What is ls Largest Number Formed by only moving two sticks in 508? Unicorn Meta Zoo #1: Why...
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What is ls Largest Number Formed by only moving two sticks in 508?
Unicorn Meta Zoo #1: Why another podcast?
Announcing the arrival of Valued Associate #679: Cesar ManaraLargest number with five 1's and five numeric operations90s Number PuzzleWhat is the largest number you can create with 0000 by moving only two sticks?Make numbers 1 - 32 using the digits 2, 0, 1, 7Yet another matchstick puzzleMathematical riddleMake numbers 1 - 30 using the digits 2, 0, 1, 8Four points with only two distancesCreate Numbers 1 - 100 using 1,9,6,8Find 108 by using 3,4,6
$begingroup$
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
mathematics matches
asked 10 hours ago
AmanSharmaAmanSharma
1496
$endgroup$
add a comment |
$begingroup$
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
mathematics matches
asked 10 hours ago
AmanSharmaAmanSharma
1496
$endgroup$
3
$begingroup$
youtube.com/watch?v=9m6S0x-AKNU
$endgroup$
– ielyamani
9 hours ago
2
$begingroup$
Waiting for Harry Potter to get a PSE account in order to answer this puzzle with an advantage :)
$endgroup$
– Brandon_J
7 hours ago
$begingroup$
@Brandon_J correct He knows magic spells but is missing math spells
$endgroup$
– AmanSharma
5 hours ago
add a comment |
$begingroup$
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
mathematics matches
asked 10 hours ago
AmanSharmaAmanSharma
1496
$endgroup$
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
mathematics matches
mathematics matches
asked 10 hours ago
AmanSharmaAmanSharma
1496
asked 10 hours ago
AmanSharmaAmanSharma
1496
edited 10 hours ago
AmanSharma
asked 10 hours ago
AmanSharmaAmanSharma
1496
asked 10 hours ago
AmanSharmaAmanSharma
1496
asked 10 hours ago
AmanSharmaAmanSharma
1496
1496
3
$begingroup$
youtube.com/watch?v=9m6S0x-AKNU
$endgroup$
– ielyamani
9 hours ago
2
$begingroup$
Waiting for Harry Potter to get a PSE account in order to answer this puzzle with an advantage :)
$endgroup$
– Brandon_J
7 hours ago
$begingroup$
@Brandon_J correct He knows magic spells but is missing math spells
$endgroup$
– AmanSharma
5 hours ago
add a comment |
3
$begingroup$
youtube.com/watch?v=9m6S0x-AKNU
$endgroup$
– ielyamani
9 hours ago
2
$begingroup$
Waiting for Harry Potter to get a PSE account in order to answer this puzzle with an advantage :)
$endgroup$
– Brandon_J
7 hours ago
$begingroup$
@Brandon_J correct He knows magic spells but is missing math spells
$endgroup$
– AmanSharma
5 hours ago
3
3
$begingroup$
youtube.com/watch?v=9m6S0x-AKNU
$endgroup$
– ielyamani
9 hours ago
$begingroup$
youtube.com/watch?v=9m6S0x-AKNU
$endgroup$
– ielyamani
9 hours ago
2
2
$begingroup$
Waiting for Harry Potter to get a PSE account in order to answer this puzzle with an advantage :)
$endgroup$
– Brandon_J
7 hours ago
$begingroup$
Waiting for Harry Potter to get a PSE account in order to answer this puzzle with an advantage :)
$endgroup$
– Brandon_J
7 hours ago
$begingroup$
@Brandon_J correct He knows magic spells but is missing math spells
$endgroup$
– AmanSharma
5 hours ago
$begingroup$
@Brandon_J correct He knows magic spells but is missing math spells
$endgroup$
– AmanSharma
5 hours ago
add a comment |
13 Answers
13
active
oldest
votes
$begingroup$
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. $5118! ge 5^{16762}$
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), and that is also a debatable action !
Which gives :
a. $8115^{11}$
b. $8$^$115$ $= 8^{115}$
c. $11^{8115}$
d. $8115!$
answered 10 hours ago
J.KhamphousoneJ.Khamphousone
31518
$endgroup$
2
$begingroup$
That was out of universe . We are not changing size of digits here Bro. Also infinity isn't a number.
