Find 108 by using 3,4,6Use 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
SFDX - Create Objects with Custom Properties
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Find 108 by using 3,4,6
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
4
$begingroup$
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 Apr 18 at 17:56
JonMark Perry
20.9k64199
asked Apr 18 at 17:05
OrayOray
16.2k437159
$endgroup$
|
show 3 more comments
4
$begingroup$
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 Apr 18 at 17:56
JonMark Perry
20.9k64199
asked Apr 18 at 17:05
OrayOray
16.2k437159
$endgroup$
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
Apr 18 at 17:44
6
$begingroup$
@user477343 or even $$3 times 4 times 9 = 108$$ since the upside down $6$ is very much like a $9$.
$endgroup$
– Weather Vane
Apr 18 at 18:12
1
$begingroup$
@WeatherVane Of course! Huh... funny how I saw what was obscure as opposed to what was actually just plain simple xD
$endgroup$
– user477343
Apr 18 at 18:19
3
$begingroup$
@user477343 ah but it took yours to make me think of it.
$endgroup$
– Weather Vane
Apr 18 at 18:19
1
$begingroup$
Based on Vilx's comment, and the fact someone's now posted an answer that uses 6 twice and doesn't use 4, I'm voting to close until it's clarified whether a) you have to use all three numbers, and b) you have to use them only once.
$endgroup$
– F1Krazy
Apr 19 at 13:32
|
show 3 more comments
4
4
4
2
$begingroup$
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 Apr 18 at 17:56
JonMark Perry
20.9k64199
asked Apr 18 at 17:05
OrayOray
16.2k437159
$endgroup$
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 Apr 18 at 17:56
JonMark Perry
20.9k64199
asked Apr 18 at 17:05
OrayOray
16.2k437159
edited Apr 18 at 17:56
JonMark Perry
20.9k64199
asked Apr 18 at 17:05
OrayOray
16.2k437159
edited Apr 18 at 17:56
JonMark Perry
20.9k64199
edited Apr 18 at 17:56
JonMark Perry
20.9k64199
edited Apr 18 at 17:56
JonMark Perry
20.9k64199
20.9k64199
asked Apr 18 at 17:05
OrayOray
16.2k437159
asked Apr 18 at 17:05
OrayOray
16.2k437159
asked Apr 18 at 17:05
OrayOray
16.2k437159
16.2k437159
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
Apr 18 at 17:44
6
$begingroup$
@user477343 or even $$3 times 4 times 9 = 108$$ since the upside down $6$ is very much like a $9$.
$endgroup$
– Weather Vane
Apr 18 at 18:12
1
$begingroup$
@WeatherVane Of course! Huh... funny how I saw what was obscure as opposed to what was actually just plain simple xD
$endgroup$
– user477343
Apr 18 at 18:19
3
$begingroup$
@user477343 ah but it took yours to make me think of it.
$endgroup$
– Weather Vane
Apr 18 at 18:19
1
$begingroup$
Based on Vilx's comment, and the fact someone's now posted an answer that uses 6 twice and doesn't use 4, I'm voting to close until it's clarified whether a) you have to use all three numbers, and b) you have to use them only once.
$endgroup$
– F1Krazy
Apr 19 at 13:32
|
show 3 more comments
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
Apr 18 at 17:44
6
$begingroup$
@user477343 or even $$3 times 4 times 9 = 108$$ since the upside down $6$ is very much like a $9$.
$endgroup$
– Weather Vane
Apr 18 at 18:12
1
$begingroup$
@WeatherVane Of course! Huh... funny how I saw what was obscure as opposed to what was actually just plain simple xD
$endgroup$
– user477343
Apr 18 at 18:19
3
$begingroup$
@user477343 ah but it took yours to make me think of it.
