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
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Find 108 by using 3,4,6
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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 4 hours ago
JonMark Perry
20.8k64199
asked 5 hours ago
OrayOray
16.2k437157
$endgroup$
|
show 1 more 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 4 hours ago
JonMark Perry
20.8k64199
asked 5 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
4 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
4 hours 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
4 hours ago
2
$begingroup$
@user477343 ah but it took yours to make me think of it.
$endgroup$
– Weather Vane
4 hours ago
$begingroup$
@WeatherVane :)
$endgroup$
– user477343
4 hours ago
|
show 1 more 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 4 hours ago
JonMark Perry
20.8k64199
asked 5 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 4 hours ago
JonMark Perry
20.8k64199
asked 5 hours ago
OrayOray
16.2k437157
edited 4 hours ago
JonMark Perry
20.8k64199
asked 5 hours ago
OrayOray
16.2k437157
edited 4 hours ago
JonMark Perry
20.8k64199
edited 4 hours ago
JonMark Perry
20.8k64199
edited 4 hours ago
JonMark Perry
20.8k64199
20.8k64199
asked 5 hours ago
OrayOray
16.2k437157
asked 5 hours ago
OrayOray
16.2k437157
asked 5 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
4 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
4 hours 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
4 hours ago
2
$begingroup$
@user477343 ah but it took yours to make me think of it.
$endgroup$
– Weather Vane
4 hours ago
$begingroup$
@WeatherVane :)
$endgroup$
– user477343
4 hours ago
|
show 1 more 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
4 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
4 hours 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
4 hours ago
2
$begingroup$
@user477343 ah but it took yours to make me think of it.
$endgroup$
– Weather Vane
4 hours ago
$begingroup$
@WeatherVane :)
$endgroup$
– user477343
4 hours 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
4 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
4 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
4 hours 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
4 hours 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
4 hours 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
4 hours ago
2
2
$begingroup$
@user477343 ah but it took yours to make me think of it.
$endgroup$
– Weather Vane
4 hours ago
$begingroup$
@user477343 ah but it took yours to make me think of it.
$endgroup$
– Weather Vane
4 hours ago
$begingroup$
@WeatherVane :)
$endgroup$
– user477343
4 hours ago
$begingroup$
@WeatherVane :)
$endgroup$
– user477343
4 hours ago
|
show 1 more 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 5 hours ago
El-GuestEl-Guest
21.8k35092
$endgroup$
$begingroup$
good finding! mine was different but this seems right too :)
$endgroup$
– Oray
5 hours ago
$begingroup$
Thank you, @Oray!!
$endgroup$
– El-Guest
5 hours ago
$begingroup$
@Oray: was this second one the one that you found?
$endgroup$
– El-Guest
5 hours ago
$begingroup$
no actually :D it was a bit more complicated.
$endgroup$
– Oray
4 hours ago
add a comment |
$begingroup$
I have found this solution
$6 times (4! - 3!) = 6 times (24 - 6) = 6 times 18 = 108$
answered 4 hours ago
Weather VaneWeather Vane
2,407112
$endgroup$
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 + 3 + 3) * (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 3 hours ago
Vilx-Vilx-
1315
$endgroup$
$begingroup$
The second answer here is not "using the numbers $3$, $4$ and $6$".
$endgroup$
– Weather Vane
3 hours ago
$begingroup$
@WeatherVane - Do you mean that I had to use all of them? It's not really clear from the question. I understood that those are the only allowed digits, but that I can use (or not use) as many of each as I want.
$endgroup$
– Vilx-
2 hours ago
$begingroup$
@WeatherVane - Anyways, updated it to use all the numbers.
$endgroup$
– Vilx-
2 hours ago
add a comment |
Your Answer
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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 5 hours ago
El-GuestEl-Guest
21.8k35092
$endgroup$
$begingroup$
good finding! mine was different but this seems right too :)
$endgroup$
– Oray
5 hours ago
$begingroup$
Thank you, @Oray!!