$endgroup$
– AmanSharma
10 hours ago
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@AmanSharma I know ! That's why I wrote it's debetable, but my wisest answer still remains
$endgroup$
– J.Khamphousone
9 hours ago
$begingroup$
@AmanSharma I've added another debatable solution using the Caret Symbol
$endgroup$
– J.Khamphousone
9 hours ago
1
$begingroup$
rot13(svsgl-avar bire mreb vf n ovttre vasvavgl guna svsgl-fvk bire mreb)
$endgroup$
– ielyamani
9 hours ago
1
$begingroup$
Haha @EricDuminil you don't have to be sorry at all, that didn't bother me ! That's one of the points of StackExchange, help, debate and exchange ;p
$endgroup$
– J.Khamphousone
9 hours ago
|
show 8 more comments
$begingroup$
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.
answered 10 hours ago
SconibulusSconibulus
14.8k128102
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2
$begingroup$
Legal But Still Bro You are long long long long .... way behind the largest Number
$endgroup$
– AmanSharma
10 hours ago
$begingroup$
@AmanSharma Cheated a bit harder
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– Sconibulus
10 hours ago
19
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@AmanSharma: You realize you might not be talking to a "Bro" or to someone who doesn't like being called "Bro"?
$endgroup$
– Eric Duminil
9 hours ago
5
$begingroup$
@EricDuminil I agree with the sentiment of your comment (upvoted it), but FWIW I know several people who say "Bro" in every other sentence in real-life, day-to-day conversations, whether speaking to a female or not; I doubt that Aman is assuming genders in this scenario.
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– Brandon_J
7 hours ago
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@EricDuminil Chillax brah
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– BlueRaja - Danny Pflughoeft
5 hours ago
add a comment |
$begingroup$
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.
answered 9 hours ago
StilezStilez
1,314211
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2
$begingroup$
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
8 hours ago
1
$begingroup$
Not in this case, I fear ;-)
$endgroup$
– Stilez
8 hours ago
$begingroup$
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.
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– Gilad M
6 hours ago
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nice +1! yet for the summation, the two horizontal lines would touch the endpoints of the 1, making it more like an inverted B...
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– Omega Krypton
5 hours ago
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This seems to beat the intended and currently accepted answer.
$endgroup$
– Jay
2 hours ago
|
show 2 more comments
$begingroup$
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.
answered 10 hours ago
PugmonkeyPugmonkey
3,6401220
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1
$begingroup$
Bro You are long long long long .... way behind the largest Number
$endgroup$
– AmanSharma
10 hours ago
add a comment |
$begingroup$
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...
answered 10 hours ago
Dr tDr t
1,258313
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1
$begingroup$
Bro You are long long long long .... way behind the largest Number
$endgroup$
– AmanSharma
10 hours ago
1
$begingroup$
You can place the one at the end. But I agree with @AmanSharma remains a sub-optimal solution
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– J.Khamphousone
9 hours ago
add a comment |
$begingroup$
Without adding extra digits and keeping with the digital-like format,
938
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.
answered 10 hours ago
SensoraySensoray
4,57311246
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add a comment |
$begingroup$
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
answered 8 hours ago
visualnotsobasicvisualnotsobasic
3239
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add a comment |
$begingroup$
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. 😋
answered 5 hours ago
Hand-E-FoodHand-E-Food
1492
$endgroup$
add a comment |
$begingroup$
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.
answered 7 hours ago
Lorenzo MLorenzo M
1
New contributor
Lorenzo M is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
2
How I got it:
I took two sticks since that's all I could move.
Subsequently,
I made a '1' out of them...
answered 7 hours ago
Tseug EmanTseug Eman
1
New contributor
Tseug Eman is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
1
$begingroup$
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.
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– Brandon_J
7 hours ago
add a comment |
$begingroup$
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.
answered 5 hours ago
OM PrakashOM Prakash
1
New contributor
OM Prakash is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
407.7 (but how... ;)
150e, where e is Euler's constant; approximately 2.71828.
answered 3 hours ago
MazuraMazura
865514
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I have already marked the correct answer See if it helps YOu.
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– AmanSharma
2 hours ago
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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
2 hours ago
add a comment |
$begingroup$
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.
answered 1 min ago
Andrew900460Andrew900460
1
New contributor
Andrew900460 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
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13 Answers
13
active
oldest
votes
13 Answers
13
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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. $5118! ge 5^{16762}$
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), and that is also a debatable action !
Which gives :
a. $8115^{11}$
b. $8$^$115$ $= 8^{115}$
c. $11^{8115}$
d. $8115!$
answered 10 hours ago
J.KhamphousoneJ.Khamphousone
31518
$endgroup$
2
$begingroup$
That was out of universe . We are not changing size of digits here Bro. Also infinity isn't a number.
