Create Numbers 1 - 100 using 1,9,6,8












1












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Create all numbers 1 - 100 using equations made up of 1,9,6,8.



Rules:




  • Use all four digits exactly once

  • Allowed operations: +, -, x, ÷, ! (factorial), exponentiation, square root.

  • Parentheses and grouping (e.g. "21") are also allowed.

  • You have to keep the order 1,9,6,8 for all numbers.

  • Exponentiation can only be used in the number order with the numbers provided. Eg. 1^9 + 6 + 8 is allowed. Not 1^6 + 9 + 8.

  • The modulus operator is not allowed.

  • Rounding is not allowed (e.g. 201/8=25).

  • Decimal point is allowed.


Credit to Fitch496 for the idea.










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$endgroup$












  • $begingroup$
    it is obvious it is impossible to find all numbers from 1 to 100 with these rules :)
    $endgroup$
    – Oray
    20 hours ago










  • $begingroup$
    @Oray can you please prove it then? thanks!
    $endgroup$
    – Omega Krypton
    16 hours ago










  • $begingroup$
    @OmegaKrypton i put it into my code many numbers cannot be achieved.
    $endgroup$
    – Oray
    16 hours ago










  • $begingroup$
    @Oray I think the only number we have not found now is $79$.
    $endgroup$
    – hexomino
    12 hours ago
















1












$begingroup$


Create all numbers 1 - 100 using equations made up of 1,9,6,8.



Rules:




  • Use all four digits exactly once

  • Allowed operations: +, -, x, ÷, ! (factorial), exponentiation, square root.

  • Parentheses and grouping (e.g. "21") are also allowed.

  • You have to keep the order 1,9,6,8 for all numbers.

  • Exponentiation can only be used in the number order with the numbers provided. Eg. 1^9 + 6 + 8 is allowed. Not 1^6 + 9 + 8.

  • The modulus operator is not allowed.

  • Rounding is not allowed (e.g. 201/8=25).

  • Decimal point is allowed.


Credit to Fitch496 for the idea.










share|improve this question









New contributor




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







$endgroup$












  • $begingroup$
    it is obvious it is impossible to find all numbers from 1 to 100 with these rules :)
    $endgroup$
    – Oray
    20 hours ago










  • $begingroup$
    @Oray can you please prove it then? thanks!
    $endgroup$
    – Omega Krypton
    16 hours ago










  • $begingroup$
    @OmegaKrypton i put it into my code many numbers cannot be achieved.
    $endgroup$
    – Oray
    16 hours ago










  • $begingroup$
    @Oray I think the only number we have not found now is $79$.
    $endgroup$
    – hexomino
    12 hours ago














1












1








1





$begingroup$


Create all numbers 1 - 100 using equations made up of 1,9,6,8.



Rules:




  • Use all four digits exactly once

  • Allowed operations: +, -, x, ÷, ! (factorial), exponentiation, square root.

  • Parentheses and grouping (e.g. "21") are also allowed.

  • You have to keep the order 1,9,6,8 for all numbers.

  • Exponentiation can only be used in the number order with the numbers provided. Eg. 1^9 + 6 + 8 is allowed. Not 1^6 + 9 + 8.

  • The modulus operator is not allowed.

  • Rounding is not allowed (e.g. 201/8=25).

  • Decimal point is allowed.


Credit to Fitch496 for the idea.










share|improve this question









New contributor




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







$endgroup$




Create all numbers 1 - 100 using equations made up of 1,9,6,8.



Rules:




  • Use all four digits exactly once

  • Allowed operations: +, -, x, ÷, ! (factorial), exponentiation, square root.

  • Parentheses and grouping (e.g. "21") are also allowed.

  • You have to keep the order 1,9,6,8 for all numbers.

  • Exponentiation can only be used in the number order with the numbers provided. Eg. 1^9 + 6 + 8 is allowed. Not 1^6 + 9 + 8.

  • The modulus operator is not allowed.

