What is ls Largest Number Formed by only moving two sticks in 508? Announcing the arrival of...
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Variable does not exist: sObjectType (Task.sObjectType)
What is ls Largest Number Formed by only moving two sticks in 508?
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Largest 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 2 hours ago
AmanSharmaAmanSharma
1344
$endgroup$
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 2 hours ago
AmanSharmaAmanSharma
1344
$endgroup$
$begingroup$
youtube.com/watch?v=9m6S0x-AKNU
$endgroup$
– ielyamani
48 mins 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 2 hours ago
AmanSharmaAmanSharma
1344
$endgroup$
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 2 hours ago
AmanSharmaAmanSharma
1344
asked 2 hours ago
AmanSharmaAmanSharma
1344
edited 1 hour ago
AmanSharma
asked 2 hours ago
AmanSharmaAmanSharma
1344
asked 2 hours ago
AmanSharmaAmanSharma
1344
asked 2 hours ago
AmanSharmaAmanSharma
1344
1344
$begingroup$
youtube.com/watch?v=9m6S0x-AKNU
$endgroup$
– ielyamani
48 mins ago
add a comment |
$begingroup$
youtube.com/watch?v=9m6S0x-AKNU
$endgroup$
– ielyamani
48 mins ago
$begingroup$
youtube.com/watch?v=9m6S0x-AKNU
$endgroup$
– ielyamani
48 mins ago
$begingroup$
youtube.com/watch?v=9m6S0x-AKNU
$endgroup$
– ielyamani
48 mins ago
add a comment |
7 Answers
7
active
oldest
votes
$begingroup$
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 1 hour ago
SconibulusSconibulus
14.7k128101
$endgroup$
$begingroup$
Legal But Still Bro You are long long long long .... way behind the largest Number
$endgroup$
– AmanSharma
1 hour ago
$begingroup$
@AmanSharma Cheated a bit harder
$endgroup$
– Sconibulus
1 hour ago
6
$begingroup$
@AmanSharma: You realize you might not be talking to a "Bro" or to someone who doesn't like being called "Bro"?
$endgroup$
– Eric Duminil
59 mins 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 1 hour ago
Dr tDr t
1,248313
$endgroup$
$begingroup$
Bro You are long long long long .... way behind the largest Number
$endgroup$
– AmanSharma
1 hour ago
$begingroup$
You can place the one at the end. But I agree with @AmanSharma remains a sub-optimal solution
$endgroup$
– J.Khamphousone
1 hour ago
add a comment |
$begingroup$
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 2 hours ago
J.KhamphousoneJ.Khamphousone
2056
$endgroup$
1
$begingroup$
That was out of universe . We are not changing size of digits here Bro. Also infinity isn't a number.
$endgroup$
– AmanSharma
1 hour ago
$begingroup$
@AmanSharma I know ! That's why I wrote it's debetable, but my wisest answer still remains
$endgroup$
– J.Khamphousone
1 hour ago
$begingroup$
@AmanSharma I've added another debatable solution using the Caret Symbol
$endgroup$
– J.Khamphousone
1 hour ago
$begingroup$
About infinity it was more of a joke rather than a solution ;p
$endgroup$
– J.Khamphousone
1 hour ago
1
$begingroup$
rot13(svsgl-avar bire mreb vf n ovttre vasvavgl guna svsgl-fvk bire mreb)
$endgroup$
– ielyamani
44 mins ago
|
show 6 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 1 hour ago
PugmonkeyPugmonkey
3,5901220
$endgroup$
$begingroup$
Bro You are long long long long .... way behind the largest Number
$endgroup$
– AmanSharma
1 hour 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 something else.....
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. Hint: start at "inconceivably vast" and then scale up an inconceivable number of times, or something like that.
answered 25 mins ago
StilezStilez
1,234211
$endgroup$
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 1 hour ago
SensoraySensoray
4,54311246
$endgroup$
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 14 mins ago
visualnotsobasicvisualnotsobasic
1839
$endgroup$
add a comment |
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7 Answers
7
active
oldest
votes
7 Answers
7
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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 1 hour ago
SconibulusSconibulus
14.7k128101
$endgroup$
$begingroup$
Legal But Still Bro You are long long long long .... way behind the largest Number
$endgroup$
– AmanSharma
1 hour ago
$begingroup$
@AmanSharma Cheated a bit harder
$endgroup$
– Sconibulus
1 hour ago
6
$begingroup$
@AmanSharma: You realize you might not be talking to a "Bro" or to someone who doesn't like being called "Bro"?