$endgroup$
– Weather Vane
Apr 18 at 18:19
1
$begingroup$
Based on Vilx's comment, and the fact someone's now posted an answer that uses 6 twice and doesn't use 4, I'm voting to close until it's clarified whether a) you have to use all three numbers, and b) you have to use them only once.
$endgroup$
– F1Krazy
Apr 19 at 13:32
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
Apr 18 at 17:44
$begingroup$
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
Apr 18 at 17:44
6
6
$begingroup$
@user477343 or even $$3 times 4 times 9 = 108$$ since the upside down $6$ is very much like a $9$.
$endgroup$
– Weather Vane
Apr 18 at 18:12
$begingroup$
@user477343 or even $$3 times 4 times 9 = 108$$ since the upside down $6$ is very much like a $9$.
$endgroup$
– Weather Vane
Apr 18 at 18:12
1
1
$begingroup$
@WeatherVane Of course! Huh... funny how I saw what was obscure as opposed to what was actually just plain simple xD
$endgroup$
– user477343
Apr 18 at 18:19
$begingroup$
@WeatherVane Of course! Huh... funny how I saw what was obscure as opposed to what was actually just plain simple xD
$endgroup$
– user477343
Apr 18 at 18:19
3
3
$begingroup$
@user477343 ah but it took yours to make me think of it.
$endgroup$
– Weather Vane
Apr 18 at 18:19
$begingroup$
@user477343 ah but it took yours to make me think of it.
$endgroup$
– Weather Vane
Apr 18 at 18:19
1
1
$begingroup$
Based on Vilx's comment, and the fact someone's now posted an answer that uses 6 twice and doesn't use 4, I'm voting to close until it's clarified whether a) you have to use all three numbers, and b) you have to use them only once.
$endgroup$
– F1Krazy
Apr 19 at 13:32
$begingroup$
Based on Vilx's comment, and the fact someone's now posted an answer that uses 6 twice and doesn't use 4, I'm voting to close until it's clarified whether a) you have to use all three numbers, and b) you have to use them only once.
$endgroup$
– F1Krazy
Apr 19 at 13:32
|
show 3 more comments
5 Answers
5
active
oldest
votes
11
$begingroup$
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 Apr 18 at 17:08
El-GuestEl-Guest
22.1k35193
$endgroup$
1
$begingroup$
good finding! mine was different but this seems right too :)
$endgroup$
– Oray
Apr 18 at 17:17
$begingroup$
Thank you, @Oray!!
$endgroup$
– El-Guest
Apr 18 at 17:26
$begingroup$
@Oray: was this second one the one that you found?
$endgroup$
– El-Guest
Apr 18 at 17:29
$begingroup$
no actually :D it was a bit more complicated.
$endgroup$
– Oray
Apr 18 at 17:40
add a comment |
14
$begingroup$
I have found this solution
$6 times (4! - 3!) = 6 times (24 - 6) = 6 times 18 = 108$
answered Apr 18 at 17:30
Weather VaneWeather Vane
2,708114
$endgroup$
add a comment |
2
$begingroup$
In Excel:
(6^3)/sqrt(4)
or as Word equation:
edited Apr 19 at 6:43
Glorfindel
14.7k45687
answered Apr 19 at 6:10
Barbara DebowskiBarbara Debowski
211
New contributor
Barbara Debowski is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
5
$begingroup$
Hi and welcome to puzzling SE! Good, but your answer is same as the accepted answer above. Please avoid posting similar answers, and also have a look at the tour page to familiarise yourselves with the workings of this site. Happy Puzzling!
$endgroup$
– Eagle
Apr 19 at 9:14
add a comment |
0
$begingroup$
No frills...
((6 x 6) x 3) + (4 x 0)
answered Apr 22 at 21:47
JonJon
1
New contributor
Jon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
1
$begingroup$
No dice... using 0 is expressly not allowed.
