1984 - take the digits 1,9, 8 and 4 and make 369

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP











up vote
1
down vote

favorite












Part II... 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 ...



Warm up



Can you assemble a formula using the numbers $1$, $9$, $8$, and $4$ in any order so that the results equals $369$. 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



If you used concatenation above then make a formula using the numbers $1$, $9$, $8$, and $4$ in any order so that the results equals $369$ without using any concatenation. so, 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.



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.



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



This is part II after the first in this series was solved



  • 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





mathematics formation-of-numbers







share|improve this question

























    up vote
    1
    down vote

    favorite












    Part II... 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 ...



    Warm up



    Can you assemble a formula using the numbers $1$, $9$, $8$, and $4$ in any order so that the results equals $369$. 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



    If you used concatenation above then make a formula using the numbers $1$, $9$, $8$, and $4$ in any order so that the results equals $369$ without using any concatenation. so, 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.



    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.



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



    This is part II after the first in this series was solved



    • 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





    mathematics formation-of-numbers







    share|improve this question























      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite











      Part II... 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 ...



      Warm up



      Can you assemble a formula using the numbers $1$, $9$, $8$, and $4$ in any order so that the results equals $369$. 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



      If you used concatenation above then make a formula using the numbers $1$, $9$, $8$, and $4$ in any order so that the results equals $369$ without using any concatenation. so, 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.



      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.



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



      This is part II after the first in this series was solved



      • 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





      mathematics formation-of-numbers







      share|improve this question













      Part II... 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 ...



      Warm up



      Can you assemble a formula using the numbers $1$, $9$, $8$, and $4$ in any order so that the results equals $369$. 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



      If you used concatenation above then make a formula using the numbers $1$, $9$, $8$, and $4$ in any order so that the results equals $369$ without using any concatenation. so, 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.



      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.



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



      This is part II after the first in this series was solved



      • 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





      mathematics formation-of-numbers




      mathematics formation-of-numbers






      share|improve this question













      share|improve this question











      share|improve this question




      share|improve this question










      asked 5 hours ago









      tom

      1,718324




      1,718324




















          3 Answers
          3






          active

          oldest

          votes

















          up vote
          1
          down vote



          accepted











          $frac(sqrt9 ! )!sqrt4+8+1=frac(3!)!2+9=frac6!2+9=frac7202+9=360+9=369$




          Or, with an infinite number of functions:




          $frac(sqrt9 ! )!sqrt4+8+sqrtsqrt...sqrt1$, which simplifies to the expression above







          share|improve this answer
















          • 1




            brilliant good job, plus one - this looks like the first answer to the main event :-)
            – tom
            4 hours ago











          • If only I didn't use the spare $0$... Eh well, you deserved it :P
            – user477343
            4 hours ago


















          up vote
          3
          down vote













          Well, for the warm-up




          $$beginalign19^sqrt4+8&= 19^2+8 \ &= 361+8 \ &=boxed369endalign$$




          Fun puzzle! And, since $1984=01984$... if I could use a spare $0$...




          $$big((8-1-0!)!div sqrt4big)+9=boxed369$$




          :D






          share|improve this answer






















          • Good job - looks like this is the first correct answer to the warm up - great job
            – tom
            4 hours ago











          • @tom Now to not use any concatenation... :P
            – user477343
            4 hours ago


















          up vote
          2
          down vote













          Perhaps, for the warm-up




          $41times 9times sqrtsqrtsqrt...8$, with infinite roots
          $=369times sqrtsqrtsqrt...8$.







          share|improve this answer










          New contributor




          binu kurian is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.

















