Find 108 by using 3,4,6 Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern) Announcing the arrival of Valued Associate #679: Cesar Manara Unicorn Meta Zoo #1: Why another podcast?Use 2, 0, 1 and 8 to make 109Use 2, 0, 1 and 8 to make 1991984 - take the digits 1,9, 8 and 4 and make 2461984 - take the digits 1,9, 8 and 4 and make 3691984 - take the digits 1,9, 8 and 4 and Hard Challenges!1984 - take the digits 1,9, 8 and 4 and make 123 - Part IIIPalindromic number puzzle - make 505 from 20202Use 0, 5, 7 and 1 to make 89Use 6, 5 and 3 to make 57Using only 1s, make 29 with the minimum number of digits
Disembodied hand growing fangs
How to compare two different files line by line in unix?
Sum letters are not two different
Do Fourier frequencies actually exist in real life in form of "instantaneous frequency"?
Is grep documentation about ignoring case wrong, since it doesn't ignore case in filenames?
NumericArray versus PackedArray in MMA12
Does the Weapon Master feat grant you a fighting style?
How do I use the new nonlinear finite element in Mathematica 12 for this equation?
How to write the following sign?
Time to Settle Down!
How would a mousetrap for use in space work?
Is there any word for a place full of confusion?
How to tell that you are a giant?
How do living politicians protect their readily obtainable signatures from misuse?
Does lack of seasonality imply random time series?
Using et al. for a last / senior author rather than for a first author
Why aren't air breathing engines used as small first stages?
Should I follow up with an employee I believe overracted to a mistake I made?
Trademark violation for app?
Benefits of using sObject.clone versus creating a new record
How fail-safe is nr as stop bytes?
How were pictures turned from film to a big picture in a picture frame before digital scanning?
What is the font for "b" letter?
Multiple OR (||) Conditions in If Statement
Find 108 by using 3,4,6
Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)
Announcing the arrival of Valued Associate #679: Cesar Manara
Unicorn Meta Zoo #1: Why another podcast?Use 2, 0, 1 and 8 to make 109Use 2, 0, 1 and 8 to make 1991984 - take the digits 1,9, 8 and 4 and make 2461984 - take the digits 1,9, 8 and 4 and make 3691984 - take the digits 1,9, 8 and 4 and Hard Challenges!1984 - take the digits 1,9, 8 and 4 and make 123 - Part IIIPalindromic number puzzle - make 505 from 20202Use 0, 5, 7 and 1 to make 89Use 6, 5 and 3 to make 57Using only 1s, make 29 with the minimum number of digits
$begingroup$
Assemble a formula using the numbers $3$, $4$ and $6$ in any order to make 108.
You may use the operations;
$x + y$
$x - y$
$x times y$
$x div y$
$x!$
$sqrtx$
$sqrt[leftroot-2uproot2x]y$
$x^y$
- Brackets to clarify order of operations "(",")"
- Concatenate two or more of the three digits you start with (concatenation of numbers from calculations is not permitted)
as long as all operands are either $3$, $4$ and $6$.
Note that double, triple, etc. factorials (n-druple-factorials) are not allowed though factorials of factorials are fine, such as $((6-3)!)! = 6!$.
mathematics logical-deduction
edited 1 hour ago
JonMark Perry
20.8k64199
asked 2 hours ago
OrayOray
16.2k437157
$endgroup$
add a comment |
$begingroup$
Assemble a formula using the numbers $3$, $4$ and $6$ in any order to make 108.
You may use the operations;
$x + y$
$x - y$
$x times y$
$x div y$
$x!$
$sqrtx$
$sqrt[leftroot-2uproot2x]y$
$x^y$
- Brackets to clarify order of operations "(",")"
- Concatenate two or more of the three digits you start with (concatenation of numbers from calculations is not permitted)
as long as all operands are either $3$, $4$ and $6$.
Note that double, triple, etc. factorials (n-druple-factorials) are not allowed though factorials of factorials are fine, such as $((6-3)!)! = 6!$.
mathematics logical-deduction
edited 1 hour ago
JonMark Perry
20.8k64199
asked 2 hours ago
OrayOray
16.2k437157
$endgroup$
1
$begingroup$
If I could, I would have done something like $$3times styledisplay: inline-block; transform: rotate(180deg)4times 6=108$$ since the upside down $4$ looks a bit like a $6$.
$endgroup$
– user477343
2 hours ago
2
$begingroup$
@user477343 or even $$3 times 4 times 9 = 108$$ since the upside down $6$ is very much like a $9$.