$endgroup$
– Rubio♦
Apr 22 at 22:55
$begingroup$
You're right. Thanks
$endgroup$
– Jon
2 days ago
add a comment |
-1
$begingroup$
The most simple answer would appear to be
3x6x6=108
Or this, for those who think all three numbers need to be used
(6!/3√4)-6√4=108
answered Apr 19 at 13:25
fluffy eragonfluffy eragon
11
New contributor
fluffy eragon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
2
$begingroup$
Welcome to Puzzling.SE! First of all, you should hide your answer using a spoiler tag>!, to avoid spoiling the solution for anyone who wants to have a go at the puzzle themselves. Secondly, it's unclear, but it seems like you have to use all three numbers, not just 3 and 6.
$endgroup$
– F1Krazy
Apr 19 at 13:30
1
$begingroup$
Sorry I should have used a spoiler tag will try to remember in future. Nothing was said about having to use 4 or only using each number once, my answer to the question as it was defined is correct. Someone has down-voted presumably because my answer doesn't fit their interpretation of the problem but that is an issue with how the question was worded not with my answer. I posted this trivial solution primarily to highlight the fact that the question people seem to be answering is not the one that was asked.
$endgroup$
– fluffy eragon
Apr 19 at 13:50
2
$begingroup$
It does say “using the numbers 3, 4 and 6” which seems clear enough to me to require all three. (It doesn’t say only one of each. I assume that was intended, but it’s certainly not stated.).
$endgroup$
– Rubio♦
Apr 19 at 14:37
2
$begingroup$
It also says "as long as all operands are either 3, 4 and 6". It does not say "using all of the numbers 3, 4 and 6 at least once". If that was the intention of the op then they phrased it very badly. Perhaps my answer should have been 3x6x6+4-4=108
$endgroup$
– fluffy eragon
Apr 19 at 15:07
add a comment |
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
11
$begingroup$
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 Apr 18 at 17:08
El-GuestEl-Guest
22.1k35193
$endgroup$
1
$begingroup$
good finding! mine was different but this seems right too :)
$endgroup$
– Oray
Apr 18 at 17:17
$begingroup$
Thank you, @Oray!!
$endgroup$
– El-Guest
Apr 18 at 17:26
$begingroup$
@Oray: was this second one the one that you found?
$endgroup$
– El-Guest
Apr 18 at 17:29
$begingroup$
no actually :D it was a bit more complicated.
$endgroup$
– Oray
Apr 18 at 17:40
add a comment |
11
$begingroup$
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 Apr 18 at 17:08
El-GuestEl-Guest
22.1k35193
$endgroup$
1
$begingroup$
good finding! mine was different but this seems right too :)
$endgroup$
– Oray
Apr 18 at 17:17
$begingroup$
Thank you, @Oray!!
$endgroup$
– El-Guest
Apr 18 at 17:26
$begingroup$
@Oray: was this second one the one that you found?
$endgroup$
– El-Guest
Apr 18 at 17:29
$begingroup$
no actually :D it was a bit more complicated.
$endgroup$
– Oray
Apr 18 at 17:40
add a comment |
11
11
11
$begingroup$
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 Apr 18 at 17:08
El-GuestEl-Guest
22.1k35193
$endgroup$
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 Apr 18 at 17:08
El-GuestEl-Guest
22.1k35193
edited Apr 18 at 17:28
answered Apr 18 at 17:08
El-GuestEl-Guest
22.1k35193
answered Apr 18 at 17:08
El-GuestEl-Guest
22.1k35193
answered Apr 18 at 17:08
El-GuestEl-Guest
22.1k35193
22.1k35193
1
$begingroup$
good finding! mine was different but this seems right too :)
$endgroup$
– Oray
Apr 18 at 17:17
$begingroup$
Thank you, @Oray!!
$endgroup$
– El-Guest
Apr 18 at 17:26
$begingroup$
@Oray: was this second one the one that you found?
$endgroup$
– El-Guest
Apr 18 at 17:29
$begingroup$
no actually :D it was a bit more complicated.
