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
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What is ls Largest Number Formed by only moving two sticks in 508?
How to begin with a paragraph in latex
Variable does not exist: sObjectType (Task.sObjectType)
/bin/ls sorts differently than just ls
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 3 hours ago
AmanSharmaAmanSharma
1494
$endgroup$
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 3 hours ago
AmanSharmaAmanSharma
1494
$endgroup$
3
$begingroup$
youtube.com/watch?v=9m6S0x-AKNU
$endgroup$
– ielyamani
2 hours ago
$begingroup$
Waiting for Harry Potter to get a PSE account in order to answer this puzzle with an advantage :)
$endgroup$
– Brandon_J
19 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 3 hours ago
AmanSharmaAmanSharma
1494
$endgroup$
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 3 hours ago
AmanSharmaAmanSharma
1494
asked 3 hours ago
AmanSharmaAmanSharma
1494
edited 3 hours ago
AmanSharma
asked 3 hours ago
AmanSharmaAmanSharma
1494
asked 3 hours ago
AmanSharmaAmanSharma
1494
asked 3 hours ago
AmanSharmaAmanSharma
1494
1494
3
$begingroup$
youtube.com/watch?v=9m6S0x-AKNU
$endgroup$
– ielyamani
2 hours ago
$begingroup$
Waiting for Harry Potter to get a PSE account in order to answer this puzzle with an advantage :)
$endgroup$
– Brandon_J
19 mins ago
add a comment |
3
$begingroup$
youtube.com/watch?v=9m6S0x-AKNU
$endgroup$
– ielyamani
2 hours ago
$begingroup$
Waiting for Harry Potter to get a PSE account in order to answer this puzzle with an advantage :)
$endgroup$
– Brandon_J
19 mins ago
3
3
$begingroup$
youtube.com/watch?v=9m6S0x-AKNU
$endgroup$
– ielyamani
2 hours ago
$begingroup$
youtube.com/watch?v=9m6S0x-AKNU
$endgroup$
– ielyamani
2 hours ago
$begingroup$
Waiting for Harry Potter to get a PSE account in order to answer this puzzle with an advantage :)
$endgroup$
– Brandon_J
19 mins ago
$begingroup$
Waiting for Harry Potter to get a PSE account in order to answer this puzzle with an advantage :)
$endgroup$
– Brandon_J
19 mins ago
add a comment |
8 Answers
8
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 2 hours ago
SconibulusSconibulus
14.8k128101
$endgroup$
$begingroup$
Legal But Still Bro You are long long long long .... way behind the largest Number
$endgroup$
– AmanSharma
2 hours ago
$begingroup$
@AmanSharma Cheated a bit harder
$endgroup$
– Sconibulus
2 hours ago
14
$begingroup$
@AmanSharma: You realize you might not be talking to a "Bro" or to someone who doesn't like being called "Bro"?
$endgroup$
– Eric Duminil
2 hours ago
3
$begingroup$
@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
21 mins 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 3 hours ago
J.KhamphousoneJ.Khamphousone
2557
$endgroup$
2
$begingroup$
That was out of universe . We are not changing size of digits here Bro. Also infinity isn't a number.
$endgroup$
– AmanSharma
3 hours ago
$begingroup$
@AmanSharma I know ! That's why I wrote it's debetable, but my wisest answer still remains
$endgroup$
– J.Khamphousone
2 hours ago
$begingroup$
@AmanSharma I've added another debatable solution using the Caret Symbol
$endgroup$
– J.Khamphousone
2 hours ago
1
$begingroup$
rot13(svsgl-avar bire mreb vf n ovttre vasvavgl guna svsgl-fvk bire mreb)
$endgroup$
– ielyamani
2 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
1 hour ago
|
show 7 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 3 hours ago
PugmonkeyPugmonkey
3,6201220
$endgroup$
1
$begingroup$
Bro You are long long long long .... way behind the largest Number
$endgroup$
– AmanSharma
2 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 3 hours ago
Dr tDr t
1,258313
$endgroup$
1
$begingroup$
Bro You are long long long long .... way behind the largest Number
$endgroup$
– AmanSharma
2 hours ago
$begingroup$
You can place the one at the end. But I agree with @AmanSharma remains a sub-optimal solution
$endgroup$
– J.Khamphousone
2 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 1 hour ago
StilezStilez
1,274211
$endgroup$
1
$begingroup$
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
1 hour ago
$begingroup$
Not in this case, I fear ;-)
$endgroup$
– Stilez
1 hour 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 2 hours ago
SensoraySensoray
4,56311246
$endgroup$
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 1 hour ago
visualnotsobasicvisualnotsobasic
3139
$endgroup$
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 26 mins 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 |
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8 Answers
8
active
oldest
votes
8 Answers
8
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 2 hours ago
SconibulusSconibulus
14.8k128101
$endgroup$
$begingroup$
Legal But Still Bro You are long long long long .... way behind the largest Number
$endgroup$
– AmanSharma
2 hours ago
$begingroup$
@AmanSharma Cheated a bit harder
$endgroup$
– Sconibulus
2 hours ago
14
$begingroup$
@AmanSharma: You realize you might not be talking to a "Bro" or to someone who doesn't like being called "Bro"?