$endgroup$
– Weather Vane
1 hour ago
$begingroup$
@WeatherVane Of course! Huh... funny how I saw what was obscure as opposed to what was actually just plain simple xD
$endgroup$
– user477343
1 hour ago
2
$begingroup$
@user477343 ah but it took yours to make me think of it.
$endgroup$
– Weather Vane
1 hour ago
$begingroup$
@WeatherVane :)
$endgroup$
– user477343
1 hour ago
add a comment |
$begingroup$
Assemble a formula using the numbers $3$, $4$ and $6$ in any order to make 108.
You may use the operations;
$x + y$
$x - y$
$x times y$
$x div y$
$x!$
$sqrtx$
$sqrt[leftroot-2uproot2x]y$
$x^y$
- Brackets to clarify order of operations "(",")"
- Concatenate two or more of the three digits you start with (concatenation of numbers from calculations is not permitted)
as long as all operands are either $3$, $4$ and $6$.
Note that double, triple, etc. factorials (n-druple-factorials) are not allowed though factorials of factorials are fine, such as $((6-3)!)! = 6!$.
mathematics logical-deduction
edited 1 hour ago
JonMark Perry
20.8k64199
asked 2 hours ago
OrayOray
16.2k437157
$endgroup$
Assemble a formula using the numbers $3$, $4$ and $6$ in any order to make 108.
You may use the operations;
$x + y$
$x - y$
$x times y$
$x div y$
$x!$
$sqrtx$
$sqrt[leftroot-2uproot2x]y$
$x^y$
- Brackets to clarify order of operations "(",")"
- Concatenate two or more of the three digits you start with (concatenation of numbers from calculations is not permitted)
as long as all operands are either $3$, $4$ and $6$.
Note that double, triple, etc. factorials (n-druple-factorials) are not allowed though factorials of factorials are fine, such as $((6-3)!)! = 6!$.
mathematics logical-deduction
mathematics logical-deduction
edited 1 hour ago
JonMark Perry
20.8k64199
asked 2 hours ago
OrayOray
16.2k437157
edited 1 hour ago
JonMark Perry
20.8k64199
asked 2 hours ago
OrayOray
16.2k437157
edited 1 hour ago
JonMark Perry
20.8k64199
edited 1 hour ago
JonMark Perry
20.8k64199
edited 1 hour ago
JonMark Perry
20.8k64199
20.8k64199
asked 2 hours ago
OrayOray
16.2k437157
asked 2 hours ago
OrayOray
16.2k437157
asked 2 hours ago
OrayOray
16.2k437157
16.2k437157
1
$begingroup$
If I could, I would have done something like $$3times styledisplay: inline-block; transform: rotate(180deg)4times 6=108$$ since the upside down $4$ looks a bit like a $6$.
$endgroup$
– user477343
2 hours ago
2
$begingroup$
@user477343 or even $$3 times 4 times 9 = 108$$ since the upside down $6$ is very much like a $9$.
$endgroup$
– Weather Vane
1 hour ago
$begingroup$
@WeatherVane Of course! Huh... funny how I saw what was obscure as opposed to what was actually just plain simple xD
$endgroup$
– user477343
1 hour ago
2
$begingroup$
@user477343 ah but it took yours to make me think of it.
$endgroup$
– Weather Vane
1 hour ago
$begingroup$
@WeatherVane :)
$endgroup$
– user477343
1 hour ago
add a comment |
1
$begingroup$
If I could, I would have done something like $$3times styledisplay: inline-block; transform: rotate(180deg)4times 6=108$$ since the upside down $4$ looks a bit like a $6$.
$endgroup$
– user477343
2 hours ago
2
$begingroup$
@user477343 or even $$3 times 4 times 9 = 108$$ since the upside down $6$ is very much like a $9$.
$endgroup$
– Weather Vane
1 hour ago
$begingroup$
@WeatherVane Of course! Huh... funny how I saw what was obscure as opposed to what was actually just plain simple xD
$endgroup$
– user477343
1 hour ago
2
$begingroup$
@user477343 ah but it took yours to make me think of it.
$endgroup$
– Weather Vane
1 hour ago
$begingroup$
@WeatherVane :)
$endgroup$
– user477343
1 hour ago
1
1
$begingroup$
If I could, I would have done something like $$3times styledisplay: inline-block; transform: rotate(180deg)4times 6=108$$ since the upside down $4$ looks a bit like a $6$.
$endgroup$
– user477343
2 hours ago
$begingroup$
If I could, I would have done something like $$3times styledisplay: inline-block; transform: rotate(180deg)4times 6=108$$ since the upside down $4$ looks a bit like a $6$.