          • Great this is what I was thinking about for the warm up... as the second part could not have infinite functions, but, therefore, the first part could :-) +1
            – tom
            4 hours ago






          • 1




            I added roll over spoiler hider - so that people have to roll over to see your answer
            – tom
            4 hours ago










          • Thanks @tom. As a newbie I am still learning how to format my answers
            – binu kurian
            3 hours ago










          • no problem, but I think @user477343 sorted it out.
            – tom
            3 hours ago










          Your Answer




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          3 Answers
          3






          active

          oldest

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          3 Answers
          3






          active

          oldest

          votes









          active

          oldest

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          active

          oldest

          votes








          up vote
          1
          down vote



          accepted











          $frac(sqrt9 ! )!sqrt4+8+1=frac(3!)!2+9=frac6!2+9=frac7202+9=360+9=369$




          Or, with an infinite number of functions:




          $frac(sqrt9 ! )!sqrt4+8+sqrtsqrt...sqrt1$, which simplifies to the expression above







          share|improve this answer
















          • 1




            brilliant good job, plus one - this looks like the first answer to the main event :-)
            – tom
            4 hours ago











          • If only I didn't use the spare $0$... Eh well, you deserved it :P
            – user477343
            4 hours ago















          up vote
          1
          down vote



          accepted











          $frac(sqrt9 ! )!sqrt4+8+1=frac(3!)!2+9=frac6!2+9=frac7202+9=360+9=369$




          Or, with an infinite number of functions:




          $frac(sqrt9 ! )!sqrt4+8+sqrtsqrt...sqrt1$, which simplifies to the expression above







          share|improve this answer
















          • 1




            brilliant good job, plus one - this looks like the first answer to the main event :-)
            – tom
            4 hours ago











          • If only I didn't use the spare $0$... Eh well, you deserved it :P
            – user477343
            4 hours ago













          up vote
          1
          down vote



          accepted







          up vote
          1
          down vote



          accepted







          $frac(sqrt9 ! )!sqrt4+8+1=frac(3!)!2+9=frac6!2+9=frac7202+9=360+9=369$




          Or, with an infinite number of functions:




          $frac(sqrt9 ! )!sqrt4+8+sqrtsqrt...sqrt1$, which simplifies to the expression above







          share|improve this answer













          $frac(sqrt9 ! )!sqrt4+8+1=frac(3!)!2+9=frac6!2+9=frac7202+9=360+9=369$




          Or, with an infinite number of functions:




          $frac(sqrt9 ! )!sqrt4+8+sqrtsqrt...sqrt1$, which simplifies to the expression above








          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered 4 hours ago









          elias

          7,71232051




          7,71232051







          • 1




            brilliant good job, plus one - this looks like the first answer to the main event :-)
            – tom
            4 hours ago











          • If only I didn't use the spare $0$... Eh well, you deserved it :P
            – user477343
            4 hours ago













          • 1




            brilliant good job, plus one - this looks like the first answer to the main event :-)
            – tom
            4 hours ago











          • If only I didn't use the spare $0$... Eh well, you deserved it :P
            – user477343
            4 hours ago








          1




          1




          brilliant good job, plus one - this looks like the first answer to the main event :-)
          – tom
          4 hours ago





          brilliant good job, plus one - this looks like the first answer to the main event :-)
          – tom
          4 hours ago













          If only I didn't use the spare $0$... Eh well, you deserved it :P
          – user477343
          4 hours ago





          If only I didn't use the spare $0$... Eh well, you deserved it :P
          – user477343
          4 hours ago











          up vote
          3
          down vote













          Well, for the warm-up




          $$beginalign19^sqrt4+8&= 19^2+8 \ &= 361+8 \ &=boxed369endalign$$




          Fun puzzle! And, since $1984=01984$... if I could use a spare $0$...




          $$big((8-1-0!)!div sqrt4big)+9=boxed369$$




          :D






          share|improve this answer






















          • Good job - looks like this is the first correct answer to the warm up - great job
            – tom
            4 hours ago











          • @tom Now to not use any concatenation... :P
            – user477343
            4 hours ago















          up vote
          3
          down vote













          Well, for the warm-up




          $$beginalign19^sqrt4+8&= 19^2+8 \ &= 361+8 \ &=boxed369endalign$$




          Fun puzzle! And, since $1984=01984$... if I could use a spare $0$...