$endgroup$
– Eric Duminil
59 mins 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 1 hour ago
SconibulusSconibulus
14.7k128101
$endgroup$
$begingroup$
Legal But Still Bro You are long long long long .... way behind the largest Number
$endgroup$
– AmanSharma
1 hour ago
$begingroup$
@AmanSharma Cheated a bit harder
$endgroup$
– Sconibulus
1 hour ago
6
$begingroup$
@AmanSharma: You realize you might not be talking to a "Bro" or to someone who doesn't like being called "Bro"?
$endgroup$
– Eric Duminil
59 mins 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 1 hour ago
SconibulusSconibulus
14.7k128101
$endgroup$
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 1 hour ago
SconibulusSconibulus
14.7k128101
edited 1 hour ago
answered 1 hour ago
SconibulusSconibulus
14.7k128101
answered 1 hour ago
SconibulusSconibulus
14.7k128101
answered 1 hour ago
SconibulusSconibulus
14.7k128101
14.7k128101
$begingroup$
Legal But Still Bro You are long long long long .... way behind the largest Number
$endgroup$
– AmanSharma
1 hour ago
$begingroup$
@AmanSharma Cheated a bit harder
$endgroup$
– Sconibulus
1 hour ago
6
$begingroup$
@AmanSharma: You realize you might not be talking to a "Bro" or to someone who doesn't like being called "Bro"?
$endgroup$
– Eric Duminil
59 mins ago
add a comment |
$begingroup$
Legal But Still Bro You are long long long long .... way behind the largest Number
$endgroup$
– AmanSharma
1 hour ago
$begingroup$
@AmanSharma Cheated a bit harder
$endgroup$
– Sconibulus
1 hour ago
6
$begingroup$
@AmanSharma: You realize you might not be talking to a "Bro" or to someone who doesn't like being called "Bro"?
$endgroup$
– Eric Duminil
59 mins ago
$begingroup$
Legal But Still Bro You are long long long long .... way behind the largest Number
$endgroup$
– AmanSharma
1 hour ago
$begingroup$
Legal But Still Bro You are long long long long .... way behind the largest Number
$endgroup$
– AmanSharma
1 hour ago
$begingroup$
@AmanSharma Cheated a bit harder
$endgroup$
– Sconibulus
1 hour ago
$begingroup$
@AmanSharma Cheated a bit harder
$endgroup$
– Sconibulus
1 hour ago
6
6
$begingroup$
@AmanSharma: You realize you might not be talking to a "Bro" or to someone who doesn't like being called "Bro"?
$endgroup$
– Eric Duminil
59 mins 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
59 mins 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 1 hour ago
Dr tDr t
1,248313
$endgroup$
$begingroup$
Bro You are long long long long .... way behind the largest Number
$endgroup$
– AmanSharma
1 hour ago
$begingroup$
You can place the one at the end. But I agree with @AmanSharma remains a sub-optimal solution
$endgroup$
– J.Khamphousone
1 hour 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 1 hour ago
Dr tDr t
1,248313
$endgroup$
$begingroup$
Bro You are long long long long .... way behind the largest Number
$endgroup$
– AmanSharma
1 hour ago
$begingroup$
You can place the one at the end. But I agree with @AmanSharma remains a sub-optimal solution
$endgroup$
– J.Khamphousone
1 hour 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 1 hour ago
Dr tDr t
1,248313
$endgroup$
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 1 hour ago
Dr tDr t
1,248313
answered 1 hour ago
Dr tDr t
1,248313
answered 1 hour ago
Dr tDr t
1,248313
answered 1 hour ago
Dr tDr t
1,248313
1,248313
$begingroup$
Bro You are long long long long .... way behind the largest Number
$endgroup$
– AmanSharma
1 hour ago
$begingroup$
You can place the one at the end. But I agree with @AmanSharma remains a sub-optimal solution
$endgroup$
– J.Khamphousone
1 hour ago
add a comment |
$begingroup$
Bro You are long long long long .... way behind the largest Number
$endgroup$
– AmanSharma
1 hour ago
$begingroup$
You can place the one at the end. But I agree with @AmanSharma remains a sub-optimal solution
$endgroup$
– J.Khamphousone
1 hour ago
$begingroup$
Bro You are long long long long .... way behind the largest Number
$endgroup$
– AmanSharma
1 hour ago
$begingroup$
Bro You are long long long long .... way behind the largest Number
$endgroup$
– AmanSharma
1 hour ago
$begingroup$
You can place the one at the end. But I agree with @AmanSharma remains a sub-optimal solution
$endgroup$
– J.Khamphousone
1 hour ago
$begingroup$
You can place the one at the end. But I agree with @AmanSharma remains a sub-optimal solution
$endgroup$
– J.Khamphousone
1 hour ago
add a comment |
$begingroup$
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 2 hours ago
J.KhamphousoneJ.Khamphousone
2056
$endgroup$
1
$begingroup$
That was out of universe . We are not changing size of digits here Bro. Also infinity isn't a number.
