Find 108 by using 3,4,6 Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern) Announcing the arrival of Valued Associate #679: Cesar Manara Unicorn Meta Zoo #1: Why another podcast?Use 2, 0, 1 and 8 to make 109Use 2, 0, 1 and 8 to make 1991984 - take the digits 1,9, 8 and 4 and make 2461984 - take the digits 1,9, 8 and 4 and make 3691984 - take the digits 1,9, 8 and 4 and Hard Challenges!1984 - take the digits 1,9, 8 and 4 and make 123 - Part IIIPalindromic number puzzle - make 505 from 20202Use 0, 5, 7 and 1 to make 89Use 6, 5 and 3 to make 57Using only 1s, make 29 with the minimum number of digits

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Find 108 by using 3,4,6



Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)
Announcing the arrival of Valued Associate #679: Cesar Manara
Unicorn Meta Zoo #1: Why another podcast?Use 2, 0, 1 and 8 to make 109Use 2, 0, 1 and 8 to make 1991984 - take the digits 1,9, 8 and 4 and make 2461984 - take the digits 1,9, 8 and 4 and make 3691984 - take the digits 1,9, 8 and 4 and Hard Challenges!1984 - take the digits 1,9, 8 and 4 and make 123 - Part IIIPalindromic number puzzle - make 505 from 20202Use 0, 5, 7 and 1 to make 89Use 6, 5 and 3 to make 57Using only 1s, make 29 with the minimum number of digits










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











$endgroup$







  • 1




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







  • 2




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











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







  • 2




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










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















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











$endgroup$







  • 1




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







  • 2




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











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







  • 2




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










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













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;



  • $x + y$


  • $x - y$


  • $x times y$


  • $x div y$


  • $x!$


  • $sqrtx$


  • $sqrt[leftroot-2uproot2x]y$


  • $x^y$


  • 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 1 hour 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 $$3times styledisplay: inline-block; transform: rotate(180deg)4times 6=108$$ since the upside down $4$ looks a bit like a $6$.
    $endgroup$
    – user477343
    2 hours ago







  • 2




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











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







  • 2




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










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












  • 1




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







  • 2




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











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







  • 2




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










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







1




1




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





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





2




2




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





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













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





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





2




2




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




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












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




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










3 Answers
3






active

oldest

votes


















3












$begingroup$

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











$endgroup$












  • $begingroup$
    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 + 6) * (4 + 6) + 4 + 4 = 108




    Or maybe:




    (4*6+3)*4 = 108




    I could go on and on, and I haven't even touched on the advanced operators...






    share|improve this answer











    $endgroup$












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












    Your Answer








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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




    $frac6^3sqrt4 = frac2162 = 108$?




    @Oray found another one, which might possibly be




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







    share|improve this answer











    $endgroup$












    • $begingroup$
      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




    $frac6^3sqrt4 = frac2162 = 108$?




    @Oray found another one, which might possibly be




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







    share|improve this answer











    $endgroup$












    • $begingroup$
      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




    $frac6^3sqrt4 = frac2162 = 108$?




    @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 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 + 6) * (4 + 6) + 4 + 4 = 108




            Or maybe:




            (4*6+3)*4 = 108




            I could go on and on, and I haven't even touched on the advanced operators...






            share|improve this answer











            $endgroup$












            • $begingroup$
              The second answer here is not "using the numbers $3$, $4$ and $6$".
              $endgroup$
              – Weather Vane
              29 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 + 6) * (4 + 6) + 4 + 4 = 108




            Or maybe:




            (4*6+3)*4 = 108




            I could go on and on, and I haven't even touched on the advanced operators...






            share|improve this answer











            $endgroup$












            • $begingroup$
              The second answer here is not "using the numbers $3$, $4$ and $6$".
              $endgroup$
              – Weather Vane
              29 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 + 6) * (4 + 6) + 4 + 4 = 108




            Or maybe:




            (4*6+3)*4 = 108




            I could go on and on, and I haven't even touched on the advanced operators...






            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 + 6) * (4 + 6) + 4 + 4 = 108




            Or maybe:




            (4*6+3)*4 = 108




            I could go on and on, and I haven't even touched on the advanced operators...







            share|improve this answer














            share|improve this answer



            share|improve this answer








            edited 54 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
              29 mins ago

















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
















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





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


















            draft saved

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