          $$big((8-1-0!)!div sqrt4big)+9=boxed369$$




          :D






          share|improve this answer






















          • Good job - looks like this is the first correct answer to the warm up - great job
            – tom
            4 hours ago











          • @tom Now to not use any concatenation... :P
            – user477343
            4 hours ago













          up vote
          3
          down vote










          up vote
          3
          down vote









          Well, for the warm-up




          $$beginalign19^sqrt4+8&= 19^2+8 \ &= 361+8 \ &=boxed369endalign$$




          Fun puzzle! And, since $1984=01984$... if I could use a spare $0$...




          $$big((8-1-0!)!div sqrt4big)+9=boxed369$$




          :D






          share|improve this answer














          Well, for the warm-up




          $$beginalign19^sqrt4+8&= 19^2+8 \ &= 361+8 \ &=boxed369endalign$$




          Fun puzzle! And, since $1984=01984$... if I could use a spare $0$...




          $$big((8-1-0!)!div sqrt4big)+9=boxed369$$




          :D







          share|improve this answer














          share|improve this answer



          share|improve this answer








          edited 4 hours ago

























          answered 4 hours ago









          user477343

          3,4471745




          3,4471745











          • Good job - looks like this is the first correct answer to the warm up - great job
            – tom
            4 hours ago











          • @tom Now to not use any concatenation... :P
            – user477343
            4 hours ago

















          • Good job - looks like this is the first correct answer to the warm up - great job
            – tom
            4 hours ago











          • @tom Now to not use any concatenation... :P
            – user477343
            4 hours ago
















          Good job - looks like this is the first correct answer to the warm up - great job
          – tom
          4 hours ago





          Good job - looks like this is the first correct answer to the warm up - great job
          – tom
          4 hours ago













          @tom Now to not use any concatenation... :P
          – user477343
          4 hours ago





          @tom Now to not use any concatenation... :P
          – user477343
          4 hours ago











          up vote
          2
          down vote













          Perhaps, for the warm-up




          $41times 9times sqrtsqrtsqrt...8$, with infinite roots
          $=369times sqrtsqrtsqrt...8$.







          share|improve this answer










          New contributor




          binu kurian is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.

















          • Great this is what I was thinking about for the warm up... as the second part could not have infinite functions, but, therefore, the first part could :-) +1
            – tom
            4 hours ago






          • 1




            I added roll over spoiler hider - so that people have to roll over to see your answer
            – tom
            4 hours ago










          • Thanks @tom. As a newbie I am still learning how to format my answers
            – binu kurian
            3 hours ago










          • no problem, but I think @user477343 sorted it out.
            – tom
            3 hours ago














          up vote
          2
          down vote













          Perhaps, for the warm-up




          $41times 9times sqrtsqrtsqrt...8$, with infinite roots
          $=369times sqrtsqrtsqrt...8$.







          share|improve this answer










          New contributor




          binu kurian is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.

















          • Great this is what I was thinking about for the warm up... as the second part could not have infinite functions, but, therefore, the first part could :-) +1
            – tom
            4 hours ago






          • 1




            I added roll over spoiler hider - so that people have to roll over to see your answer
            – tom
            4 hours ago










          • Thanks @tom. As a newbie I am still learning how to format my answers
            – binu kurian
            3 hours ago










          • no problem, but I think @user477343 sorted it out.
            – tom
            3 hours ago












          up vote
          2
          down vote










          up vote
          2
          down vote









          Perhaps, for the warm-up




          $41times 9times sqrtsqrtsqrt...8$, with infinite roots
          $=369times sqrtsqrtsqrt...8$.







          share|improve this answer










          New contributor




          binu kurian is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.









          Perhaps, for the warm-up




          $41times 9times sqrtsqrtsqrt...8$, with infinite roots
          $=369times sqrtsqrtsqrt...8$.








          share|improve this answer










          New contributor




          binu kurian 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 4 hours ago









          user477343

          3,4471745




          3,4471745






          New contributor




          binu kurian is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.









          answered 4 hours ago









          binu kurian

          312




          312




          New contributor




          binu kurian is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.