  • Rounding is not allowed (e.g. 201/8=25).

  • Decimal point is allowed.


Credit to Fitch496 for the idea.







mathematics






share|improve this question









New contributor




Auroxa is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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share|improve this question









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share|improve this question








edited 20 hours ago









Aryaman

1078




1078






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asked 21 hours ago









AuroxaAuroxa

141




141




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New contributor





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






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Check out our Code of Conduct.












  • $begingroup$
    it is obvious it is impossible to find all numbers from 1 to 100 with these rules :)
    $endgroup$
    – Oray
    20 hours ago










  • $begingroup$
    @Oray can you please prove it then? thanks!
    $endgroup$
    – Omega Krypton
    16 hours ago










  • $begingroup$
    @OmegaKrypton i put it into my code many numbers cannot be achieved.
    $endgroup$
    – Oray
    16 hours ago










  • $begingroup$
    @Oray I think the only number we have not found now is $79$.
    $endgroup$
    – hexomino
    12 hours ago


















  • $begingroup$
    it is obvious it is impossible to find all numbers from 1 to 100 with these rules :)
    $endgroup$
    – Oray
    20 hours ago










  • $begingroup$
    @Oray can you please prove it then? thanks!
    $endgroup$
    – Omega Krypton
    16 hours ago










  • $begingroup$
    @OmegaKrypton i put it into my code many numbers cannot be achieved.
    $endgroup$
    – Oray
    16 hours ago










  • $begingroup$
    @Oray I think the only number we have not found now is $79$.
    $endgroup$
    – hexomino
    12 hours ago
















$begingroup$
it is obvious it is impossible to find all numbers from 1 to 100 with these rules :)
$endgroup$
– Oray
20 hours ago




$begingroup$
it is obvious it is impossible to find all numbers from 1 to 100 with these rules :)
$endgroup$
– Oray
20 hours ago












$begingroup$
@Oray can you please prove it then? thanks!
$endgroup$
– Omega Krypton
16 hours ago




$begingroup$
@Oray can you please prove it then? thanks!
$endgroup$
– Omega Krypton
16 hours ago












$begingroup$
@OmegaKrypton i put it into my code many numbers cannot be achieved.
$endgroup$
– Oray
16 hours ago




$begingroup$
@OmegaKrypton i put it into my code many numbers cannot be achieved.
$endgroup$
– Oray
16 hours ago












$begingroup$
@Oray I think the only number we have not found now is $79$.
$endgroup$
– hexomino
12 hours ago




$begingroup$
@Oray I think the only number we have not found now is $79$.
$endgroup$
– hexomino
12 hours ago










4 Answers
4






active

oldest

votes


















1












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Partial answer (the most obvious solutions):




1 = -1^9 - 6 + 8 = 1^968

2 = 1 + sqrt(9) + 6 - 8

5 = 19 - 6 - 8

6 = -1 + 9 + 6 - 8 = 1 - 9 + 6 + 8

7 = 1 * (9 + 6) - 8

8 = 1 + 9 + 6 - 8

9 = 1^96 + 8

11 = -1 + 96 / 8

12 = 1 * 96 / 8 = 1 + 9 - 6 + 8

13 = 1 + 96 / 8

16 = (1 - 9) * (6 - 8)