$endgroup$
– Oray
Apr 18 at 17:40
add a comment |
1
$begingroup$
good finding! mine was different but this seems right too :)
$endgroup$
– Oray
Apr 18 at 17:17
$begingroup$
Thank you, @Oray!!
$endgroup$
– El-Guest
Apr 18 at 17:26
$begingroup$
@Oray: was this second one the one that you found?
$endgroup$
– El-Guest
Apr 18 at 17:29
$begingroup$
no actually :D it was a bit more complicated.
$endgroup$
– Oray
Apr 18 at 17:40
1
1
$begingroup$
good finding! mine was different but this seems right too :)
$endgroup$
– Oray
Apr 18 at 17:17
$begingroup$
good finding! mine was different but this seems right too :)
$endgroup$
– Oray
Apr 18 at 17:17
$begingroup$
Thank you, @Oray!!
$endgroup$
– El-Guest
Apr 18 at 17:26
$begingroup$
Thank you, @Oray!!
$endgroup$
– El-Guest
Apr 18 at 17:26
$begingroup$
@Oray: was this second one the one that you found?
$endgroup$
– El-Guest
Apr 18 at 17:29
$begingroup$
@Oray: was this second one the one that you found?
$endgroup$
– El-Guest
Apr 18 at 17:29
$begingroup$
no actually :D it was a bit more complicated.
$endgroup$
– Oray
Apr 18 at 17:40
$begingroup$
no actually :D it was a bit more complicated.
$endgroup$
– Oray
Apr 18 at 17:40
add a comment |
14
$begingroup$
I have found this solution
$6 times (4! - 3!) = 6 times (24 - 6) = 6 times 18 = 108$
answered Apr 18 at 17:30
Weather VaneWeather Vane
2,708114
$endgroup$
add a comment |
14
$begingroup$
I have found this solution
$6 times (4! - 3!) = 6 times (24 - 6) = 6 times 18 = 108$
answered Apr 18 at 17:30
Weather VaneWeather Vane
2,708114
$endgroup$
add a comment |
14
14
14
$begingroup$
I have found this solution
$6 times (4! - 3!) = 6 times (24 - 6) = 6 times 18 = 108$
answered Apr 18 at 17:30
Weather VaneWeather Vane
2,708114
$endgroup$
I have found this solution
$6 times (4! - 3!) = 6 times (24 - 6) = 6 times 18 = 108$
answered Apr 18 at 17:30
Weather VaneWeather Vane
2,708114
answered Apr 18 at 17:30
Weather VaneWeather Vane
2,708114
answered Apr 18 at 17:30
Weather VaneWeather Vane
2,708114
answered Apr 18 at 17:30
Weather VaneWeather Vane
2,708114
2,708114
add a comment |
add a comment |
2
$begingroup$
In Excel:
(6^3)/sqrt(4)
or as Word equation:
edited Apr 19 at 6:43
Glorfindel
14.7k45687
answered Apr 19 at 6:10
Barbara DebowskiBarbara Debowski
211
New contributor
Barbara Debowski is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
5
$begingroup$
Hi and welcome to puzzling SE! Good, but your answer is same as the accepted answer above. Please avoid posting similar answers, and also have a look at the tour page to familiarise yourselves with the workings of this site. Happy Puzzling!
$endgroup$
– Eagle
Apr 19 at 9:14
add a comment |
2
$begingroup$
In Excel:
(6^3)/sqrt(4)
or as Word equation:
edited Apr 19 at 6:43
Glorfindel
14.7k45687
answered Apr 19 at 6:10
Barbara DebowskiBarbara Debowski
211
New contributor
Barbara Debowski is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
5
$begingroup$
Hi and welcome to puzzling SE! Good, but your answer is same as the accepted answer above. Please avoid posting similar answers, and also have a look at the tour page to familiarise yourselves with the workings of this site. Happy Puzzling!