$endgroup$
– user477343
2 hours ago
2
2
$begingroup$
@user477343 or even $$3 times 4 times 9 = 108$$ since the upside down $6$ is very much like a $9$.
$endgroup$
– Weather Vane
1 hour ago
$begingroup$
@user477343 or even $$3 times 4 times 9 = 108$$ since the upside down $6$ is very much like a $9$.
$endgroup$
– Weather Vane
1 hour ago
$begingroup$
@WeatherVane Of course! Huh... funny how I saw what was obscure as opposed to what was actually just plain simple xD
$endgroup$
– user477343
1 hour ago
$begingroup$
@WeatherVane Of course! Huh... funny how I saw what was obscure as opposed to what was actually just plain simple xD
$endgroup$
– user477343
1 hour ago
2
2
$begingroup$
@user477343 ah but it took yours to make me think of it.
$endgroup$
– Weather Vane
1 hour ago
$begingroup$
@user477343 ah but it took yours to make me think of it.
$endgroup$
– Weather Vane
1 hour ago
$begingroup$
@WeatherVane :)
$endgroup$
– user477343
1 hour ago
$begingroup$
@WeatherVane :)
$endgroup$
– user477343
1 hour ago
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Could this be
$frac6^3sqrt4 = frac2162 = 108$?
@Oray found another one, which might possibly be
$6^sqrt4 times 3 = 6^2 times 3 = 36 times 3 = 108$.
answered 2 hours ago
El-GuestEl-Guest
21.8k35092
$endgroup$
$begingroup$
good finding! mine was different but this seems right too :)
$endgroup$
– Oray
2 hours ago
$begingroup$
Thank you, @Oray!!
$endgroup$
– El-Guest
2 hours ago
$begingroup$
@Oray: was this second one the one that you found?
$endgroup$
– El-Guest
2 hours ago
$begingroup$
no actually :D it was a bit more complicated.
$endgroup$
– Oray
2 hours ago
add a comment |
$begingroup$
I have found this solution
$6 times (4! - 3!) = 6 times (24 - 6) = 6 times 18 = 108$
answered 2 hours ago
Weather VaneWeather Vane
2,387112
$endgroup$
add a comment |
$begingroup$
In high school I used to play a game with my buddies where we tried to make valid arithmetic expressions out of licence plates of passing cars. The rules were pretty much similar except that you couldn't rearrange or duplicate (or omit) any of the digits. In time we got pretty good at it, basically solving any licence plate in under 10 seconds; most under 5. Except for the ones that couldn't be solved, of course.
Anyways, with your constraints (limitless numbers, can be rearranged, can be combined) there are like a zillion possible solutions.
Here's the simplest thing I could think of:
34 + 66 + 4 + 4 = 108
Here's another one without concatenation:
(4 + 6) * (4 + 6) + 4 + 4 = 108
Or maybe:
(4*6+3)*4 = 108
I could go on and on, and I haven't even touched on the advanced operators...
answered 1 hour ago
Vilx-Vilx-
1315
$endgroup$
$begingroup$
The second answer here is not "using the numbers $3$, $4$ and $6$".
$endgroup$
– Weather Vane
29 mins ago
add a comment |
Your Answer
StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "559"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);
else
createEditor();
);
function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);
);
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fpuzzling.stackexchange.com%2fquestions%2f81947%2ffind-108-by-using-3-4-6%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Could this be
$frac6^3sqrt4 = frac2162 = 108$?
@Oray found another one, which might possibly be
$6^sqrt4 times 3 = 6^2 times 3 = 36 times 3 = 108$.
answered 2 hours ago
El-GuestEl-Guest
21.8k35092
$endgroup$
$begingroup$
good finding! mine was different but this seems right too :)
$endgroup$
– Oray
2 hours ago
$begingroup$
Thank you, @Oray!!
$endgroup$
– El-Guest
2 hours ago
$begingroup$
@Oray: was this second one the one that you found?
$endgroup$
– El-Guest
2 hours ago
$begingroup$
no actually :D it was a bit more complicated.
$endgroup$
– Oray
2 hours ago
add a comment |
$begingroup$
Could this be
$frac6^3sqrt4 = frac2162 = 108$?
@Oray found another one, which might possibly be
$6^sqrt4 times 3 = 6^2 times 3 = 36 times 3 = 108$.
answered 2 hours ago
El-GuestEl-Guest
21.8k35092
$endgroup$
$begingroup$
good finding! mine was different but this seems right too :)
$endgroup$
– Oray
2 hours ago
$begingroup$
Thank you, @Oray!!
$endgroup$
– El-Guest
2 hours ago
$begingroup$
@Oray: was this second one the one that you found?