$endgroup$
– AmanSharma
10 hours ago
$begingroup$
@AmanSharma I know ! That's why I wrote it's debetable, but my wisest answer still remains
$endgroup$
– J.Khamphousone
9 hours ago
$begingroup$
@AmanSharma I've added another debatable solution using the Caret Symbol
$endgroup$
– J.Khamphousone
9 hours ago
1
$begingroup$
rot13(svsgl-avar bire mreb vf n ovttre vasvavgl guna svsgl-fvk bire mreb)
$endgroup$
– ielyamani
9 hours ago
1
$begingroup$
Haha @EricDuminil you don't have to be sorry at all, that didn't bother me ! That's one of the points of StackExchange, help, debate and exchange ;p
$endgroup$
– J.Khamphousone
9 hours ago
|
show 8 more comments
$begingroup$
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. $5118! ge 5^{16762}$
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), and that is also a debatable action !
Which gives :
a. $8115^{11}$
b. $8$^$115$ $= 8^{115}$
c. $11^{8115}$
d. $8115!$
answered 10 hours ago
J.KhamphousoneJ.Khamphousone
31518
$endgroup$
2
$begingroup$
That was out of universe . We are not changing size of digits here Bro. Also infinity isn't a number.
$endgroup$
– AmanSharma
10 hours ago
$begingroup$
@AmanSharma I know ! That's why I wrote it's debetable, but my wisest answer still remains
$endgroup$
– J.Khamphousone
9 hours ago
$begingroup$
@AmanSharma I've added another debatable solution using the Caret Symbol
$endgroup$
– J.Khamphousone
9 hours ago
1
$begingroup$
rot13(svsgl-avar bire mreb vf n ovttre vasvavgl guna svsgl-fvk bire mreb)
$endgroup$
– ielyamani
9 hours ago
1
$begingroup$
Haha @EricDuminil you don't have to be sorry at all, that didn't bother me ! That's one of the points of StackExchange, help, debate and exchange ;p
$endgroup$
– J.Khamphousone
9 hours ago
|
show 8 more comments
$begingroup$
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. $5118! ge 5^{16762}$
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), and that is also a debatable action !
Which gives :
a. $8115^{11}$
b. $8$^$115$ $= 8^{115}$
c. $11^{8115}$
d. $8115!$
answered 10 hours ago
J.KhamphousoneJ.Khamphousone
31518
$endgroup$
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. $5118! ge 5^{16762}$
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), and that is also a debatable action !
Which gives :
a. $8115^{11}$
b. $8$^$115$ $= 8^{115}$
c. $11^{8115}$
d. $8115!$
answered 10 hours ago
J.KhamphousoneJ.Khamphousone
31518
edited 9 hours ago
answered 10 hours ago
J.KhamphousoneJ.Khamphousone
31518
answered 10 hours ago
J.KhamphousoneJ.Khamphousone
31518
answered 10 hours ago
J.KhamphousoneJ.Khamphousone
31518
31518
2
$begingroup$
That was out of universe . We are not changing size of digits here Bro. Also infinity isn't a number.
$endgroup$
– AmanSharma
10 hours ago
$begingroup$
@AmanSharma I know ! That's why I wrote it's debetable, but my wisest answer still remains
$endgroup$
– J.Khamphousone
9 hours ago
$begingroup$
@AmanSharma I've added another debatable solution using the Caret Symbol
$endgroup$
– J.Khamphousone
9 hours ago
1
$begingroup$
rot13(svsgl-avar bire mreb vf n ovttre vasvavgl guna svsgl-fvk bire mreb)
$endgroup$
– ielyamani
9 hours ago
1
$begingroup$
Haha @EricDuminil you don't have to be sorry at all, that didn't bother me ! That's one of the points of StackExchange, help, debate and exchange ;p
$endgroup$
– J.Khamphousone
9 hours ago
|
show 8 more comments
2
$begingroup$
That was out of universe . We are not changing size of digits here Bro. Also infinity isn't a number.
$endgroup$
– AmanSharma
10 hours ago
$begingroup$
@AmanSharma I know ! That's why I wrote it's debetable, but my wisest answer still remains
$endgroup$
– J.Khamphousone
9 hours ago
$begingroup$
@AmanSharma I've added another debatable solution using the Caret Symbol
$endgroup$
– J.Khamphousone
9 hours ago
1
$begingroup$
rot13(svsgl-avar bire mreb vf n ovttre vasvavgl guna svsgl-fvk bire mreb)
$endgroup$
– ielyamani
9 hours ago
1
$begingroup$
Haha @EricDuminil you don't have to be sorry at all, that didn't bother me ! That's one of the points of StackExchange, help, debate and exchange ;p
$endgroup$
– J.Khamphousone
9 hours ago
2
2
$begingroup$
That was out of universe . We are not changing size of digits here Bro. Also infinity isn't a number.
$endgroup$
– AmanSharma
10 hours ago
$begingroup$
That was out of universe . We are not changing size of digits here Bro. Also infinity isn't a number.