$endgroup$
– Eagle
Apr 19 at 9:14
add a comment |
2
2
2
$begingroup$
In Excel:
(6^3)/sqrt(4)
or as Word equation:
edited Apr 19 at 6:43
Glorfindel
14.7k45687
answered Apr 19 at 6:10
Barbara DebowskiBarbara Debowski
211
New contributor
Barbara Debowski is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
In Excel:
(6^3)/sqrt(4)
or as Word equation:
edited Apr 19 at 6:43
Glorfindel
14.7k45687
answered Apr 19 at 6:10
Barbara DebowskiBarbara Debowski
211
New contributor
Barbara Debowski is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Apr 19 at 6:43
Glorfindel
14.7k45687
edited Apr 19 at 6:43
Glorfindel
14.7k45687
edited Apr 19 at 6:43
Glorfindel
14.7k45687
14.7k45687
answered Apr 19 at 6:10
Barbara DebowskiBarbara Debowski
211
New contributor
Barbara Debowski is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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answered Apr 19 at 6:10
Barbara DebowskiBarbara Debowski
211
answered Apr 19 at 6:10
Barbara DebowskiBarbara Debowski
211
211
New contributor
Barbara Debowski is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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New contributor
Barbara Debowski is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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Barbara Debowski is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
5
$begingroup$
Hi and welcome to puzzling SE! Good, but your answer is same as the accepted answer above. Please avoid posting similar answers, and also have a look at the tour page to familiarise yourselves with the workings of this site. Happy Puzzling!
$endgroup$
– Eagle
Apr 19 at 9:14
add a comment |
5
$begingroup$
Hi and welcome to puzzling SE! Good, but your answer is same as the accepted answer above. Please avoid posting similar answers, and also have a look at the tour page to familiarise yourselves with the workings of this site. Happy Puzzling!
$endgroup$
– Eagle
Apr 19 at 9:14
5
5
$begingroup$
Hi and welcome to puzzling SE! Good, but your answer is same as the accepted answer above. Please avoid posting similar answers, and also have a look at the tour page to familiarise yourselves with the workings of this site. Happy Puzzling!
$endgroup$
– Eagle
Apr 19 at 9:14
$begingroup$
Hi and welcome to puzzling SE! Good, but your answer is same as the accepted answer above. Please avoid posting similar answers, and also have a look at the tour page to familiarise yourselves with the workings of this site. Happy Puzzling!
$endgroup$
– Eagle
Apr 19 at 9:14
add a comment |
0
$begingroup$
No frills...
((6 x 6) x 3) + (4 x 0)
answered Apr 22 at 21:47
JonJon
1
New contributor
Jon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$
1
$begingroup$
No dice... using 0 is expressly not allowed.
$endgroup$
– Rubio♦
Apr 22 at 22:55
$begingroup$
You're right. Thanks
$endgroup$
– Jon
2 days ago
add a comment |
0
$begingroup$
No frills...
((6 x 6) x 3) + (4 x 0)
answered Apr 22 at 21:47
JonJon
1
New contributor
Jon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
1
$begingroup$
No dice... using 0 is expressly not allowed.
$endgroup$
– Rubio♦
Apr 22 at 22:55
$begingroup$
You're right. Thanks
$endgroup$
– Jon
2 days ago
add a comment |
0
0
0
$begingroup$
No frills...
((6 x 6) x 3) + (4 x 0)
answered Apr 22 at 21:47
JonJon
1
New contributor
Jon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
No frills...
((6 x 6) x 3) + (4 x 0)
answered Apr 22 at 21:47
JonJon
1
New contributor
Jon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered Apr 22 at 21:47
JonJon
1
New contributor
Jon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered Apr 22 at 21:47
JonJon
1
answered Apr 22 at 21:47
JonJon
1
1
New contributor
Jon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Jon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Jon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
$begingroup$
No dice... using 0 is expressly not allowed.
