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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Why is the underscore command _ useful?

Drawing a german abacus as in the books of Adam Ries

Can I criticise the more senior developers around me for not writing clean code?

As an international instructor, should I openly talk about my accent?



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 $((6-3)!)! = 6!$.










share|improve this question











$endgroup$







  • 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















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 $((6-3)!)! = 6!$.










share|improve this question











$endgroup$







  • 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













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 $((6-3)!)! = 6!$.










share|improve this question











$endgroup$




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






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Apr 18 at 17:56









JonMark Perry

20.9k64199




20.9k64199










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












  • 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










5 Answers
5






active

oldest

votes


















11












$begingroup$

Could this be




$frac6^3sqrt4 = frac2162 = 108$?




@Oray found another one, which might possibly be




$6^sqrt4 times 3 = 6^2 times 3 = 36 times 3 = 108$.







share|improve this answer











$endgroup$








  • 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


















14












$begingroup$

I have found this solution




$6 times (4! - 3!) = 6 times (24 - 6) = 6 times 18 = 108$







share|improve this answer









$endgroup$




















    2












    $begingroup$

    In Excel:




    (6^3)/sqrt(4)




    or as Word equation:










    share|improve this answer










    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


















    0












    $begingroup$

    No frills...




    ((6 x 6) x 3) + (4 x 0)







    share|improve this answer








    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


















    -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







    share|improve this answer










    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











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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




    $6^sqrt4 times 3 = 6^2 times 3 = 36 times 3 = 108$.







    share|improve this answer











    $endgroup$








    • 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















    11












    $begingroup$

    Could this be




    $frac6^3sqrt4 = frac2162 = 108$?




    @Oray found another one, which might possibly be




    $6^sqrt4 times 3 = 6^2 times 3 = 36 times 3 = 108$.







    share|improve this answer











    $endgroup$








    • 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













    11












    11








    11





    $begingroup$

    Could this be




    $frac6^3sqrt4 = frac2162 = 108$?




    @Oray found another one, which might possibly be




    $6^sqrt4 times 3 = 6^2 times 3 = 36 times 3 = 108$.







    share|improve this answer











    $endgroup$



    Could this be




    $frac6^3sqrt4 = frac2162 = 108$?




    @Oray found another one, which might possibly be




    $6^sqrt4 times 3 = 6^2 times 3 = 36 times 3 = 108$.








    share|improve this answer














    share|improve this answer



    share|improve this answer








    edited Apr 18 at 17:28

























    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












    • 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











    14












    $begingroup$

    I have found this solution




    $6 times (4! - 3!) = 6 times (24 - 6) = 6 times 18 = 108$







    share|improve this answer









    $endgroup$

















      14












      $begingroup$

      I have found this solution




      $6 times (4! - 3!) = 6 times (24 - 6) = 6 times 18 = 108$







      share|improve this answer









      $endgroup$















        14












        14








        14





        $begingroup$

        I have found this solution




        $6 times (4! - 3!) = 6 times (24 - 6) = 6 times 18 = 108$







        share|improve this answer









        $endgroup$



        I have found this solution




        $6 times (4! - 3!) = 6 times (24 - 6) = 6 times 18 = 108$








        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered Apr 18 at 17:30









        Weather VaneWeather Vane

        2,708114




        2,708114





















            2












            $begingroup$

            In Excel:




            (6^3)/sqrt(4)




            or as Word equation:










            share|improve this answer










            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















            2












            $begingroup$

            In Excel:




            (6^3)/sqrt(4)




            or as Word equation:










            share|improve this answer










            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













            2












            2








            2





            $begingroup$

            In Excel:




            (6^3)/sqrt(4)




            or as Word equation:










            share|improve this answer










            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:











            share|improve this answer










            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.









            share|improve this answer



            share|improve this answer








            edited Apr 19 at 6:43









            Glorfindel

            14.7k45687




            14.7k45687






            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.









            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.
            Check out our Code of Conduct.





            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.






            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












            • 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











            0












            $begingroup$

            No frills...




            ((6 x 6) x 3) + (4 x 0)







            share|improve this answer








            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















            0












            $begingroup$

            No frills...




            ((6 x 6) x 3) + (4 x 0)







            share|improve this answer








            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













            0












            0








            0





            $begingroup$

            No frills...




            ((6 x 6) x 3) + (4 x 0)







            share|improve this answer








            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)








            share|improve this answer








            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.









            share|improve this answer



            share|improve this answer






            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




            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












            • 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











            -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







            share|improve this answer










            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















            -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







            share|improve this answer










            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













            -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







            share|improve this answer










            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








            share|improve this answer










            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.









            share|improve this answer



            share|improve this answer








            edited Apr 19 at 15:21





















            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




            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












            • 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

















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