17 = 19 + 6 - 8

18 = 1 + sqrt(9) + 6 + 8

21 = 19 - 6 + 8

22 = -1 + 9 + 6 + 8

23 = 1 * (9 + 6) + 8

24 = 1 + 9 + 6 + 8

25 = -1 + sqrt(9) * 6 + 8

26 = 1 * sqrt(9) * 6 + 8

27 = 1 + sqrt(9) * 6 + 8

29 = -19 + 6 * 8

31 = 19 + 6 + 8

32 = (1 + sqrt(9)) * 6 + 8

52 = (1 + 9) * 6 - 8

57 = 1 * 9 + 6 * 8

58 = -1 - 9 + 68

59 = -1 * 9 + 68

60 = 1 - 9 + 68

62 = 1 * 9 * 6 + 8

63 = 1 + 9 * 6 + 8

67 = 19 + 6 * 8

68 = (1 + 9) * 6 + 8

76 = -1 + 9 + 68

77 = 1 * 9 + 68

78 = 1 + 9 + 68

80 = -1 - 9 + 6! / 8

81 = -1 * 9 + 6! / 8

82 = 1 - 9 + 6! / 8

87 = -1 + 96 - 8

88 = 1 * 96 - 8

89 = 1 + 96 - 8

98 = -1 + 9 + 6! / 8

99 = 1 * 9 + 6! / 8

100 = 1 + 9 + 6! / 8







share|improve this answer









$endgroup$





















    1












    $begingroup$

    Partial Answer (too lazy to type them out :P)




    enter image description here







    share|improve this answer









    $endgroup$





















      1












      $begingroup$

      Here are all the numbers which have not been obtained already by trolley813, Omega Krypton and ppgdev




      $28 = 1 times ((sqrt{9})! times 6) - 8$
      $30 =-(1times(sqrt{9})!) + sqrt{sqrt{6^8}} $
      $33 = 19 + 6 + 8$
      $34 = ((1+(sqrt{9})!) times 6) - 8 $
      $35 = (1 times sqrt{(sqrt{9})^6}) + 8$
      $36 = (1 + sqrt{(sqrt{9})^6}) + 8$
      $37 = 1^9 + sqrt{sqrt{6^8}}$
      $38 = (1+sqrt{9})! + 6 + 8$
      $41 = -(1 + (sqrt{9})!) + (6 times 8)$
      $42 = -(1 times (sqrt{9})!) + (6 times 8)$
      $43 = -1 + ((sqrt{9})! times 6) + 8$
      $44 = 1 times ((sqrt{9})! times 6) + 8$
      $45 = 1 + ((sqrt{9})! times 6) + 8$
      $ 55 = 1 + (sqrt{9})! + (6 times 8)$
      $56 = (1^9 + 6) times 8 $
      $ 64 = ((-1 + sqrt{9}) + 6) times 8$
      $65 = -(1 times sqrt{9}) + 68 $
      $66 = 1 - sqrt{9} + 68 $
      $69 = 1^9 + 68$
      $70 = -1 + sqrt{9} + 68 $
      $71 = -1 + ((sqrt{9} + 6)times 8)$
      $72 = ((1 times sqrt{9}) + 6) times 8$
      $73 = 1 + ((sqrt{9} + 6)times 8) $
      $74 = (1 times (sqrt{9})!) + 68$
      $75 = 1 + (sqrt{9})!) + 68$
      $ 79 = -1 + ((9 + sqrt{sqrt{ldots sqrt{6}}}) times 8)$
      $ 86 = -1 - sqrt{9} + (6!/8)$
      $90 = (1^9 times 6!)/8 $
      $91 = 1^9 + (6!/8)$







      share|improve this answer











      $endgroup$













      • $begingroup$
        @Oray which numbers are impossible?
        $endgroup$
        – Auroxa
        7 hours ago










      • $begingroup$
        nice one @Auroxa
        $endgroup$
        – Omega Krypton
        7 hours ago



















      1












      $begingroup$

      Here is an answer for 39 that was missing so far:




      39 = -(1*9) + (6*8)




      I also got 79




      79 = ((-1 + (sqrt(9)!))!! + (8!!)/6







      share|improve this answer










      New contributor




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






      $endgroup$













      • $begingroup$
        @Auroxa for my 79 answer I assumed double factorial (!!) as defined on Wolfram Mathworld is allowed.
        $endgroup$
        – ppgdev
        7 hours ago











      Your Answer





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      4 Answers
      4






      active

      oldest

      votes








      4 Answers
      4






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      1












      $begingroup$

      Partial answer (the most obvious solutions):




