1984 - take the digits 1,9, 8 and 4 and make 123 - Part III

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Part III ... Next year is the 70th anniversary of the publication of the book 1984 by George Orwell. Here is a puzzle to start the anniversary celebrations off a bit early ... I think this is the final one from me on 1984



Warm up



Can you assemble a formula using the numbers $1$, $9$, $8$, and $4$ in any order so that the results equals $123$. You may use the operations $x + y$, $x - y$, $x times y$, $x div y$, $x!$, $sqrtx$, $sqrt[leftroot-2uproot2x]y$ and $x^y$, as long as all operands are either $1$, $9$, $8$, or $4$. Operands may of course also be derived from calculations e.g. $19*8*(sqrt4)$. You may also use brackets to clarify order of operations, and you may concatenate two or more of the four digits you start with (such as $8$ and $4$ to make the number $84$) if you wish. You may only use each of the starting digits once and you must use all four of them. I'm afraid that concatenation of numbers from calculations is not permitted, but answers with concatenations will get plus one from me.



Main Event



The main event is to find two different methods to find 123 without using concatenation. If you did not use concatenation above in the warm up then you only have one more formula to find using the numbers $1$, $9$, $8$, and $4$ in any order so that the results equals $123$ without using any concatenation. so now, for example, you cannot put $8$ and $4$ together to make the number $84$. The rest of the rules above apply, but concatenation is not allowed. If you didn't use any concatenation above then you have solved the puzzle, but you could try to solve it with concatenation, that is concatenation of the initial numbers only. The two methods must be different; for example, $(9+4)*(1+8)$ and $(8+1)*(4+9)$ would not be considered to be two different methods.



Note (and perhaps hint): For this second part any finite number of functions can be used, though ingenious solutions with infinite numbers of functions will get plus one from me.



Note that in all the puzzles above
Double, triple, etc. factorials (n-druple-factorials), such as $4!! = 4 times 2$ are not allowed, but factorials of factorials are fine, such as $(4!)! = 24!$. I will upvote answers with double, triple and n-druple-factorials which get the required answers, but will not mark them as correct - particularly because a general method was developed by @Carl Schildkraut to solve these puzzles provided that you can make a single number greater than $2$ two times from the numbers you start with - here it could be $9+1$ and $8 + sqrt4$ to get two $10$s, for example.



many thanks to the authors of the similar questions below for inspiring this question.



  • Use 2 0 1 and 8 to make 67

  • Make numbers 93 using the digits 2, 0, 1, 8

  • Make numbers 1 - 30 using the digits 2, 0, 1, 8

This is part III after the first and second in this series were solved






mathematics formation-of-numbers







share|improve this question



























    up vote
    1
    down vote

    favorite












    Part III ... Next year is the 70th anniversary of the publication of the book 1984 by George Orwell. Here is a puzzle to start the anniversary celebrations off a bit early ... I think this is the final one from me on 1984



    Warm up



    Can you assemble a formula using the numbers $1$, $9$, $8$, and $4$ in any order so that the results equals $123$. You may use the operations $x + y$, $x - y$, $x times y$, $x div y$, $x!$, $sqrtx$, $sqrt[leftroot-2uproot2x]y$ and $x^y$, as long as all operands are either $1$, $9$, $8$, or $4$. Operands may of course also be derived from calculations e.g. $19*8*(sqrt4)$. You may also use brackets to clarify order of operations, and you may concatenate two or more of the four digits you start with (such as $8$ and $4$ to make the number $84$) if you wish. You may only use each of the starting digits once and you must use all four of them. I'm afraid that concatenation of numbers from calculations is not permitted, but answers with concatenations will get plus one from me.



    Main Event



    The main event is to find two different methods to find 123 without using concatenation. If you did not use concatenation above in the warm up then you only have one more formula to find using the numbers $1$, $9$, $8$, and $4$ in any order so that the results equals $123$ without using any concatenation. so now, for example, you cannot put $8$ and $4$ together to make the number $84$. The rest of the rules above apply, but concatenation is not allowed. If you didn't use any concatenation above then you have solved the puzzle, but you could try to solve it with concatenation, that is concatenation of the initial numbers only. The two methods must be different; for example, $(9+4)*(1+8)$ and $(8+1)*(4+9)$ would not be considered to be two different methods.



    Note (and perhaps hint): For this second part any finite number of functions can be used, though ingenious solutions with infinite numbers of functions will get plus one from me.



    Note that in all the puzzles above
    Double, triple, etc. factorials (n-druple-factorials), such as $4!! = 4 times 2$ are not allowed, but factorials of factorials are fine, such as $(4!)! = 24!$. I will upvote answers with double, triple and n-druple-factorials which get the required answers, but will not mark them as correct - particularly because a general method was developed by @Carl Schildkraut to solve these puzzles provided that you can make a single number greater than $2$ two times from the numbers you start with - here it could be $9+1$ and $8 + sqrt4$ to get two $10$s, for example.



    many thanks to the authors of the similar questions below for inspiring this question.