          New contributor





          binu kurian is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.






          binu kurian is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.











          • Great this is what I was thinking about for the warm up... as the second part could not have infinite functions, but, therefore, the first part could :-) +1
            – tom
            4 hours ago






          • 1




            I added roll over spoiler hider - so that people have to roll over to see your answer
            – tom
            4 hours ago










          • Thanks @tom. As a newbie I am still learning how to format my answers
            – binu kurian
            3 hours ago










          • no problem, but I think @user477343 sorted it out.
            – tom
            3 hours ago
















          • Great this is what I was thinking about for the warm up... as the second part could not have infinite functions, but, therefore, the first part could :-) +1
            – tom
            4 hours ago






          • 1




            I added roll over spoiler hider - so that people have to roll over to see your answer
            – tom
            4 hours ago










          • Thanks @tom. As a newbie I am still learning how to format my answers
            – binu kurian
            3 hours ago










          • no problem, but I think @user477343 sorted it out.
            – tom
            3 hours ago















          Great this is what I was thinking about for the warm up... as the second part could not have infinite functions, but, therefore, the first part could :-) +1
          – tom
          4 hours ago




          Great this is what I was thinking about for the warm up... as the second part could not have infinite functions, but, therefore, the first part could :-) +1
          – tom
          4 hours ago




          1




          1




          I added roll over spoiler hider - so that people have to roll over to see your answer
          – tom
          4 hours ago




          I added roll over spoiler hider - so that people have to roll over to see your answer
          – tom
          4 hours ago












          Thanks @tom. As a newbie I am still learning how to format my answers
          – binu kurian
          3 hours ago




          Thanks @tom. As a newbie I am still learning how to format my answers
          – binu kurian
          3 hours ago












          no problem, but I think @user477343 sorted it out.
          – tom
          3 hours ago




          no problem, but I think @user477343 sorted it out.
          – tom
          3 hours ago

















           

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          Installing NextGIS Connect into QGIS 3?

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          Clash Royale CLAN TAG #URR8PPP .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty margin-bottom:0; up vote 1 down vote favorite Just acquired free version of NEXTGIS and trying to install the plugin for QGIS. It tells me it was not designed for version 3 of QGIS. How do I set up the QGIS-NEXTGIS interface then? qgis qgis-plugins nextgis share | improve this question edited 3 hours ago PolyGeo ♦ 51.9k 17 77 233 asked 4 hours ago IanGS 16 4 New contributor IanGS is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct. add a comment  |  up vote 1 down vote favorite Just acquired free version of NEXTGIS and trying to install the plugin for QGIS. It tells me it was not designed for version 3 of QGIS. How do I set up the QGIS-NEXTGIS interface then? qgis qgis-plugin...
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          One-line joke

          A one-liner is a joke that is delivered in a single line. A good one-liner is said to be pithy - concise and meaningful. [1] Comedians and actors use this comedic method as part of their act, e.g. Jimmy Carr, Tommy Cooper, Rodney Dangerfield, Norm Macdonald, Ken Dodd, Stewart Francis, Zach Galifianakis, Mitch Hedberg, Anthony Jeselnik, Milton Jones, Shappi Khorsandi, Jay London, Mark Linn-Baker, Demetri Martin, Groucho Marx, Dan Mintz, Emo Philips, Tim Vine, Steven Wright and Henny Youngman and Arnold Schwarzenegger. [ citation needed ] Many fictional characters are also known to deliver one-liners, including James Bond, who usually includes pithy and laconic quips after disposing of a villain. [ citation needed ] Examples "A baby seal walks into a club." "A dyslexic man walks into a bra." (Anonymous) "There are three types of people, those who can count and those who can't." "Venison's dear, isn't it?" (Jimmy Carr) "An escala...
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