      1 = -1^9 - 6 + 8 = 1^968

      2 = 1 + sqrt(9) + 6 - 8

      5 = 19 - 6 - 8

      6 = -1 + 9 + 6 - 8 = 1 - 9 + 6 + 8

      7 = 1 * (9 + 6) - 8

      8 = 1 + 9 + 6 - 8

      9 = 1^96 + 8

      11 = -1 + 96 / 8

      12 = 1 * 96 / 8 = 1 + 9 - 6 + 8

      13 = 1 + 96 / 8

      16 = (1 - 9) * (6 - 8)

      17 = 19 + 6 - 8

      18 = 1 + sqrt(9) + 6 + 8

      21 = 19 - 6 + 8

      22 = -1 + 9 + 6 + 8

      23 = 1 * (9 + 6) + 8

      24 = 1 + 9 + 6 + 8

      25 = -1 + sqrt(9) * 6 + 8

      26 = 1 * sqrt(9) * 6 + 8

      27 = 1 + sqrt(9) * 6 + 8

      29 = -19 + 6 * 8

      31 = 19 + 6 + 8

      32 = (1 + sqrt(9)) * 6 + 8

      52 = (1 + 9) * 6 - 8

      57 = 1 * 9 + 6 * 8

      58 = -1 - 9 + 68

      59 = -1 * 9 + 68

      60 = 1 - 9 + 68

      62 = 1 * 9 * 6 + 8

      63 = 1 + 9 * 6 + 8

      67 = 19 + 6 * 8

      68 = (1 + 9) * 6 + 8

      76 = -1 + 9 + 68

      77 = 1 * 9 + 68

      78 = 1 + 9 + 68

      80 = -1 - 9 + 6! / 8

      81 = -1 * 9 + 6! / 8

      82 = 1 - 9 + 6! / 8

      87 = -1 + 96 - 8

      88 = 1 * 96 - 8

      89 = 1 + 96 - 8

      98 = -1 + 9 + 6! / 8

      99 = 1 * 9 + 6! / 8

      100 = 1 + 9 + 6! / 8







      share|improve this answer









      $endgroup$


















        1












        $begingroup$

        Partial answer (the most obvious solutions):




        1 = -1^9 - 6 + 8 = 1^968

        2 = 1 + sqrt(9) + 6 - 8

        5 = 19 - 6 - 8

        6 = -1 + 9 + 6 - 8 = 1 - 9 + 6 + 8

        7 = 1 * (9 + 6) - 8

        8 = 1 + 9 + 6 - 8

        9 = 1^96 + 8

        11 = -1 + 96 / 8

        12 = 1 * 96 / 8 = 1 + 9 - 6 + 8

        13 = 1 + 96 / 8

        16 = (1 - 9) * (6 - 8)

        17 = 19 + 6 - 8

        18 = 1 + sqrt(9) + 6 + 8

        21 = 19 - 6 + 8

        22 = -1 + 9 + 6 + 8

        23 = 1 * (9 + 6) + 8

        24 = 1 + 9 + 6 + 8

        25 = -1 + sqrt(9) * 6 + 8

        26 = 1 * sqrt(9) * 6 + 8

        27 = 1 + sqrt(9) * 6 + 8

        29 = -19 + 6 * 8

        31 = 19 + 6 + 8

        32 = (1 + sqrt(9)) * 6 + 8

        52 = (1 + 9) * 6 - 8

        57 = 1 * 9 + 6 * 8

        58 = -1 - 9 + 68

        59 = -1 * 9 + 68

        60 = 1 - 9 + 68

        62 = 1 * 9 * 6 + 8

        63 = 1 + 9 * 6 + 8

        67 = 19 + 6 * 8

        68 = (1 + 9) * 6 + 8

        76 = -1 + 9 + 68

        77 = 1 * 9 + 68

        78 = 1 + 9 + 68

        80 = -1 - 9 + 6! / 8

        81 = -1 * 9 + 6! / 8

        82 = 1 - 9 + 6! / 8

        87 = -1 + 96 - 8

        88 = 1 * 96 - 8

        89 = 1 + 96 - 8

        98 = -1 + 9 + 6! / 8

        99 = 1 * 9 + 6! / 8

        100 = 1 + 9 + 6! / 8







        share|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          Partial answer (the most obvious solutions):