$endgroup$
– Eric Duminil
2 hours ago
3
$begingroup$
@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
21 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 2 hours ago
SconibulusSconibulus
14.8k128101
$endgroup$
$begingroup$
Legal But Still Bro You are long long long long .... way behind the largest Number
$endgroup$
– AmanSharma
2 hours ago
$begingroup$
@AmanSharma Cheated a bit harder
$endgroup$
– Sconibulus
2 hours ago
14
$begingroup$
@AmanSharma: You realize you might not be talking to a "Bro" or to someone who doesn't like being called "Bro"?
$endgroup$
– Eric Duminil
2 hours ago
3
$begingroup$
@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
21 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 2 hours ago
SconibulusSconibulus
14.8k128101
$endgroup$
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 2 hours ago
SconibulusSconibulus
14.8k128101
edited 2 hours ago
answered 2 hours ago
SconibulusSconibulus
14.8k128101
answered 2 hours ago
SconibulusSconibulus
14.8k128101
answered 2 hours ago
SconibulusSconibulus
14.8k128101
14.8k128101
$begingroup$
Legal But Still Bro You are long long long long .... way behind the largest Number
$endgroup$
– AmanSharma
2 hours ago
$begingroup$
@AmanSharma Cheated a bit harder
$endgroup$
– Sconibulus
2 hours ago
14
$begingroup$
@AmanSharma: You realize you might not be talking to a "Bro" or to someone who doesn't like being called "Bro"?
$endgroup$
– Eric Duminil
2 hours ago
3
$begingroup$
@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
21 mins ago
add a comment |
$begingroup$
Legal But Still Bro You are long long long long .... way behind the largest Number
$endgroup$
– AmanSharma
2 hours ago
$begingroup$
@AmanSharma Cheated a bit harder
$endgroup$
– Sconibulus
2 hours ago
14
$begingroup$
@AmanSharma: You realize you might not be talking to a "Bro" or to someone who doesn't like being called "Bro"?
$endgroup$
– Eric Duminil
2 hours ago
3
$begingroup$
@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
21 mins ago
$begingroup$
Legal But Still Bro You are long long long long .... way behind the largest Number
$endgroup$
– AmanSharma
2 hours ago
$begingroup$
Legal But Still Bro You are long long long long .... way behind the largest Number
$endgroup$
– AmanSharma
2 hours ago
$begingroup$
@AmanSharma Cheated a bit harder
$endgroup$
– Sconibulus
2 hours ago
$begingroup$
@AmanSharma Cheated a bit harder
$endgroup$
– Sconibulus
2 hours ago
14
14
$begingroup$
@AmanSharma: You realize you might not be talking to a "Bro" or to someone who doesn't like being called "Bro"?
$endgroup$
– Eric Duminil
2 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
2 hours ago
3
3
$begingroup$
@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
21 mins 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
21 mins 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 3 hours ago
J.KhamphousoneJ.Khamphousone
2557
$endgroup$
2
$begingroup$
That was out of universe . We are not changing size of digits here Bro. Also infinity isn't a number.
$endgroup$
– AmanSharma
3 hours ago
$begingroup$
@AmanSharma I know ! That's why I wrote it's debetable, but my wisest answer still remains
$endgroup$
– J.Khamphousone
2 hours ago
$begingroup$
@AmanSharma I've added another debatable solution using the Caret Symbol
$endgroup$
– J.Khamphousone
2 hours ago
1
$begingroup$
rot13(svsgl-avar bire mreb vf n ovttre vasvavgl guna svsgl-fvk bire mreb)
$endgroup$
– ielyamani
2 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
1 hour ago
|
show 7 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 3 hours ago
J.KhamphousoneJ.Khamphousone
2557
$endgroup$
2
$begingroup$
That was out of universe . We are not changing size of digits here Bro. Also infinity isn't a number.
