't even exist. $\endgroup$ – Mazura Apr 24 at 3:10
  • 1
    $\begingroup$ Or you could move 0 sticks and have a larger number... $\endgroup$ – Draco18s Apr 24 at 19:44
0
$\begingroup$

I was going to say move 2 sticks in 5 to turn it into an F which would give you the number "F88" in hexadecimal. Which is 3976 in decimal, but then I noticed others were saying 15118. So I could just easily say that as my answer, but that number is hexadecimal. and 15118 in hexadecimal is 86296.

So.

$\endgroup$
0
$\begingroup$

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

that is

$\beth_{88}$,

rather large

transfinite number,

but unlike

1/0

and similar, it is well defined.

$\endgroup$
  • $\begingroup$ You are not allowed to change size of Numbers bro. But thanks for Information $\endgroup$ – AmanSharma Apr 25 at 15:14

protected by Community Apr 24 at 5:25

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

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