$endgroup$
– Rubio♦
Apr 22 at 22:55
$begingroup$
You're right. Thanks
$endgroup$
– Jon
2 days ago
add a comment |
1
$begingroup$
No dice... using 0 is expressly not allowed.
$endgroup$
– Rubio♦
Apr 22 at 22:55
$begingroup$
You're right. Thanks
$endgroup$
– Jon
2 days ago
1
1
$begingroup$
No dice... using 0 is expressly not allowed.
$endgroup$
– Rubio♦
Apr 22 at 22:55
$begingroup$
No dice... using 0 is expressly not allowed.
$endgroup$
– Rubio♦
Apr 22 at 22:55
$begingroup$
You're right. Thanks
$endgroup$
– Jon
2 days ago
$begingroup$
You're right. Thanks
$endgroup$
– Jon
2 days ago
add a comment |
-1
$begingroup$
The most simple answer would appear to be
3x6x6=108
Or this, for those who think all three numbers need to be used
(6!/3√4)-6√4=108
answered Apr 19 at 13:25
fluffy eragonfluffy eragon
11
New contributor
fluffy eragon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$
2
$begingroup$
Welcome to Puzzling.SE! First of all, you should hide your answer using a spoiler tag>!, to avoid spoiling the solution for anyone who wants to have a go at the puzzle themselves. Secondly, it's unclear, but it seems like you have to use all three numbers, not just 3 and 6.
$endgroup$
– F1Krazy
Apr 19 at 13:30
1
$begingroup$
Sorry I should have used a spoiler tag will try to remember in future. Nothing was said about having to use 4 or only using each number once, my answer to the question as it was defined is correct. Someone has down-voted presumably because my answer doesn't fit their interpretation of the problem but that is an issue with how the question was worded not with my answer. I posted this trivial solution primarily to highlight the fact that the question people seem to be answering is not the one that was asked.
$endgroup$
– fluffy eragon
Apr 19 at 13:50
2
$begingroup$
It does say “using the numbers 3, 4 and 6” which seems clear enough to me to require all three. (It doesn’t say only one of each. I assume that was intended, but it’s certainly not stated.).
$endgroup$
– Rubio♦
Apr 19 at 14:37
2
$begingroup$
It also says "as long as all operands are either 3, 4 and 6". It does not say "using all of the numbers 3, 4 and 6 at least once". If that was the intention of the op then they phrased it very badly. Perhaps my answer should have been 3x6x6+4-4=108
$endgroup$
– fluffy eragon
Apr 19 at 15:07
add a comment |
-1
$begingroup$
The most simple answer would appear to be
3x6x6=108
Or this, for those who think all three numbers need to be used
(6!/3√4)-6√4=108
answered Apr 19 at 13:25
fluffy eragonfluffy eragon
11
New contributor
fluffy eragon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
2
$begingroup$
Welcome to Puzzling.SE! First of all, you should hide your answer using a spoiler tag>!, to avoid spoiling the solution for anyone who wants to have a go at the puzzle themselves. Secondly, it's unclear, but it seems like you have to use all three numbers, not just 3 and 6.
$endgroup$
– F1Krazy
Apr 19 at 13:30
1
$begingroup$
Sorry I should have used a spoiler tag will try to remember in future. Nothing was said about having to use 4 or only using each number once, my answer to the question as it was defined is correct. Someone has down-voted presumably because my answer doesn't fit their interpretation of the problem but that is an issue with how the question was worded not with my answer. I posted this trivial solution primarily to highlight the fact that the question people seem to be answering is not the one that was asked.
$endgroup$
– fluffy eragon
Apr 19 at 13:50
2
$begingroup$
It does say “using the numbers 3, 4 and 6” which seems clear enough to me to require all three. (It doesn’t say only one of each. I assume that was intended, but it’s certainly not stated.).