$endgroup$
– AmanSharma
3 hours ago
$begingroup$
@AmanSharma I know ! That's why I wrote it's debetable, but my wisest answer still remains
$endgroup$
– J.Khamphousone
2 hours ago
$begingroup$
@AmanSharma I've added another debatable solution using the Caret Symbol
$endgroup$
– J.Khamphousone
2 hours ago
1
$begingroup$
rot13(svsgl-avar bire mreb vf n ovttre vasvavgl guna svsgl-fvk bire mreb)
$endgroup$
– ielyamani
2 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
1 hour ago
|
show 7 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 3 hours ago
J.KhamphousoneJ.Khamphousone
2557
$endgroup$
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 3 hours ago
J.KhamphousoneJ.Khamphousone
2557
edited 1 hour ago
answered 3 hours ago
J.KhamphousoneJ.Khamphousone
2557
answered 3 hours ago
J.KhamphousoneJ.Khamphousone
2557
answered 3 hours ago
J.KhamphousoneJ.Khamphousone
2557
2557
2
$begingroup$
That was out of universe . We are not changing size of digits here Bro. Also infinity isn't a number.
$endgroup$
– AmanSharma
3 hours ago
$begingroup$
@AmanSharma I know ! That's why I wrote it's debetable, but my wisest answer still remains
$endgroup$
– J.Khamphousone
2 hours ago
$begingroup$
@AmanSharma I've added another debatable solution using the Caret Symbol
$endgroup$
– J.Khamphousone
2 hours ago
1
$begingroup$
rot13(svsgl-avar bire mreb vf n ovttre vasvavgl guna svsgl-fvk bire mreb)
$endgroup$
– ielyamani
2 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
1 hour ago
|
show 7 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
3 hours ago
$begingroup$
@AmanSharma I know ! That's why I wrote it's debetable, but my wisest answer still remains
$endgroup$
– J.Khamphousone
2 hours ago
$begingroup$
@AmanSharma I've added another debatable solution using the Caret Symbol
$endgroup$
– J.Khamphousone
2 hours ago
1
$begingroup$
rot13(svsgl-avar bire mreb vf n ovttre vasvavgl guna svsgl-fvk bire mreb)
$endgroup$
– ielyamani
2 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
1 hour 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
3 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
3 hours ago
$begingroup$
@AmanSharma I know ! That's why I wrote it's debetable, but my wisest answer still remains
$endgroup$
– J.Khamphousone
2 hours ago
$begingroup$
@AmanSharma I know ! That's why I wrote it's debetable, but my wisest answer still remains
$endgroup$
– J.Khamphousone
2 hours ago
$begingroup$
@AmanSharma I've added another debatable solution using the Caret Symbol
$endgroup$
– J.Khamphousone
2 hours ago
$begingroup$
@AmanSharma I've added another debatable solution using the Caret Symbol
$endgroup$
– J.Khamphousone
2 hours ago
1
1
$begingroup$
rot13(svsgl-avar bire mreb vf n ovttre vasvavgl guna svsgl-fvk bire mreb)
$endgroup$
– ielyamani
2 hours ago
$begingroup$
rot13(svsgl-avar bire mreb vf n ovttre vasvavgl guna svsgl-fvk bire mreb)
$endgroup$
– ielyamani
2 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
1 hour 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
1 hour ago
|
show 7 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 3 hours ago
PugmonkeyPugmonkey
3,6201220
$endgroup$
1
$begingroup$
Bro You are long long long long .... way behind the largest Number
$endgroup$
– AmanSharma
2 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 3 hours ago
PugmonkeyPugmonkey
3,6201220
$endgroup$
1
$begingroup$
Bro You are long long long long .... way behind the largest Number
$endgroup$
– AmanSharma
2 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 3 hours ago
PugmonkeyPugmonkey
3,6201220
$endgroup$
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 3 hours ago
PugmonkeyPugmonkey
3,6201220
edited 2 hours ago
answered 3 hours ago
PugmonkeyPugmonkey
3,6201220
answered 3 hours ago
PugmonkeyPugmonkey
3,6201220
answered 3 hours ago
PugmonkeyPugmonkey
3,6201220
3,6201220
1
$begingroup$
Bro You are long long long long .... way behind the largest Number
$endgroup$
– AmanSharma
2 hours ago
add a comment |
1
$begingroup$
Bro You are long long long long .... way behind the largest Number
$endgroup$
– AmanSharma
2 hours ago
1
1
$begingroup$
Bro You are long long long long .... way behind the largest Number
$endgroup$
– AmanSharma
2 hours ago
$begingroup$
Bro You are long long long long .... way behind the largest Number
$endgroup$
– AmanSharma
2 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 3 hours ago
Dr tDr t
1,258313
$endgroup$
1
$begingroup$
Bro You are long long long long .... way behind the largest Number
$endgroup$
– AmanSharma
2 hours ago
$begingroup$
You can place the one at the end. But I agree with @AmanSharma remains a sub-optimal solution
$endgroup$
– J.Khamphousone
2 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 3 hours ago
Dr tDr t
1,258313
$endgroup$
1
$begingroup$
Bro You are long long long long .... way behind the largest Number
$endgroup$
– AmanSharma
2 hours ago
$begingroup$
You can place the one at the end. But I agree with @AmanSharma remains a sub-optimal solution
$endgroup$
– J.Khamphousone