$endgroup$
– El-Guest
5 hours ago
$begingroup$
@Oray: was this second one the one that you found?
$endgroup$
– El-Guest
5 hours ago
$begingroup$
no actually :D it was a bit more complicated.
$endgroup$
– Oray
4 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 5 hours ago
El-GuestEl-Guest
21.8k35092
$endgroup$
$begingroup$
good finding! mine was different but this seems right too :)
$endgroup$
– Oray
5 hours ago
$begingroup$
Thank you, @Oray!!
$endgroup$
– El-Guest
5 hours ago
$begingroup$
@Oray: was this second one the one that you found?
$endgroup$
– El-Guest
5 hours ago
$begingroup$
no actually :D it was a bit more complicated.
$endgroup$
– Oray
4 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 5 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 5 hours ago
El-GuestEl-Guest
21.8k35092
edited 5 hours ago
answered 5 hours ago
El-GuestEl-Guest
21.8k35092
answered 5 hours ago
El-GuestEl-Guest
21.8k35092
answered 5 hours ago
El-GuestEl-Guest
21.8k35092
21.8k35092
$begingroup$
good finding! mine was different but this seems right too :)
$endgroup$
– Oray
5 hours ago
$begingroup$
Thank you, @Oray!!
$endgroup$
– El-Guest
5 hours ago
$begingroup$
@Oray: was this second one the one that you found?
$endgroup$
– El-Guest
5 hours ago
$begingroup$
no actually :D it was a bit more complicated.
$endgroup$
– Oray
4 hours ago
add a comment |
$begingroup$
good finding! mine was different but this seems right too :)
$endgroup$
– Oray
5 hours ago
$begingroup$
Thank you, @Oray!!
$endgroup$
– El-Guest
5 hours ago
$begingroup$
@Oray: was this second one the one that you found?
$endgroup$
– El-Guest
5 hours ago
$begingroup$
no actually :D it was a bit more complicated.
$endgroup$
– Oray
4 hours ago
$begingroup$
good finding! mine was different but this seems right too :)
$endgroup$
– Oray
5 hours ago
$begingroup$
good finding! mine was different but this seems right too :)
$endgroup$
– Oray
5 hours ago
$begingroup$
Thank you, @Oray!!
$endgroup$
– El-Guest
5 hours ago
$begingroup$
Thank you, @Oray!!
$endgroup$
– El-Guest
5 hours ago
$begingroup$
@Oray: was this second one the one that you found?
$endgroup$
– El-Guest
5 hours ago
$begingroup$
@Oray: was this second one the one that you found?
$endgroup$
– El-Guest
5 hours ago
$begingroup$
no actually :D it was a bit more complicated.
$endgroup$
– Oray
4 hours ago
$begingroup$
no actually :D it was a bit more complicated.
$endgroup$
– Oray
4 hours ago
add a comment |
$begingroup$
I have found this solution
$6 times (4! - 3!) = 6 times (24 - 6) = 6 times 18 = 108$
answered 4 hours ago
Weather VaneWeather Vane
2,407112
$endgroup$
add a comment |
$begingroup$
I have found this solution
$6 times (4! - 3!) = 6 times (24 - 6) = 6 times 18 = 108$
answered 4 hours ago
Weather VaneWeather Vane
2,407112
$endgroup$
add a comment |
$begingroup$
I have found this solution
$6 times (4! - 3!) = 6 times (24 - 6) = 6 times 18 = 108$
answered 4 hours ago
Weather VaneWeather Vane
2,407112
$endgroup$
I have found this solution
$6 times (4! - 3!) = 6 times (24 - 6) = 6 times 18 = 108$
answered 4 hours ago
Weather VaneWeather Vane
2,407112
answered 4 hours ago
Weather VaneWeather Vane
2,407112
answered 4 hours ago
Weather VaneWeather Vane
2,407112
answered 4 hours ago
Weather VaneWeather Vane
2,407112
2,407112
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 + 3 + 3) * (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 3 hours ago
Vilx-Vilx-
1315
$endgroup$
$begingroup$
The second answer here is not "using the numbers $3$, $4$ and $6$".