$endgroup$
– AmanSharma
10 hours ago
$begingroup$
@AmanSharma I know ! That's why I wrote it's debetable, but my wisest answer still remains
$endgroup$
– J.Khamphousone
9 hours ago
$begingroup$
@AmanSharma I know ! That's why I wrote it's debetable, but my wisest answer still remains
$endgroup$
– J.Khamphousone
9 hours ago
$begingroup$
@AmanSharma I've added another debatable solution using the Caret Symbol
$endgroup$
– J.Khamphousone
9 hours ago
$begingroup$
@AmanSharma I've added another debatable solution using the Caret Symbol
$endgroup$
– J.Khamphousone
9 hours ago
1
1
$begingroup$
rot13(svsgl-avar bire mreb vf n ovttre vasvavgl guna svsgl-fvk bire mreb)
$endgroup$
– ielyamani
9 hours ago
$begingroup$
rot13(svsgl-avar bire mreb vf n ovttre vasvavgl guna svsgl-fvk bire mreb)
$endgroup$
– ielyamani
9 hours ago
1
1
$begingroup$
Haha @EricDuminil you don't have to be sorry at all, that didn't bother me ! That's one of the points of StackExchange, help, debate and exchange ;p
$endgroup$
– J.Khamphousone
9 hours ago
$begingroup$
Haha @EricDuminil you don't have to be sorry at all, that didn't bother me ! That's one of the points of StackExchange, help, debate and exchange ;p
$endgroup$
– J.Khamphousone
9 hours ago
|
show 8 more comments
$begingroup$
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.
answered 10 hours ago
SconibulusSconibulus
14.8k128102
$endgroup$
2
$begingroup$
Legal But Still Bro You are long long long long .... way behind the largest Number
$endgroup$
– AmanSharma
10 hours ago
$begingroup$
@AmanSharma Cheated a bit harder
$endgroup$
– Sconibulus
10 hours ago
19
$begingroup$
@AmanSharma: You realize you might not be talking to a "Bro" or to someone who doesn't like being called "Bro"?
$endgroup$
– Eric Duminil
9 hours ago
5
$begingroup$
@EricDuminil I agree with the sentiment of your comment (upvoted it), but FWIW I know several people who say "Bro" in every other sentence in real-life, day-to-day conversations, whether speaking to a female or not; I doubt that Aman is assuming genders in this scenario.
$endgroup$
– Brandon_J
7 hours ago
$begingroup$
@EricDuminil Chillax brah
$endgroup$
– BlueRaja - Danny Pflughoeft
5 hours ago
add a comment |
$begingroup$
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.
answered 10 hours ago
SconibulusSconibulus
14.8k128102
$endgroup$
2
$begingroup$
Legal But Still Bro You are long long long long .... way behind the largest Number
$endgroup$
– AmanSharma
10 hours ago
$begingroup$
@AmanSharma Cheated a bit harder
$endgroup$
– Sconibulus
10 hours ago
19
$begingroup$
@AmanSharma: You realize you might not be talking to a "Bro" or to someone who doesn't like being called "Bro"?
$endgroup$
– Eric Duminil
9 hours ago
5
$begingroup$
@EricDuminil I agree with the sentiment of your comment (upvoted it), but FWIW I know several people who say "Bro" in every other sentence in real-life, day-to-day conversations, whether speaking to a female or not; I doubt that Aman is assuming genders in this scenario.
$endgroup$
– Brandon_J
7 hours ago
$begingroup$
@EricDuminil Chillax brah
$endgroup$
– BlueRaja - Danny Pflughoeft
5 hours ago
add a comment |
$begingroup$
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.
answered 10 hours ago
SconibulusSconibulus
14.8k128102
$endgroup$
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.
answered 10 hours ago
SconibulusSconibulus
14.8k128102
edited 10 hours ago
answered 10 hours ago
SconibulusSconibulus
14.8k128102
answered 10 hours ago
SconibulusSconibulus
14.8k128102
answered 10 hours ago
SconibulusSconibulus
14.8k128102
14.8k128102
2
$begingroup$
Legal But Still Bro You are long long long long .... way behind the largest Number
$endgroup$
– AmanSharma
10 hours ago
$begingroup$
@AmanSharma Cheated a bit harder
$endgroup$
– Sconibulus
10 hours ago
19
$begingroup$
@AmanSharma: You realize you might not be talking to a "Bro" or to someone who doesn't like being called "Bro"?
$endgroup$
– Eric Duminil
9 hours ago
5
$begingroup$
@EricDuminil I agree with the sentiment of your comment (upvoted it), but FWIW I know several people who say "Bro" in every other sentence in real-life, day-to-day conversations, whether speaking to a female or not; I doubt that Aman is assuming genders in this scenario.