2 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 3 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 3 hours ago
Dr tDr t
1,258313
answered 3 hours ago
Dr tDr t
1,258313
answered 3 hours ago
Dr tDr t
1,258313
answered 3 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
2 hours ago
$begingroup$
You can place the one at the end. But I agree with @AmanSharma remains a sub-optimal solution
$endgroup$
– J.Khamphousone
2 hours ago
add a comment |
1
$begingroup$
Bro You are long long long long .... way behind the largest Number
$endgroup$
– AmanSharma
2 hours ago
$begingroup$
You can place the one at the end. But I agree with @AmanSharma remains a sub-optimal solution
$endgroup$
– J.Khamphousone
2 hours ago
1
1
$begingroup$
Bro You are long long long long .... way behind the largest Number
$endgroup$
– AmanSharma
2 hours ago
$begingroup$
Bro You are long long long long .... way behind the largest Number
$endgroup$
– AmanSharma
2 hours ago
$begingroup$
You can place the one at the end. But I agree with @AmanSharma remains a sub-optimal solution
$endgroup$
– J.Khamphousone
2 hours ago
$begingroup$
You can place the one at the end. But I agree with @AmanSharma remains a sub-optimal solution
$endgroup$
– J.Khamphousone
2 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 1 hour ago
StilezStilez
1,274211
$endgroup$
1
$begingroup$
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
1 hour ago
$begingroup$
Not in this case, I fear ;-)
$endgroup$
– Stilez
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 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 1 hour ago
StilezStilez
1,274211
$endgroup$
1
$begingroup$
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
1 hour ago
$begingroup$
Not in this case, I fear ;-)
$endgroup$
– Stilez
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 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 1 hour ago
StilezStilez
1,274211
$endgroup$
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 1 hour ago
StilezStilez
1,274211
edited 1 hour ago
answered 1 hour ago
StilezStilez
1,274211
answered 1 hour ago
StilezStilez
1,274211
answered 1 hour ago
StilezStilez
1,274211
1,274211
1
$begingroup$
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
1 hour ago
$begingroup$
Not in this case, I fear ;-)
$endgroup$
– Stilez
1 hour ago
add a comment |
1
$begingroup$
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
1 hour ago
$begingroup$
Not in this case, I fear ;-)
$endgroup$
– Stilez
1 hour ago
1
1
$begingroup$
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
1 hour 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
1 hour ago
$begingroup$
Not in this case, I fear ;-)
$endgroup$
– Stilez
1 hour ago
$begingroup$
Not in this case, I fear ;-)
$endgroup$
– Stilez
1 hour 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 2 hours ago
SensoraySensoray
4,56311246
$endgroup$
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 2 hours ago
SensoraySensoray
4,56311246
$endgroup$
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 2 hours ago
SensoraySensoray
4,56311246
$endgroup$
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 2 hours ago
SensoraySensoray
4,56311246
answered 2 hours ago
SensoraySensoray
4,56311246
answered 2 hours ago
SensoraySensoray
4,56311246
answered 2 hours ago
SensoraySensoray
4,56311246
4,56311246
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 1 hour ago
visualnotsobasicvisualnotsobasic
3139
$endgroup$
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 1 hour ago
visualnotsobasicvisualnotsobasic
3139
$endgroup$
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 1 hour ago
visualnotsobasicvisualnotsobasic
3139
$endgroup$
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 1 hour ago
visualnotsobasicvisualnotsobasic
3139
answered 1 hour ago
visualnotsobasicvisualnotsobasic
3139
answered 1 hour ago
visualnotsobasicvisualnotsobasic
3139
answered 1 hour ago
visualnotsobasicvisualnotsobasic
3139
3139
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 26 mins 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 |
$begingroup$
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 26 mins 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 |
$begingroup$
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 26 mins 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.
$endgroup$
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 26 mins 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.
answered 26 mins 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.
answered 26 mins ago
Lorenzo MLorenzo M
1
answered 26 mins 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.
Check out our Code of Conduct.
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.
Check out our Code of Conduct.
add a comment |
add a comment |
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3
$begingroup$
youtube.com/watch?v=9m6S0x-AKNU
$endgroup$
– ielyamani
2 hours ago
$begingroup$
Waiting for Harry Potter to get a PSE account in order to answer this puzzle with an advantage :)
$endgroup$
– Brandon_J
19 mins ago