$endgroup$
– Rubio♦
Apr 19 at 14:37
2
$begingroup$
It also says "as long as all operands are either 3, 4 and 6". It does not say "using all of the numbers 3, 4 and 6 at least once". If that was the intention of the op then they phrased it very badly. Perhaps my answer should have been 3x6x6+4-4=108
$endgroup$
– fluffy eragon
Apr 19 at 15:07
add a comment |
-1
-1
-1
$begingroup$
The most simple answer would appear to be
3x6x6=108
Or this, for those who think all three numbers need to be used
(6!/3√4)-6√4=108
answered Apr 19 at 13:25
fluffy eragonfluffy eragon
11
New contributor
fluffy eragon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
The most simple answer would appear to be
3x6x6=108
Or this, for those who think all three numbers need to be used
(6!/3√4)-6√4=108
answered Apr 19 at 13:25
fluffy eragonfluffy eragon
11
New contributor
fluffy eragon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Apr 19 at 15:21
answered Apr 19 at 13:25
fluffy eragonfluffy eragon
11
New contributor
fluffy eragon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered Apr 19 at 13:25
fluffy eragonfluffy eragon
11
answered Apr 19 at 13:25
fluffy eragonfluffy eragon
11
11
New contributor
fluffy eragon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
fluffy eragon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
fluffy eragon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
2
$begingroup$
Welcome to Puzzling.SE! First of all, you should hide your answer using a spoiler tag>!, to avoid spoiling the solution for anyone who wants to have a go at the puzzle themselves. Secondly, it's unclear, but it seems like you have to use all three numbers, not just 3 and 6.
$endgroup$
– F1Krazy
Apr 19 at 13:30
1
$begingroup$
Sorry I should have used a spoiler tag will try to remember in future. Nothing was said about having to use 4 or only using each number once, my answer to the question as it was defined is correct. Someone has down-voted presumably because my answer doesn't fit their interpretation of the problem but that is an issue with how the question was worded not with my answer. I posted this trivial solution primarily to highlight the fact that the question people seem to be answering is not the one that was asked.
$endgroup$
– fluffy eragon
Apr 19 at 13:50
2
$begingroup$
It does say “using the numbers 3, 4 and 6” which seems clear enough to me to require all three. (It doesn’t say only one of each. I assume that was intended, but it’s certainly not stated.).
$endgroup$
– Rubio♦
Apr 19 at 14:37
2
$begingroup$
It also says "as long as all operands are either 3, 4 and 6". It does not say "using all of the numbers 3, 4 and 6 at least once". If that was the intention of the op then they phrased it very badly. Perhaps my answer should have been 3x6x6+4-4=108
$endgroup$
– fluffy eragon
Apr 19 at 15:07
add a comment |
2
$begingroup$
Welcome to Puzzling.SE! First of all, you should hide your answer using a spoiler tag>!, to avoid spoiling the solution for anyone who wants to have a go at the puzzle themselves. Secondly, it's unclear, but it seems like you have to use all three numbers, not just 3 and 6.
$endgroup$
– F1Krazy
Apr 19 at 13:30
1
$begingroup$
Sorry I should have used a spoiler tag will try to remember in future. Nothing was said about having to use 4 or only using each number once, my answer to the question as it was defined is correct. Someone has down-voted presumably because my answer doesn't fit their interpretation of the problem but that is an issue with how the question was worded not with my answer. I posted this trivial solution primarily to highlight the fact that the question people seem to be answering is not the one that was asked.
$endgroup$
– fluffy eragon
Apr 19 at 13:50
2
$begingroup$
It does say “using the numbers 3, 4 and 6” which seems clear enough to me to require all three. (It doesn’t say only one of each. I assume that was intended, but it’s certainly not stated.).