$endgroup$
– Brandon_J
7 hours ago
$begingroup$
@EricDuminil Chillax brah
$endgroup$
– BlueRaja - Danny Pflughoeft
5 hours ago
add a comment |
2
$begingroup$
Legal But Still Bro You are long long long long .... way behind the largest Number
$endgroup$
– AmanSharma
10 hours ago
$begingroup$
@AmanSharma Cheated a bit harder
$endgroup$
– Sconibulus
10 hours ago
19
$begingroup$
@AmanSharma: You realize you might not be talking to a "Bro" or to someone who doesn't like being called "Bro"?
$endgroup$
– Eric Duminil
9 hours ago
5
$begingroup$
@EricDuminil I agree with the sentiment of your comment (upvoted it), but FWIW I know several people who say "Bro" in every other sentence in real-life, day-to-day conversations, whether speaking to a female or not; I doubt that Aman is assuming genders in this scenario.
$endgroup$
– Brandon_J
7 hours ago
$begingroup$
@EricDuminil Chillax brah
$endgroup$
– BlueRaja - Danny Pflughoeft
5 hours ago
2
2
$begingroup$
Legal But Still Bro You are long long long long .... way behind the largest Number
$endgroup$
– AmanSharma
10 hours ago
$begingroup$
Legal But Still Bro You are long long long long .... way behind the largest Number
$endgroup$
– AmanSharma
10 hours ago
$begingroup$
@AmanSharma Cheated a bit harder
$endgroup$
– Sconibulus
10 hours ago
$begingroup$
@AmanSharma Cheated a bit harder
$endgroup$
– Sconibulus
10 hours ago
19
19
$begingroup$
@AmanSharma: You realize you might not be talking to a "Bro" or to someone who doesn't like being called "Bro"?
$endgroup$
– Eric Duminil
9 hours ago
$begingroup$
@AmanSharma: You realize you might not be talking to a "Bro" or to someone who doesn't like being called "Bro"?
$endgroup$
– Eric Duminil
9 hours ago
5
5
$begingroup$
@EricDuminil I agree with the sentiment of your comment (upvoted it), but FWIW I know several people who say "Bro" in every other sentence in real-life, day-to-day conversations, whether speaking to a female or not; I doubt that Aman is assuming genders in this scenario.
$endgroup$
– Brandon_J
7 hours ago
$begingroup$
@EricDuminil I agree with the sentiment of your comment (upvoted it), but FWIW I know several people who say "Bro" in every other sentence in real-life, day-to-day conversations, whether speaking to a female or not; I doubt that Aman is assuming genders in this scenario.
$endgroup$
– Brandon_J
7 hours ago
$begingroup$
@EricDuminil Chillax brah
$endgroup$
– BlueRaja - Danny Pflughoeft
5 hours ago
$begingroup$
@EricDuminil Chillax brah
$endgroup$
– BlueRaja - Danny Pflughoeft
5 hours ago
add a comment |
$begingroup$
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.
answered 9 hours ago
StilezStilez
1,314211
$endgroup$
2
$begingroup$
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
8 hours ago
1
$begingroup$
Not in this case, I fear ;-)
$endgroup$
– Stilez
8 hours ago
$begingroup$
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
6 hours ago
$begingroup$
nice +1! yet for the summation, the two horizontal lines would touch the endpoints of the 1, making it more like an inverted B...
$endgroup$
– Omega Krypton
5 hours ago
$begingroup$
This seems to beat the intended and currently accepted answer.
$endgroup$
– Jay
2 hours ago
|
show 2 more comments
$begingroup$
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.
answered 9 hours ago
StilezStilez
1,314211
$endgroup$
2
$begingroup$
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
8 hours ago
1
$begingroup$
Not in this case, I fear ;-)
$endgroup$
– Stilez
8 hours ago
$begingroup$
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
6 hours ago
$begingroup$
nice +1! yet for the summation, the two horizontal lines would touch the endpoints of the 1, making it more like an inverted B...
$endgroup$
– Omega Krypton
5 hours ago
$begingroup$
This seems to beat the intended and currently accepted answer.
$endgroup$
– Jay
2 hours ago
|
show 2 more comments
$begingroup$
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.
answered 9 hours ago
StilezStilez
1,314211
$endgroup$
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.
answered 9 hours ago
StilezStilez
1,314211
edited 8 hours ago
answered 9 hours ago
StilezStilez
1,314211
answered 9 hours ago
StilezStilez
1,314211
answered 9 hours ago
StilezStilez
1,314211
1,314211
2
$begingroup$
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
8 hours ago
1
$begingroup$
Not in this case, I fear ;-)
$endgroup$
– Stilez
8 hours ago
$begingroup$
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
6 hours ago
$begingroup$
nice +1! yet for the summation, the two horizontal lines would touch the endpoints of the 1, making it more like an inverted B...