$endgroup$
– El-Guest
2 hours ago
$begingroup$
no actually :D it was a bit more complicated.
$endgroup$
– Oray
2 hours ago
add a comment |
$begingroup$
Could this be
$frac6^3sqrt4 = frac2162 = 108$?
@Oray found another one, which might possibly be
$6^sqrt4 times 3 = 6^2 times 3 = 36 times 3 = 108$.
answered 2 hours ago
El-GuestEl-Guest
21.8k35092
$endgroup$
Could this be
$frac6^3sqrt4 = frac2162 = 108$?
@Oray found another one, which might possibly be
$6^sqrt4 times 3 = 6^2 times 3 = 36 times 3 = 108$.
answered 2 hours ago
El-GuestEl-Guest
21.8k35092
edited 2 hours ago
answered 2 hours ago
El-GuestEl-Guest
21.8k35092
answered 2 hours ago
El-GuestEl-Guest
21.8k35092
answered 2 hours ago
El-GuestEl-Guest
21.8k35092
21.8k35092
$begingroup$
good finding! mine was different but this seems right too :)
$endgroup$
– Oray
2 hours ago
$begingroup$
Thank you, @Oray!!
$endgroup$
– El-Guest
2 hours ago
$begingroup$
@Oray: was this second one the one that you found?
$endgroup$
– El-Guest
2 hours ago
$begingroup$
no actually :D it was a bit more complicated.
$endgroup$
– Oray
2 hours ago
add a comment |
$begingroup$
good finding! mine was different but this seems right too :)
$endgroup$
– Oray
2 hours ago
$begingroup$
Thank you, @Oray!!
$endgroup$
– El-Guest
2 hours ago
$begingroup$
@Oray: was this second one the one that you found?
$endgroup$
– El-Guest
2 hours ago
$begingroup$
no actually :D it was a bit more complicated.
$endgroup$
– Oray
2 hours ago
$begingroup$
good finding! mine was different but this seems right too :)
$endgroup$
– Oray
2 hours ago
$begingroup$
good finding! mine was different but this seems right too :)
$endgroup$
– Oray
2 hours ago
$begingroup$
Thank you, @Oray!!
$endgroup$
– El-Guest
2 hours ago
$begingroup$
Thank you, @Oray!!
$endgroup$
– El-Guest
2 hours ago
$begingroup$
@Oray: was this second one the one that you found?
$endgroup$
– El-Guest
2 hours ago
$begingroup$
@Oray: was this second one the one that you found?
$endgroup$
– El-Guest
2 hours ago
$begingroup$
no actually :D it was a bit more complicated.
$endgroup$
– Oray
2 hours ago
$begingroup$
no actually :D it was a bit more complicated.
$endgroup$
– Oray
2 hours ago
add a comment |
$begingroup$
I have found this solution
$6 times (4! - 3!) = 6 times (24 - 6) = 6 times 18 = 108$
answered 2 hours ago
Weather VaneWeather Vane
2,387112
$endgroup$
add a comment |
$begingroup$
I have found this solution
$6 times (4! - 3!) = 6 times (24 - 6) = 6 times 18 = 108$
answered 2 hours ago
Weather VaneWeather Vane
2,387112
$endgroup$
add a comment |
$begingroup$
I have found this solution
$6 times (4! - 3!) = 6 times (24 - 6) = 6 times 18 = 108$
answered 2 hours ago
Weather VaneWeather Vane
2,387112
$endgroup$
I have found this solution
$6 times (4! - 3!) = 6 times (24 - 6) = 6 times 18 = 108$
answered 2 hours ago
Weather VaneWeather Vane
2,387112
answered 2 hours ago
Weather VaneWeather Vane
2,387112
answered 2 hours ago
Weather VaneWeather Vane
2,387112
answered 2 hours ago
Weather VaneWeather Vane
2,387112
2,387112
add a comment |
add a comment |
$begingroup$
In high school I used to play a game with my buddies where we tried to make valid arithmetic expressions out of licence plates of passing cars. The rules were pretty much similar except that you couldn't rearrange or duplicate (or omit) any of the digits. In time we got pretty good at it, basically solving any licence plate in under 10 seconds; most under 5. Except for the ones that couldn't be solved, of course.
Anyways, with your constraints (limitless numbers, can be rearranged, can be combined) there are like a zillion possible solutions.
Here's the simplest thing I could think of:
34 + 66 + 4 + 4 = 108
Here's another one without concatenation:
(4 + 6) * (4 + 6) + 4 + 4 = 108
Or maybe:
(4*6+3)*4 = 108
I could go on and on, and I haven't even touched on the advanced operators...
answered 1 hour ago
Vilx-Vilx-
1315
$endgroup$
$begingroup$
The second answer here is not "using the numbers $3$, $4$ and $6$".
