Find 108 by using 3,4,6












2












$begingroup$


Assemble a formula using the numbers $3$, $4$ and $6$ in any order to make 108.



You may use the operations;




  • $x + y$


  • $x - y$


  • $x times y$


  • $x div y$


  • $x!$


  • $sqrt{x}$


  • $sqrt[leftroot{-2}uproot{2}x]{y}$


  • $x^y$


  • 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











$endgroup$








  • 1




    $begingroup$
    If I could, I would have done something like $$3times style{display: inline-block; transform: rotate(180deg)}{4}times 6=108$$ since the upside down $4$ looks a bit like a $6$.
    $endgroup$
    – user477343
    2 hours ago








  • 2




    $begingroup$
    @user477343 or even $$3 times 4 times 9 = 108$$ since the upside down $6$ is very much like a $9$.
    $endgroup$
    – Weather Vane
    1 hour ago












  • $begingroup$
    @WeatherVane Of course! Huh... funny how I saw what was obscure as opposed to what was actually just plain simple xD
    $endgroup$
    – user477343
    1 hour ago








  • 2




    $begingroup$
    @user477343 ah but it took yours to make me think of it.
    $endgroup$
    – Weather Vane
    1 hour ago










  • $begingroup$
    @WeatherVane :)
    $endgroup$
    – user477343
    1 hour ago
















2












$begingroup$


Assemble a formula using the numbers $3$, $4$ and $6$ in any order to make 108.



You may use the operations;




  • $x + y$


  • $x - y$


  • $x times y$


  • $x div y$


  • $x!$


  • $sqrt{x}$


  • $sqrt[leftroot{-2}uproot{2}x]{y}$


  • $x^y$


  • 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











$endgroup$








  • 1




    $begingroup$
    If I could, I would have done something like $$3times style{display: inline-block; transform: rotate(180deg)}{4}times 6=108$$ since the upside down $4$ looks a bit like a $6$.
    $endgroup$
    – user477343
    2 hours ago








  • 2




    $begingroup$
    @user477343 or even $$3 times 4 times 9 = 108$$ since the upside down $6$ is very much like a $9$.
    $endgroup$
    – Weather Vane
    1 hour ago












  • $begingroup$
    @WeatherVane Of course! Huh... funny how I saw what was obscure as opposed to what was actually just plain simple xD
    $endgroup$
    – user477343
    1 hour ago








  • 2




    $begingroup$
    @user477343 ah but it took yours to make me think of it.
    $endgroup$
    – Weather Vane
    1 hour ago










  • $begingroup$
    @WeatherVane :)
    $endgroup$
    – user477343
    1 hour ago














2












2








2


1



$begingroup$


Assemble a formula using the numbers $3$, $4$ and $6$ in any order to make 108.



You may use the operations;




  • $x + y$


  • $x - y$


  • $x times y$


  • $x div y$


  • $x!$


  • $sqrt{x}$


  • $sqrt[leftroot{-2}uproot{2}x]{y}$


  • $x^y$


  • 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











$endgroup$




Assemble a formula using the numbers $3$, $4$ and $6$ in any order to make 108.



You may use the operations;




  • $x + y$


  • $x - y$


  • $x times y$


  • $x div y$


  • $x!$


  • $sqrt{x}$


  • $sqrt[leftroot{-2}uproot{2}x]{y}$


  • $x^y$


  • 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






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited 2 hours ago









JonMark Perry

20.8k64199




20.8k64199










asked 2 hours ago









OrayOray

16.2k437157




16.2k437157








  • 1




    $begingroup$
    If I could, I would have done something like $$3times style{display: inline-block; transform: rotate(180deg)}{4}times 6=108$$ since the upside down $4$ looks a bit like a $6$.
    $endgroup$
    – user477343
    2 hours ago








  • 2




    $begingroup$
    @user477343 or even $$3 times 4 times 9 = 108$$ since the upside down $6$ is very much like a $9$.
    $endgroup$
    – Weather Vane
    1 hour ago












  • $begingroup$
    @WeatherVane Of course! Huh... funny how I saw what was obscure as opposed to what was actually just plain simple xD
    $endgroup$
    – user477343
    1 hour ago








  • 2




    $begingroup$
    @user477343 ah but it took yours to make me think of it.
    $endgroup$
    – Weather Vane
    1 hour ago










  • $begingroup$
    @WeatherVane :)
    $endgroup$
    – user477343
    1 hour ago














  • 1




    $begingroup$
    If I could, I would have done something like $$3times style{display: inline-block; transform: rotate(180deg)}{4}times 6=108$$ since the upside down $4$ looks a bit like a $6$.
    $endgroup$
    – user477343
    2 hours ago








  • 2




    $begingroup$
    @user477343 or even $$3 times 4 times 9 = 108$$ since the upside down $6$ is very much like a $9$.
    $endgroup$
    – Weather Vane
    1 hour ago












  • $begingroup$
    @WeatherVane Of course! Huh... funny how I saw what was obscure as opposed to what was actually just plain simple xD
    $endgroup$
    – user477343
    1 hour ago