$endgroup$
– Omega Krypton
5 hours ago
$begingroup$
This seems to beat the intended and currently accepted answer.
$endgroup$
– Jay
2 hours ago
|
show 2 more comments
2
$begingroup$
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
8 hours ago
1
$begingroup$
Not in this case, I fear ;-)
$endgroup$
– Stilez
8 hours ago
$begingroup$
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
6 hours ago
$begingroup$
nice +1! yet for the summation, the two horizontal lines would touch the endpoints of the 1, making it more like an inverted B...
$endgroup$
– Omega Krypton
5 hours ago
$begingroup$
This seems to beat the intended and currently accepted answer.
$endgroup$
– Jay
2 hours ago
2
2
$begingroup$
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
8 hours ago
$begingroup$
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
8 hours ago
1
1
$begingroup$
Not in this case, I fear ;-)
$endgroup$
– Stilez
8 hours ago
$begingroup$
Not in this case, I fear ;-)
$endgroup$
– Stilez
8 hours ago
$begingroup$
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
6 hours ago
$begingroup$
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
6 hours ago
$begingroup$
nice +1! yet for the summation, the two horizontal lines would touch the endpoints of the 1, making it more like an inverted B...
$endgroup$
– Omega Krypton
5 hours ago
$begingroup$
nice +1! yet for the summation, the two horizontal lines would touch the endpoints of the 1, making it more like an inverted B...
$endgroup$
– Omega Krypton
5 hours ago
$begingroup$
This seems to beat the intended and currently accepted answer.
$endgroup$
– Jay
2 hours ago
$begingroup$
This seems to beat the intended and currently accepted answer.
$endgroup$
– Jay
2 hours ago
|
show 2 more comments
$begingroup$
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.
answered 10 hours ago
PugmonkeyPugmonkey
3,6401220
$endgroup$
1
$begingroup$
Bro You are long long long long .... way behind the largest Number
$endgroup$
– AmanSharma
10 hours ago
add a comment |
$begingroup$
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.
answered 10 hours ago
PugmonkeyPugmonkey
3,6401220
$endgroup$
1
$begingroup$
Bro You are long long long long .... way behind the largest Number
$endgroup$
– AmanSharma
10 hours ago
add a comment |
$begingroup$
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.
answered 10 hours ago
PugmonkeyPugmonkey
3,6401220
$endgroup$
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.
answered 10 hours ago
PugmonkeyPugmonkey
3,6401220
edited 10 hours ago
answered 10 hours ago
PugmonkeyPugmonkey
3,6401220
answered 10 hours ago
PugmonkeyPugmonkey
3,6401220
answered 10 hours ago
PugmonkeyPugmonkey
3,6401220
3,6401220
1
$begingroup$
Bro You are long long long long .... way behind the largest Number
$endgroup$
– AmanSharma
10 hours ago
add a comment |
1
$begingroup$
Bro You are long long long long .... way behind the largest Number
$endgroup$
– AmanSharma
10 hours ago
1
1
$begingroup$
Bro You are long long long long .... way behind the largest Number
$endgroup$
– AmanSharma
10 hours ago
$begingroup$
Bro You are long long long long .... way behind the largest Number
$endgroup$
– AmanSharma
10 hours ago
add a comment |
$begingroup$
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...
answered 10 hours ago
Dr tDr t
1,258313
$endgroup$
1
$begingroup$
Bro You are long long long long .... way behind the largest Number
$endgroup$
– AmanSharma
10 hours ago
1
$begingroup$
You can place the one at the end. But I agree with @AmanSharma remains a sub-optimal solution
$endgroup$
– J.Khamphousone
9 hours ago
add a comment |
$begingroup$
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...
answered 10 hours ago
Dr tDr t
1,258313
$endgroup$
1
$begingroup$
Bro You are long long long long .... way behind the largest Number
$endgroup$
– AmanSharma
10 hours ago
1
$begingroup$
You can place the one at the end. But I agree with @AmanSharma remains a sub-optimal solution
$endgroup$
– J.Khamphousone
9 hours ago
add a comment |
$begingroup$
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...
answered 10 hours ago
Dr tDr t
1,258313
$endgroup$
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...