$endgroup$
– AmanSharma
1 hour ago
$begingroup$
@AmanSharma I know ! That's why I wrote it's debetable, but my wisest answer still remains
$endgroup$
– J.Khamphousone
1 hour ago
$begingroup$
@AmanSharma I've added another debatable solution using the Caret Symbol
$endgroup$
– J.Khamphousone
1 hour ago
$begingroup$
About infinity it was more of a joke rather than a solution ;p
$endgroup$
– J.Khamphousone
1 hour ago
1
$begingroup$
rot13(svsgl-avar bire mreb vf n ovttre vasvavgl guna svsgl-fvk bire mreb)
$endgroup$
– ielyamani
44 mins ago
|
show 6 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 2 hours ago
J.KhamphousoneJ.Khamphousone
2056
$endgroup$
1
$begingroup$
That was out of universe . We are not changing size of digits here Bro. Also infinity isn't a number.
$endgroup$
– AmanSharma
1 hour ago
$begingroup$
@AmanSharma I know ! That's why I wrote it's debetable, but my wisest answer still remains
$endgroup$
– J.Khamphousone
1 hour ago
$begingroup$
@AmanSharma I've added another debatable solution using the Caret Symbol
$endgroup$
– J.Khamphousone
1 hour ago
$begingroup$
About infinity it was more of a joke rather than a solution ;p
$endgroup$
– J.Khamphousone
1 hour ago
1
$begingroup$
rot13(svsgl-avar bire mreb vf n ovttre vasvavgl guna svsgl-fvk bire mreb)
$endgroup$
– ielyamani
44 mins ago
|
show 6 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 2 hours ago
J.KhamphousoneJ.Khamphousone
2056
$endgroup$
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 2 hours ago
J.KhamphousoneJ.Khamphousone
2056
edited 25 mins ago
answered 2 hours ago
J.KhamphousoneJ.Khamphousone
2056
answered 2 hours ago
J.KhamphousoneJ.Khamphousone
2056
answered 2 hours ago
J.KhamphousoneJ.Khamphousone
2056
2056
1
$begingroup$
That was out of universe . We are not changing size of digits here Bro. Also infinity isn't a number.
$endgroup$
– AmanSharma
1 hour ago
$begingroup$
@AmanSharma I know ! That's why I wrote it's debetable, but my wisest answer still remains
$endgroup$
– J.Khamphousone
1 hour ago
$begingroup$
@AmanSharma I've added another debatable solution using the Caret Symbol
$endgroup$
– J.Khamphousone
1 hour ago
$begingroup$
About infinity it was more of a joke rather than a solution ;p
$endgroup$
– J.Khamphousone
1 hour ago
1
$begingroup$
rot13(svsgl-avar bire mreb vf n ovttre vasvavgl guna svsgl-fvk bire mreb)
$endgroup$
– ielyamani
44 mins ago
|
show 6 more comments
1
$begingroup$
That was out of universe . We are not changing size of digits here Bro. Also infinity isn't a number.
$endgroup$
– AmanSharma
1 hour ago
$begingroup$
@AmanSharma I know ! That's why I wrote it's debetable, but my wisest answer still remains
$endgroup$
– J.Khamphousone
1 hour ago
$begingroup$
@AmanSharma I've added another debatable solution using the Caret Symbol
$endgroup$
– J.Khamphousone
1 hour ago
$begingroup$
About infinity it was more of a joke rather than a solution ;p
$endgroup$
– J.Khamphousone
1 hour ago
1
$begingroup$
rot13(svsgl-avar bire mreb vf n ovttre vasvavgl guna svsgl-fvk bire mreb)
$endgroup$
– ielyamani
44 mins ago
1
1
$begingroup$
That was out of universe . We are not changing size of digits here Bro. Also infinity isn't a number.
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– AmanSharma
1 hour ago
$begingroup$
That was out of universe . We are not changing size of digits here Bro. Also infinity isn't a number.