    • Use 2 0 1 and 8 to make 67

    • Make numbers 93 using the digits 2, 0, 1, 8

    • Make numbers 1 - 30 using the digits 2, 0, 1, 8

    This is part III after the first and second in this series were solved






    mathematics formation-of-numbers







    share|improve this question

























      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite











      Part III ... Next year is the 70th anniversary of the publication of the book 1984 by George Orwell. Here is a puzzle to start the anniversary celebrations off a bit early ... I think this is the final one from me on 1984



      Warm up



      Can you assemble a formula using the numbers $1$, $9$, $8$, and $4$ in any order so that the results equals $123$. You may use the operations $x + y$, $x - y$, $x times y$, $x div y$, $x!$, $sqrtx$, $sqrt[leftroot-2uproot2x]y$ and $x^y$, as long as all operands are either $1$, $9$, $8$, or $4$. Operands may of course also be derived from calculations e.g. $19*8*(sqrt4)$. You may also use brackets to clarify order of operations, and you may concatenate two or more of the four digits you start with (such as $8$ and $4$ to make the number $84$) if you wish. You may only use each of the starting digits once and you must use all four of them. I'm afraid that concatenation of numbers from calculations is not permitted, but answers with concatenations will get plus one from me.



      Main Event



      The main event is to find two different methods to find 123 without using concatenation. If you did not use concatenation above in the warm up then you only have one more formula to find using the numbers $1$, $9$, $8$, and $4$ in any order so that the results equals $123$ without using any concatenation. so now, for example, you cannot put $8$ and $4$ together to make the number $84$. The rest of the rules above apply, but concatenation is not allowed. If you didn't use any concatenation above then you have solved the puzzle, but you could try to solve it with concatenation, that is concatenation of the initial numbers only. The two methods must be different; for example, $(9+4)*(1+8)$ and $(8+1)*(4+9)$ would not be considered to be two different methods.



      Note (and perhaps hint): For this second part any finite number of functions can be used, though ingenious solutions with infinite numbers of functions will get plus one from me.



      Note that in all the puzzles above
      Double, triple, etc. factorials (n-druple-factorials), such as $4!! = 4 times 2$ are not allowed, but factorials of factorials are fine, such as $(4!)! = 24!$. I will upvote answers with double, triple and n-druple-factorials which get the required answers, but will not mark them as correct - particularly because a general method was developed by @Carl Schildkraut to solve these puzzles provided that you can make a single number greater than $2$ two times from the numbers you start with - here it could be $9+1$ and $8 + sqrt4$ to get two $10$s, for example.



      many thanks to the authors of the similar questions below for inspiring this question.



      • Use 2 0 1 and 8 to make 67

      • Make numbers 93 using the digits 2, 0, 1, 8

      • Make numbers 1 - 30 using the digits 2, 0, 1, 8

      This is part III after the first and second in this series were solved






      mathematics formation-of-numbers







      share|improve this question















      Part III ... Next year is the 70th anniversary of the publication of the book 1984 by George Orwell. Here is a puzzle to start the anniversary celebrations off a bit early ... I think this is the final one from me on 1984



      Warm up



      Can you assemble a formula using the numbers $1$, $9$, $8$, and $4$ in any order so that the results equals $123$. You may use the operations $x + y$, $x - y$, $x times y$, $x div y$, $x!$, $sqrtx$, $sqrt[leftroot-2uproot2x]y$ and $x^y$, as long as all operands are either $1$, $9$, $8$, or $4$. Operands may of course also be derived from calculations e.g. $19*8*(sqrt4)$. You may also use brackets to clarify order of operations, and you may concatenate two or more of the four digits you start with (such as $8$ and $4$ to make the number $84$) if you wish. You may only use each of the starting digits once and you must use all four of them. I'm afraid that concatenation of numbers from calculations is not permitted, but answers with concatenations will get plus one from me.



      Main Event



      The main event is to find two different methods to find 123 without using concatenation. If you did not use concatenation above in the warm up then you only have one more formula to find using the numbers $1$, $9$, $8$, and $4$ in any order so that the results equals $123$ without using any concatenation. so now, for example, you cannot put $8$ and $4$ together to make the number $84$. The rest of the rules above apply, but concatenation is not allowed. If you didn't use any concatenation above then you have solved the puzzle, but you could try to solve it with concatenation, that is concatenation of the initial numbers only. The two methods must be different; for example, $(9+4)*(1+8)$ and $(8+1)*(4+9)$ would not be considered to be two different methods.