answered 10 hours ago
Dr tDr t
1,258313
answered 10 hours ago
Dr tDr t
1,258313
answered 10 hours ago
Dr tDr t
1,258313
answered 10 hours ago
Dr tDr t
1,258313
1,258313
1
$begingroup$
Bro You are long long long long .... way behind the largest Number
$endgroup$
– AmanSharma
10 hours ago
1
$begingroup$
You can place the one at the end. But I agree with @AmanSharma remains a sub-optimal solution
$endgroup$
– J.Khamphousone
9 hours ago
add a comment |
1
$begingroup$
Bro You are long long long long .... way behind the largest Number
$endgroup$
– AmanSharma
10 hours ago
1
$begingroup$
You can place the one at the end. But I agree with @AmanSharma remains a sub-optimal solution
$endgroup$
– J.Khamphousone
9 hours ago
1
1
$begingroup$
Bro You are long long long long .... way behind the largest Number
$endgroup$
– AmanSharma
10 hours ago
$begingroup$
Bro You are long long long long .... way behind the largest Number
$endgroup$
– AmanSharma
10 hours ago
1
1
$begingroup$
You can place the one at the end. But I agree with @AmanSharma remains a sub-optimal solution
$endgroup$
– J.Khamphousone
9 hours ago
$begingroup$
You can place the one at the end. But I agree with @AmanSharma remains a sub-optimal solution
$endgroup$
– J.Khamphousone
9 hours ago
add a comment |
$begingroup$
Without adding extra digits and keeping with the digital-like format,
938
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.
answered 10 hours ago
SensoraySensoray
4,57311246
$endgroup$
add a comment |
$begingroup$
Without adding extra digits and keeping with the digital-like format,
938
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.
answered 10 hours ago
SensoraySensoray
4,57311246
$endgroup$
add a comment |
$begingroup$
Without adding extra digits and keeping with the digital-like format,
938
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.
answered 10 hours ago
SensoraySensoray
4,57311246
$endgroup$
Without adding extra digits and keeping with the digital-like format,
938
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.
answered 10 hours ago
SensoraySensoray
4,57311246
answered 10 hours ago
SensoraySensoray
4,57311246
answered 10 hours ago
SensoraySensoray
4,57311246
answered 10 hours ago
SensoraySensoray
4,57311246
4,57311246
add a comment |
add a comment |
$begingroup$
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
answered 8 hours ago
visualnotsobasicvisualnotsobasic
3239
$endgroup$
add a comment |
$begingroup$
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
answered 8 hours ago
visualnotsobasicvisualnotsobasic
3239
$endgroup$
add a comment |
$begingroup$
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
answered 8 hours ago
visualnotsobasicvisualnotsobasic
3239
$endgroup$
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
answered 8 hours ago
visualnotsobasicvisualnotsobasic
3239
answered 8 hours ago
visualnotsobasicvisualnotsobasic
3239
answered 8 hours ago
visualnotsobasicvisualnotsobasic
3239
answered 8 hours ago
visualnotsobasicvisualnotsobasic
3239
3239
add a comment |
add a comment |
$begingroup$
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. 😋
answered 5 hours ago
Hand-E-FoodHand-E-Food
1492
$endgroup$
add a comment |
$begingroup$
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. 😋
answered 5 hours ago
Hand-E-FoodHand-E-Food
1492
$endgroup$
add a comment |
$begingroup$
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. 😋
answered 5 hours ago
Hand-E-FoodHand-E-Food
1492
$endgroup$
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. 😋
answered 5 hours ago
Hand-E-FoodHand-E-Food
1492
answered 5 hours ago
Hand-E-FoodHand-E-Food
1492
answered 5 hours ago
Hand-E-FoodHand-E-Food
1492
answered 5 hours ago
Hand-E-FoodHand-E-Food
1492
1492
add a comment |
add a comment |
$begingroup$
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.
answered 7 hours ago
Lorenzo MLorenzo M
1
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$endgroup$
add a comment |
$begingroup$
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.
answered 7 hours ago
Lorenzo MLorenzo M
1
New contributor
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$endgroup$
add a comment |
$begingroup$
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.
answered 7 hours ago
Lorenzo MLorenzo M
1
New contributor
Lorenzo M is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$
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.
answered 7 hours ago
Lorenzo MLorenzo M
1
New contributor
Lorenzo M is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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answered 7 hours ago
Lorenzo MLorenzo M
1
New contributor
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answered 7 hours ago
Lorenzo MLorenzo M
1
answered 7 hours ago
Lorenzo MLorenzo M
1
1
New contributor
Lorenzo M is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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New contributor
Lorenzo M is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Lorenzo M is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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add a comment |
add a comment |
$begingroup$
2
How I got it:
I took two sticks since that's all I could move.
Subsequently,
I made a '1' out of them...
answered 7 hours ago
Tseug EmanTseug Eman
1
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$endgroup$
1
$begingroup$
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
7 hours ago
add a comment |
$begingroup$
2
How I got it:
I took two sticks since that's all I could move.
Subsequently,
I made a '1' out of them...
answered 7 hours ago
Tseug EmanTseug Eman
1
New contributor
Tseug Eman is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
1
$begingroup$
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
7 hours ago
add a comment |
$begingroup$
2
How I got it:
I took two sticks since that's all I could move.