  • 2




    $begingroup$
    @user477343 ah but it took yours to make me think of it.
    $endgroup$
    – Weather Vane
    1 hour ago










  • $begingroup$
    @WeatherVane :)
    $endgroup$
    – user477343
    1 hour ago








1




1




$begingroup$
If I could, I would have done something like $$3times style{display: inline-block; transform: rotate(180deg)}{4}times 6=108$$ since the upside down $4$ looks a bit like a $6$.
$endgroup$
– user477343
2 hours ago






$begingroup$
If I could, I would have done something like $$3times style{display: inline-block; transform: rotate(180deg)}{4}times 6=108$$ since the upside down $4$ looks a bit like a $6$.
$endgroup$
– user477343
2 hours ago






2




2




$begingroup$
@user477343 or even $$3 times 4 times 9 = 108$$ since the upside down $6$ is very much like a $9$.
$endgroup$
– Weather Vane
1 hour ago






$begingroup$
@user477343 or even $$3 times 4 times 9 = 108$$ since the upside down $6$ is very much like a $9$.
$endgroup$
– Weather Vane
1 hour ago














$begingroup$
@WeatherVane Of course! Huh... funny how I saw what was obscure as opposed to what was actually just plain simple xD
$endgroup$
– user477343
1 hour ago






$begingroup$
@WeatherVane Of course! Huh... funny how I saw what was obscure as opposed to what was actually just plain simple xD
$endgroup$
– user477343
1 hour ago






2




2




$begingroup$
@user477343 ah but it took yours to make me think of it.
$endgroup$
– Weather Vane
1 hour ago




$begingroup$
@user477343 ah but it took yours to make me think of it.
$endgroup$
– Weather Vane
1 hour ago












$begingroup$
@WeatherVane :)
$endgroup$
– user477343
1 hour ago




$begingroup$
@WeatherVane :)
$endgroup$
– user477343
1 hour ago










3 Answers
3






active

oldest

votes


















3












$begingroup$

Could this be




$frac{6^3}{sqrt{4}} = frac{216}{2} = 108$?




@Oray found another one, which might possibly be




$6^{sqrt{4}} times 3 = 6^2 times 3 = 36 times 3 = 108$.







share|improve this answer











$endgroup$













  • $begingroup$
    good finding! mine was different but this seems right too :)
    $endgroup$
    – Oray
    2 hours ago










  • $begingroup$
    Thank you, @Oray!!
    $endgroup$
    – El-Guest
    2 hours ago










  • $begingroup$
    @Oray: was this second one the one that you found?
    $endgroup$
    – El-Guest
    2 hours ago










  • $begingroup$
    no actually :D it was a bit more complicated.
    $endgroup$
    – Oray
    2 hours ago



















4












$begingroup$

I have found this solution




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







share|improve this answer









$endgroup$





















    0












    $begingroup$

    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...






    share|improve this answer











    $endgroup$













    • $begingroup$
      The second answer here is not "using the numbers $3$, $4$ and $6$".
      $endgroup$
      – Weather Vane
      48 mins 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-
      4 mins ago










    • $begingroup$
      @WeatherVane - Anyways, updated it to use all the numbers.
      $endgroup$
      – Vilx-
      2 mins ago












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    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    3












    $begingroup$

    Could this be




    $frac{6^3}{sqrt{4}} = frac{216}{2} = 108$?




    @Oray found another one, which might possibly be




    $6^{sqrt{4}} times 3 = 6^2 times 3 = 36 times 3 = 108$.







    share|improve this answer











    $endgroup$













    • $begingroup$
      good finding! mine was different but this seems right too :)
      $endgroup$
      – Oray
      2 hours ago










    • $begingroup$
      Thank you, @Oray!!
      $endgroup$
      – El-Guest
      2 hours ago










    • $begingroup$
      @Oray: was this second one the one that you found?
      $endgroup$
      – El-Guest
      2 hours ago










    • $begingroup$
      no actually :D it was a bit more complicated.
      $endgroup$
      – Oray
      2 hours ago
















    3












    $begingroup$

    Could this be




    $frac{6^3}{sqrt{4}} = frac{216}{2} = 108$?




    @Oray found another one, which might possibly be




    $6^{sqrt{4}} times 3 = 6^2 times 3 = 36 times 3 = 108$.







    share|improve this answer











    $endgroup$













    • $begingroup$
      good finding! mine was different but this seems right too :)
      $endgroup$
      – Oray
      2 hours ago










    • $begingroup$
      Thank you, @Oray!!
      $endgroup$
      – El-Guest
      2 hours ago










    • $begingroup$
      @Oray: was this second one the one that you found?
      $endgroup$
      – El-Guest
      2 hours ago










    • $begingroup$
      no actually :D it was a bit more complicated.
      $endgroup$
      – Oray
      2 hours ago














    3












    3








    3





    $begingroup$

    Could this be




    $frac{6^3}{sqrt{4}} = frac{216}{2} = 108$?




    @Oray found another one, which might possibly be




    $6^{sqrt{4}} times 3 = 6^2 times 3 = 36 times 3 = 108$.







    share|improve this answer











    $endgroup$



    Could this be




    $frac{6^3}{sqrt{4}} = frac{216}{2} = 108$?