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– AmanSharma
1 hour ago
$begingroup$
@AmanSharma I know ! That's why I wrote it's debetable, but my wisest answer still remains
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– J.Khamphousone
1 hour ago
$begingroup$
@AmanSharma I know ! That's why I wrote it's debetable, but my wisest answer still remains
$endgroup$
– J.Khamphousone
1 hour ago
$begingroup$
@AmanSharma I've added another debatable solution using the Caret Symbol
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– J.Khamphousone
1 hour ago
$begingroup$
@AmanSharma I've added another debatable solution using the Caret Symbol
$endgroup$
– J.Khamphousone
1 hour ago
$begingroup$
About infinity it was more of a joke rather than a solution ;p
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– J.Khamphousone
1 hour ago
$begingroup$
About infinity it was more of a joke rather than a solution ;p
$endgroup$
– J.Khamphousone
1 hour ago
1
1
$begingroup$
rot13(svsgl-avar bire mreb vf n ovttre vasvavgl guna svsgl-fvk bire mreb)
$endgroup$
– ielyamani
44 mins ago
$begingroup$
rot13(svsgl-avar bire mreb vf n ovttre vasvavgl guna svsgl-fvk bire mreb)
$endgroup$
– ielyamani
44 mins ago
|
show 6 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 1 hour ago
PugmonkeyPugmonkey
3,5901220
$endgroup$
$begingroup$
Bro You are long long long long .... way behind the largest Number
$endgroup$
– AmanSharma
1 hour 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 1 hour ago
PugmonkeyPugmonkey
3,5901220
$endgroup$
$begingroup$
Bro You are long long long long .... way behind the largest Number
$endgroup$
– AmanSharma
1 hour 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 1 hour ago
PugmonkeyPugmonkey
3,5901220
$endgroup$
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 1 hour ago
PugmonkeyPugmonkey
3,5901220
edited 1 hour ago
answered 1 hour ago
PugmonkeyPugmonkey
3,5901220
answered 1 hour ago
PugmonkeyPugmonkey
3,5901220
answered 1 hour ago
PugmonkeyPugmonkey
3,5901220
3,5901220
$begingroup$
Bro You are long long long long .... way behind the largest Number
$endgroup$
– AmanSharma
1 hour ago
add a comment |
$begingroup$
Bro You are long long long long .... way behind the largest Number
$endgroup$
– AmanSharma
1 hour ago
$begingroup$
Bro You are long long long long .... way behind the largest Number
$endgroup$
– AmanSharma
1 hour ago
$begingroup$
Bro You are long long long long .... way behind the largest Number
$endgroup$
– AmanSharma
1 hour 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 something else.....
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. Hint: start at "inconceivably vast" and then scale up an inconceivable number of times, or something like that.
answered 25 mins ago
StilezStilez
1,234211
$endgroup$
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 something else.....
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. Hint: start at "inconceivably vast" and then scale up an inconceivable number of times, or something like that.
answered 25 mins ago
StilezStilez
1,234211
$endgroup$
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 something else.....
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. Hint: start at "inconceivably vast" and then scale up an inconceivable number of times, or something like that.
answered 25 mins ago
StilezStilez
1,234211
$endgroup$
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 something else.....
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. Hint: start at "inconceivably vast" and then scale up an inconceivable number of times, or something like that.
answered 25 mins ago
StilezStilez
1,234211
edited 6 mins ago
answered 25 mins ago
StilezStilez
1,234211
answered 25 mins ago
StilezStilez
1,234211
answered 25 mins ago
StilezStilez
1,234211
1,234211
add a comment |
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 1 hour ago
SensoraySensoray
4,54311246
$endgroup$
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 1 hour ago
SensoraySensoray
4,54311246
$endgroup$
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 1 hour ago
SensoraySensoray
4,54311246
$endgroup$
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 1 hour ago
SensoraySensoray
4,54311246
answered 1 hour ago
SensoraySensoray
4,54311246
answered 1 hour ago
SensoraySensoray
4,54311246
answered 1 hour ago
SensoraySensoray
4,54311246
4,54311246
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 14 mins ago
visualnotsobasicvisualnotsobasic
1839
$endgroup$
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 14 mins ago
visualnotsobasicvisualnotsobasic
1839
$endgroup$
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 14 mins ago
visualnotsobasicvisualnotsobasic
1839
$endgroup$
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 14 mins ago
visualnotsobasicvisualnotsobasic
1839
answered 14 mins ago
visualnotsobasicvisualnotsobasic
1839
answered 14 mins ago
visualnotsobasicvisualnotsobasic
1839
answered 14 mins ago
visualnotsobasicvisualnotsobasic
1839
1839
add a comment |
add a comment |
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$begingroup$
youtube.com/watch?v=9m6S0x-AKNU
$endgroup$
– ielyamani
48 mins ago