$endgroup$
– Weather Vane
3 hours ago
$begingroup$
@WeatherVane - Do you mean that I had to use all of them? It's not really clear from the question. I understood that those are the only allowed digits, but that I can use (or not use) as many of each as I want.
$endgroup$
– Vilx-
2 hours ago
$begingroup$
@WeatherVane - Anyways, updated it to use all the numbers.
$endgroup$
– Vilx-
2 hours 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 + 3 + 3) * (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 3 hours ago
Vilx-Vilx-
1315
$endgroup$
$begingroup$
The second answer here is not "using the numbers $3$, $4$ and $6$".
$endgroup$
– Weather Vane
3 hours ago
$begingroup$
@WeatherVane - Do you mean that I had to use all of them? It's not really clear from the question. I understood that those are the only allowed digits, but that I can use (or not use) as many of each as I want.
$endgroup$
– Vilx-
2 hours ago
$begingroup$
@WeatherVane - Anyways, updated it to use all the numbers.
$endgroup$
– Vilx-
2 hours 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 + 3 + 3) * (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 3 hours 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 + 3 + 3) * (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 3 hours ago
Vilx-Vilx-
1315
edited 2 hours ago
answered 3 hours ago
Vilx-Vilx-
1315
answered 3 hours ago
Vilx-Vilx-
1315
answered 3 hours ago
Vilx-Vilx-
1315
1315
$begingroup$
The second answer here is not "using the numbers $3$, $4$ and $6$".
$endgroup$
– Weather Vane
3 hours ago
$begingroup$
@WeatherVane - Do you mean that I had to use all of them? It's not really clear from the question. I understood that those are the only allowed digits, but that I can use (or not use) as many of each as I want.
$endgroup$
– Vilx-
2 hours ago
$begingroup$
@WeatherVane - Anyways, updated it to use all the numbers.
$endgroup$
– Vilx-
2 hours ago
add a comment |
$begingroup$
The second answer here is not "using the numbers $3$, $4$ and $6$".
$endgroup$
– Weather Vane
3 hours ago
$begingroup$
@WeatherVane - Do you mean that I had to use all of them? It's not really clear from the question. I understood that those are the only allowed digits, but that I can use (or not use) as many of each as I want.
$endgroup$
– Vilx-
2 hours ago
$begingroup$
@WeatherVane - Anyways, updated it to use all the numbers.
$endgroup$
– Vilx-
2 hours ago
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The second answer here is not "using the numbers $3$, $4$ and $6$".
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– Weather Vane
3 hours ago
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The second answer here is not "using the numbers $3$, $4$ and $6$".
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– Weather Vane
3 hours ago
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@WeatherVane - Do you mean that I had to use all of them? It's not really clear from the question. I understood that those are the only allowed digits, but that I can use (or not use) as many of each as I want.
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– Vilx-
2 hours ago
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@WeatherVane - Do you mean that I had to use all of them? It's not really clear from the question. I understood that those are the only allowed digits, but that I can use (or not use) as many of each as I want.
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– Vilx-
2 hours ago
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@WeatherVane - Anyways, updated it to use all the numbers.
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– Vilx-
2 hours ago
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@WeatherVane - Anyways, updated it to use all the numbers.
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– Vilx-
2 hours ago
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1
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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 $6$.
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– user477343
4 hours ago
2
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@user477343 or even $$3 times 4 times 9 = 108$$ since the upside down $6$ is very much like a $9$.
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– Weather Vane
4 hours ago
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@WeatherVane Of course! Huh... funny how I saw what was obscure as opposed to what was actually just plain simple xD
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– user477343
4 hours ago
2
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@user477343 ah but it took yours to make me think of it.
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– Weather Vane
4 hours ago
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@WeatherVane :)
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– user477343
4 hours ago