    @Oray found another one, which might possibly be




    $6^{sqrt{4}} times 3 = 6^2 times 3 = 36 times 3 = 108$.








    share|improve this answer














    share|improve this answer



    share|improve this answer








    edited 2 hours ago

























    answered 2 hours ago









    El-GuestEl-Guest

    21.8k35092




    21.8k35092












    • $begingroup$
      good finding! mine was different but this seems right too :)
      $endgroup$
      – Oray
      2 hours ago










    • $begingroup$
      Thank you, @Oray!!
      $endgroup$
      – El-Guest
      2 hours ago










    • $begingroup$
      @Oray: was this second one the one that you found?
      $endgroup$
      – El-Guest
      2 hours ago










    • $begingroup$
      no actually :D it was a bit more complicated.
      $endgroup$
      – Oray
      2 hours ago


















    • $begingroup$
      good finding! mine was different but this seems right too :)
      $endgroup$
      – Oray
      2 hours ago










    • $begingroup$
      Thank you, @Oray!!
      $endgroup$
      – El-Guest
      2 hours ago










    • $begingroup$
      @Oray: was this second one the one that you found?
      $endgroup$
      – El-Guest
      2 hours ago










    • $begingroup$
      no actually :D it was a bit more complicated.
      $endgroup$
      – Oray
      2 hours ago
















    $begingroup$
    good finding! mine was different but this seems right too :)
    $endgroup$
    – Oray
    2 hours ago




    $begingroup$
    good finding! mine was different but this seems right too :)
    $endgroup$
    – Oray
    2 hours ago












    $begingroup$
    Thank you, @Oray!!
    $endgroup$
    – El-Guest
    2 hours ago




    $begingroup$
    Thank you, @Oray!!
    $endgroup$
    – El-Guest
    2 hours ago












    $begingroup$
    @Oray: was this second one the one that you found?
    $endgroup$
    – El-Guest
    2 hours ago




    $begingroup$
    @Oray: was this second one the one that you found?
    $endgroup$
    – El-Guest
    2 hours ago












    $begingroup$
    no actually :D it was a bit more complicated.
    $endgroup$
    – Oray
    2 hours ago




    $begingroup$
    no actually :D it was a bit more complicated.
    $endgroup$
    – Oray
    2 hours ago











    4












    $begingroup$

    I have found this solution




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







    share|improve this answer









    $endgroup$


















      4












      $begingroup$

      I have found this solution




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







      share|improve this answer









      $endgroup$
















        4












        4








        4





        $begingroup$

        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 2 hours ago









        Weather VaneWeather Vane

        2,387112




        2,387112























            0












            $begingroup$

            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...






            share|improve this answer











            $endgroup$













            • $begingroup$
              The second answer here is not "using the numbers $3$, $4$ and $6$".
              $endgroup$
              – Weather Vane
              48 mins 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-
              4 mins ago










            • $begingroup$
              @WeatherVane - Anyways, updated it to use all the numbers.
              $endgroup$
              – Vilx-
              2 mins ago
















            0












            $begingroup$

            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...






            share|improve this answer











            $endgroup$













            • $begingroup$
              The second answer here is not "using the numbers $3$, $4$ and $6$".
              $endgroup$
              – Weather Vane
              48 mins 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-
              4 mins ago










            • $begingroup$
              @WeatherVane - Anyways, updated it to use all the numbers.
              $endgroup$
              – Vilx-
              2 mins ago














            0












            0








            0





            $begingroup$

            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...






            share|improve this answer











            $endgroup$



            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...







            share|improve this answer














            share|improve this answer



            share|improve this answer








            edited 2 mins ago

























            answered 1 hour ago









            Vilx-Vilx-

            1315




            1315












            • $begingroup$
              The second answer here is not "using the numbers $3$, $4$ and $6$".
              $endgroup$
              – Weather Vane
              48 mins 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-
              4 mins ago










            • $begingroup$
              @WeatherVane - Anyways, updated it to use all the numbers.
              $endgroup$
              – Vilx-
              2 mins ago


















            • $begingroup$
              The second answer here is not "using the numbers $3$, $4$ and $6$".
              $endgroup$
              – Weather Vane
              48 mins 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-
              4 mins ago










            • $begingroup$
              @WeatherVane - Anyways, updated it to use all the numbers.
              $endgroup$
              – Vilx-
              2 mins ago
















            $begingroup$
            The second answer here is not "using the numbers $3$, $4$ and $6$".
            $endgroup$
            – Weather Vane
            48 mins ago






            $begingroup$
            The second answer here is not "using the numbers $3$, $4$ and $6$".
            $endgroup$
            – Weather Vane
            48 mins 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-
            4 mins 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-
            4 mins ago












            $begingroup$
            @WeatherVane - Anyways, updated it to use all the numbers.
            $endgroup$
            – Vilx-
            2 mins ago




            $begingroup$
            @WeatherVane - Anyways, updated it to use all the numbers.
            $endgroup$
            – Vilx-
            2 mins ago


















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