      Note (and perhaps hint): For this second part any finite number of functions can be used, though ingenious solutions with infinite numbers of functions will get plus one from me.



      Note that in all the puzzles above
      Double, triple, etc. factorials (n-druple-factorials), such as $4!! = 4 times 2$ are not allowed, but factorials of factorials are fine, such as $(4!)! = 24!$. I will upvote answers with double, triple and n-druple-factorials which get the required answers, but will not mark them as correct - particularly because a general method was developed by @Carl Schildkraut to solve these puzzles provided that you can make a single number greater than $2$ two times from the numbers you start with - here it could be $9+1$ and $8 + sqrt4$ to get two $10$s, for example.



      many thanks to the authors of the similar questions below for inspiring this question.



      • Use 2 0 1 and 8 to make 67

      • Make numbers 93 using the digits 2, 0, 1, 8

      • Make numbers 1 - 30 using the digits 2, 0, 1, 8

      This is part III after the first and second in this series were solved






      mathematics formation-of-numbers




      mathematics formation-of-numbers






      share|improve this question















      share|improve this question













      share|improve this question




      share|improve this question








      edited 56 mins ago

























      asked 1 hour ago









      tom

      1,725324




      1,725324




















          2 Answers
          2






          active

          oldest

          votes

















          up vote
          3
          down vote













          Is it




          $(8-sqrt9)!+(4-1)$




          And




          $(8-4+1)! + sqrt9$







          share|improve this answer






















          • I like your answer and it will get plus one, but I only consider this to be one method - does that that make sense?
            – tom
            25 mins ago

















          up vote
          0
          down vote













          Here is two possibilities that I have found:




          $sqrt9times(8!/4!+1)=123$




          and the other one is (thanks to @PotatoLatte)




          $(8-sqrt9)!+(4-1)=123$




          Bonus:




          $sqrt8+1+9!/4!=123$






          share




















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            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes








            up vote
            3
            down vote













            Is it




            $(8-sqrt9)!+(4-1)$




            And




            $(8-4+1)! + sqrt9$







            share|improve this answer






















            • I like your answer and it will get plus one, but I only consider this to be one method - does that that make sense?
              – tom
              25 mins ago














            up vote
            3
            down vote













            Is it




            $(8-sqrt9)!+(4-1)$




            And




            $(8-4+1)! + sqrt9$







            share|improve this answer






















            • I like your answer and it will get plus one, but I only consider this to be one method - does that that make sense?
              – tom
              25 mins ago












            up vote
            3
            down vote










            up vote
            3
            down vote









            Is it




            $(8-sqrt9)!+(4-1)$




            And




            $(8-4+1)! + sqrt9$







            share|improve this answer














            Is it




            $(8-sqrt9)!+(4-1)$




            And




            $(8-4+1)! + sqrt9$








            share|improve this answer














            share|improve this answer



            share|improve this answer








            edited 21 mins ago

























            answered 32 mins ago









            PotatoLatte

            1,175228




            1,175228











            • I like your answer and it will get plus one, but I only consider this to be one method - does that that make sense?
              – tom
              25 mins ago
















            • I like your answer and it will get plus one, but I only consider this to be one method - does that that make sense?
              – tom
              25 mins ago















            I like your answer and it will get plus one, but I only consider this to be one method - does that that make sense?
            – tom
            25 mins ago




            I like your answer and it will get plus one, but I only consider this to be one method - does that that make sense?
            – tom
            25 mins ago










            up vote
            0
            down vote













            Here is two possibilities that I have found:




            $sqrt9times(8!/4!+1)=123$




            and the other one is (thanks to @PotatoLatte)




            $(8-sqrt9)!+(4-1)=123$




            Bonus:




            $sqrt8+1+9!/4!=123$






            share
























              up vote
              0
              down vote













              Here is two possibilities that I have found:




              $sqrt9times(8!/4!+1)=123$




              and the other one is (thanks to @PotatoLatte)




              $(8-sqrt9)!+(4-1)=123$




              Bonus:




              $sqrt8+1+9!/4!=123$






              share






















                up vote
                0
                down vote










                up vote
                0
                down vote









                Here is two possibilities that I have found:




                $sqrt9times(8!/4!+1)=123$




                and the other one is (thanks to @PotatoLatte)




                $(8-sqrt9)!+(4-1)=123$




                Bonus:




                $sqrt8+1+9!/4!=123$






                share












                Here is two possibilities that I have found:




                $sqrt9times(8!/4!+1)=123$




                and the other one is (thanks to @PotatoLatte)




                $(8-sqrt9)!+(4-1)=123$




                Bonus:




                $sqrt8+1+9!/4!=123$







                share











                share


                share










                answered 3 mins ago









                Oray

                14.4k435142




                14.4k435142



























                     

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