Subsequently,
I made a '1' out of them...
answered 7 hours ago
Tseug EmanTseug Eman
1
New contributor
Tseug Eman is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
2
How I got it:
I took two sticks since that's all I could move.
Subsequently,
I made a '1' out of them...
answered 7 hours ago
Tseug EmanTseug Eman
1
New contributor
Tseug Eman is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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answered 7 hours ago
Tseug EmanTseug Eman
1
New contributor
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answered 7 hours ago
Tseug EmanTseug Eman
1
answered 7 hours ago
Tseug EmanTseug Eman
1
1
New contributor
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New contributor
Tseug Eman is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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1
$begingroup$
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
7 hours ago
add a comment |
1
$begingroup$
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
7 hours ago
1
1
$begingroup$
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
7 hours ago
$begingroup$
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
7 hours ago
add a comment |
$begingroup$
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.
answered 5 hours ago
OM PrakashOM Prakash
1
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$endgroup$
add a comment |
$begingroup$
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.
answered 5 hours ago
OM PrakashOM Prakash
1
New contributor
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$endgroup$
add a comment |
$begingroup$
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.
answered 5 hours ago
OM PrakashOM Prakash
1
New contributor
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$endgroup$
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.
answered 5 hours ago
OM PrakashOM Prakash
1
New contributor
OM Prakash is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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answered 5 hours ago
OM PrakashOM Prakash
1
New contributor
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answered 5 hours ago
OM PrakashOM Prakash
1
answered 5 hours ago
OM PrakashOM Prakash
1
1
New contributor
OM Prakash is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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New contributor
OM Prakash is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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add a comment |
add a comment |
$begingroup$
407.7 (but how... ;)
150e, where e is Euler's constant; approximately 2.71828.
answered 3 hours ago
MazuraMazura
865514
$endgroup$
$begingroup$
I have already marked the correct answer See if it helps YOu.
$endgroup$
– AmanSharma
2 hours ago
$begingroup$
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
2 hours ago
add a comment |
$begingroup$
407.7 (but how... ;)
150e, where e is Euler's constant; approximately 2.71828.
answered 3 hours ago
MazuraMazura
865514
$endgroup$
$begingroup$
I have already marked the correct answer See if it helps YOu.
$endgroup$
– AmanSharma
2 hours ago
$begingroup$
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
2 hours ago
add a comment |
$begingroup$
407.7 (but how... ;)
150e, where e is Euler's constant; approximately 2.71828.
answered 3 hours ago
MazuraMazura
865514
$endgroup$
407.7 (but how... ;)
150e, where e is Euler's constant; approximately 2.71828.
answered 3 hours ago
MazuraMazura
865514
answered 3 hours ago
MazuraMazura
865514
answered 3 hours ago
MazuraMazura
865514
answered 3 hours ago
MazuraMazura
865514
865514
$begingroup$
I have already marked the correct answer See if it helps YOu.
$endgroup$
– AmanSharma
2 hours ago
$begingroup$
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
2 hours ago
add a comment |
$begingroup$
I have already marked the correct answer See if it helps YOu.
$endgroup$
– AmanSharma
2 hours ago
$begingroup$
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
2 hours ago
$begingroup$
I have already marked the correct answer See if it helps YOu.
$endgroup$
– AmanSharma
2 hours ago
$begingroup$
I have already marked the correct answer See if it helps YOu.
$endgroup$
– AmanSharma
2 hours ago
$begingroup$
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
2 hours ago
$begingroup$
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
2 hours ago
add a comment |
$begingroup$
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.
answered 1 min ago
Andrew900460Andrew900460
1
New contributor
Andrew900460 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
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.
answered 1 min ago
Andrew900460Andrew900460
1
New contributor
Andrew900460 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
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.
answered 1 min ago
Andrew900460Andrew900460
1
New contributor
Andrew900460 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
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.
answered 1 min ago
Andrew900460Andrew900460
1
New contributor
Andrew900460 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 1 min ago
Andrew900460Andrew900460
1
New contributor
Andrew900460 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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answered 1 min ago
Andrew900460Andrew900460
1
answered 1 min ago
Andrew900460Andrew900460
1
1
New contributor
Andrew900460 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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New contributor
Andrew900460 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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Andrew900460 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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add a comment |
add a comment |
protected by Community♦ 1 min ago
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).
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3
$begingroup$
youtube.com/watch?v=9m6S0x-AKNU
$endgroup$
– ielyamani
9 hours ago
2
$begingroup$
Waiting for Harry Potter to get a PSE account in order to answer this puzzle with an advantage :)
$endgroup$
– Brandon_J
7 hours ago
$begingroup$
@Brandon_J correct He knows magic spells but is missing math spells
$endgroup$
– AmanSharma
5 hours ago