$endgroup$
– Weather Vane
29 mins ago
add a comment |
$begingroup$
In high school I used to play a game with my buddies where we tried to make valid arithmetic expressions out of licence plates of passing cars. The rules were pretty much similar except that you couldn't rearrange or duplicate (or omit) any of the digits. In time we got pretty good at it, basically solving any licence plate in under 10 seconds; most under 5. Except for the ones that couldn't be solved, of course.
Anyways, with your constraints (limitless numbers, can be rearranged, can be combined) there are like a zillion possible solutions.
Here's the simplest thing I could think of:
34 + 66 + 4 + 4 = 108
Here's another one without concatenation:
(4 + 6) * (4 + 6) + 4 + 4 = 108
Or maybe:
(4*6+3)*4 = 108
I could go on and on, and I haven't even touched on the advanced operators...
answered 1 hour ago
Vilx-Vilx-
1315
$endgroup$
$begingroup$
The second answer here is not "using the numbers $3$, $4$ and $6$".
$endgroup$
– Weather Vane
29 mins ago
add a comment |
$begingroup$
In high school I used to play a game with my buddies where we tried to make valid arithmetic expressions out of licence plates of passing cars. The rules were pretty much similar except that you couldn't rearrange or duplicate (or omit) any of the digits. In time we got pretty good at it, basically solving any licence plate in under 10 seconds; most under 5. Except for the ones that couldn't be solved, of course.
Anyways, with your constraints (limitless numbers, can be rearranged, can be combined) there are like a zillion possible solutions.
Here's the simplest thing I could think of:
34 + 66 + 4 + 4 = 108
Here's another one without concatenation:
(4 + 6) * (4 + 6) + 4 + 4 = 108
Or maybe:
(4*6+3)*4 = 108
I could go on and on, and I haven't even touched on the advanced operators...
answered 1 hour ago
Vilx-Vilx-
1315
$endgroup$
In high school I used to play a game with my buddies where we tried to make valid arithmetic expressions out of licence plates of passing cars. The rules were pretty much similar except that you couldn't rearrange or duplicate (or omit) any of the digits. In time we got pretty good at it, basically solving any licence plate in under 10 seconds; most under 5. Except for the ones that couldn't be solved, of course.
Anyways, with your constraints (limitless numbers, can be rearranged, can be combined) there are like a zillion possible solutions.
Here's the simplest thing I could think of:
34 + 66 + 4 + 4 = 108
Here's another one without concatenation:
(4 + 6) * (4 + 6) + 4 + 4 = 108
Or maybe:
(4*6+3)*4 = 108
I could go on and on, and I haven't even touched on the advanced operators...
answered 1 hour ago
Vilx-Vilx-
1315
edited 54 mins ago
answered 1 hour ago
Vilx-Vilx-
1315
answered 1 hour ago
Vilx-Vilx-
1315
answered 1 hour ago
Vilx-Vilx-
1315
1315
$begingroup$
The second answer here is not "using the numbers $3$, $4$ and $6$".
$endgroup$
– Weather Vane
29 mins ago
add a comment |
$begingroup$
The second answer here is not "using the numbers $3$, $4$ and $6$".
$endgroup$
– Weather Vane
29 mins ago
$begingroup$
The second answer here is not "using the numbers $3$, $4$ and $6$".
$endgroup$
– Weather Vane
29 mins ago
$begingroup$
The second answer here is not "using the numbers $3$, $4$ and $6$".
$endgroup$
– Weather Vane
29 mins ago
add a comment |
Thanks for contributing an answer to Puzzling Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fpuzzling.stackexchange.com%2fquestions%2f81947%2ffind-108-by-using-3-4-6%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
1
$begingroup$
If I could, I would have done something like $$3times styledisplay: inline-block; transform: rotate(180deg)4times 6=108$$ since the upside down $4$ looks a bit like a $6$.
$endgroup$
– user477343
2 hours ago
2
$begingroup$
@user477343 or even $$3 times 4 times 9 = 108$$ since the upside down $6$ is very much like a $9$.
$endgroup$
– Weather Vane
1 hour ago
$begingroup$
@WeatherVane Of course! Huh... funny how I saw what was obscure as opposed to what was actually just plain simple xD
$endgroup$
– user477343
1 hour ago
2
$begingroup$
@user477343 ah but it took yours to make me think of it.
$endgroup$
– Weather Vane
1 hour ago
$begingroup$
@WeatherVane :)
$endgroup$
– user477343
1 hour ago