$endgroup$
– Rubio♦
Apr 19 at 14:37
2
$begingroup$
It also says "as long as all operands are either 3, 4 and 6". It does not say "using all of the numbers 3, 4 and 6 at least once". If that was the intention of the op then they phrased it very badly. Perhaps my answer should have been 3x6x6+4-4=108
$endgroup$
– fluffy eragon
Apr 19 at 15:07
2
2
$begingroup$
Welcome to Puzzling.SE! First of all, you should hide your answer using a spoiler tag
>!, to avoid spoiling the solution for anyone who wants to have a go at the puzzle themselves. Secondly, it's unclear, but it seems like you have to use all three numbers, not just 3 and 6.$endgroup$
– F1Krazy
Apr 19 at 13:30
$begingroup$
Welcome to Puzzling.SE! First of all, you should hide your answer using a spoiler tag
>!, to avoid spoiling the solution for anyone who wants to have a go at the puzzle themselves. Secondly, it's unclear, but it seems like you have to use all three numbers, not just 3 and 6.$endgroup$
– F1Krazy
Apr 19 at 13:30
1
1
$begingroup$
Sorry I should have used a spoiler tag will try to remember in future. Nothing was said about having to use 4 or only using each number once, my answer to the question as it was defined is correct. Someone has down-voted presumably because my answer doesn't fit their interpretation of the problem but that is an issue with how the question was worded not with my answer. I posted this trivial solution primarily to highlight the fact that the question people seem to be answering is not the one that was asked.
$endgroup$
– fluffy eragon
Apr 19 at 13:50
$begingroup$
Sorry I should have used a spoiler tag will try to remember in future. Nothing was said about having to use 4 or only using each number once, my answer to the question as it was defined is correct. Someone has down-voted presumably because my answer doesn't fit their interpretation of the problem but that is an issue with how the question was worded not with my answer. I posted this trivial solution primarily to highlight the fact that the question people seem to be answering is not the one that was asked.
$endgroup$
– fluffy eragon
Apr 19 at 13:50
2
2
$begingroup$
It does say “using the numbers 3, 4 and 6” which seems clear enough to me to require all three. (It doesn’t say only one of each. I assume that was intended, but it’s certainly not stated.).
$endgroup$
– Rubio♦
Apr 19 at 14:37
$begingroup$
It does say “using the numbers 3, 4 and 6” which seems clear enough to me to require all three. (It doesn’t say only one of each. I assume that was intended, but it’s certainly not stated.).
$endgroup$
– Rubio♦
Apr 19 at 14:37
2
2
$begingroup$
It also says "as long as all operands are either 3, 4 and 6". It does not say "using all of the numbers 3, 4 and 6 at least once". If that was the intention of the op then they phrased it very badly. Perhaps my answer should have been 3x6x6+4-4=108
$endgroup$
– fluffy eragon
Apr 19 at 15:07
$begingroup$
It also says "as long as all operands are either 3, 4 and 6". It does not say "using all of the numbers 3, 4 and 6 at least once". If that was the intention of the op then they phrased it very badly. Perhaps my answer should have been 3x6x6+4-4=108
$endgroup$
– fluffy eragon
Apr 19 at 15:07
add a comment |
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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
Apr 18 at 17:44
6
$begingroup$
@user477343 or even $$3 times 4 times 9 = 108$$ since the upside down $6$ is very much like a $9$.
$endgroup$
– Weather Vane
Apr 18 at 18:12
1
$begingroup$
@WeatherVane Of course! Huh... funny how I saw what was obscure as opposed to what was actually just plain simple xD
$endgroup$
– user477343
Apr 18 at 18:19
3
$begingroup$
@user477343 ah but it took yours to make me think of it.
$endgroup$
– Weather Vane
Apr 18 at 18:19
1
$begingroup$
Based on Vilx's comment, and the fact someone's now posted an answer that uses 6 twice and doesn't use 4, I'm voting to close until it's clarified whether a) you have to use all three numbers, and b) you have to use them only once.
$endgroup$
– F1Krazy
Apr 19 at 13:32