          1 = -1^9 - 6 + 8 = 1^968

          2 = 1 + sqrt(9) + 6 - 8

          5 = 19 - 6 - 8

          6 = -1 + 9 + 6 - 8 = 1 - 9 + 6 + 8

          7 = 1 * (9 + 6) - 8

          8 = 1 + 9 + 6 - 8

          9 = 1^96 + 8

          11 = -1 + 96 / 8

          12 = 1 * 96 / 8 = 1 + 9 - 6 + 8

          13 = 1 + 96 / 8

          16 = (1 - 9) * (6 - 8)

          17 = 19 + 6 - 8

          18 = 1 + sqrt(9) + 6 + 8

          21 = 19 - 6 + 8

          22 = -1 + 9 + 6 + 8

          23 = 1 * (9 + 6) + 8

          24 = 1 + 9 + 6 + 8

          25 = -1 + sqrt(9) * 6 + 8

          26 = 1 * sqrt(9) * 6 + 8

          27 = 1 + sqrt(9) * 6 + 8

          29 = -19 + 6 * 8

          31 = 19 + 6 + 8

          32 = (1 + sqrt(9)) * 6 + 8

          52 = (1 + 9) * 6 - 8

          57 = 1 * 9 + 6 * 8

          58 = -1 - 9 + 68

          59 = -1 * 9 + 68

          60 = 1 - 9 + 68

          62 = 1 * 9 * 6 + 8

          63 = 1 + 9 * 6 + 8

          67 = 19 + 6 * 8

          68 = (1 + 9) * 6 + 8

          76 = -1 + 9 + 68

          77 = 1 * 9 + 68

          78 = 1 + 9 + 68

          80 = -1 - 9 + 6! / 8

          81 = -1 * 9 + 6! / 8

          82 = 1 - 9 + 6! / 8

          87 = -1 + 96 - 8

          88 = 1 * 96 - 8

          89 = 1 + 96 - 8

          98 = -1 + 9 + 6! / 8

          99 = 1 * 9 + 6! / 8

          100 = 1 + 9 + 6! / 8







          share|improve this answer









          $endgroup$



          Partial answer (the most obvious solutions):




          1 = -1^9 - 6 + 8 = 1^968

          2 = 1 + sqrt(9) + 6 - 8

          5 = 19 - 6 - 8

          6 = -1 + 9 + 6 - 8 = 1 - 9 + 6 + 8

          7 = 1 * (9 + 6) - 8

          8 = 1 + 9 + 6 - 8

          9 = 1^96 + 8

          11 = -1 + 96 / 8

          12 = 1 * 96 / 8 = 1 + 9 - 6 + 8

          13 = 1 + 96 / 8

          16 = (1 - 9) * (6 - 8)

          17 = 19 + 6 - 8

          18 = 1 + sqrt(9) + 6 + 8

          21 = 19 - 6 + 8

          22 = -1 + 9 + 6 + 8

          23 = 1 * (9 + 6) + 8

          24 = 1 + 9 + 6 + 8

          25 = -1 + sqrt(9) * 6 + 8

          26 = 1 * sqrt(9) * 6 + 8

          27 = 1 + sqrt(9) * 6 + 8

          29 = -19 + 6 * 8

          31 = 19 + 6 + 8

          32 = (1 + sqrt(9)) * 6 + 8

          52 = (1 + 9) * 6 - 8

          57 = 1 * 9 + 6 * 8

          58 = -1 - 9 + 68

          59 = -1 * 9 + 68

          60 = 1 - 9 + 68

          62 = 1 * 9 * 6 + 8

          63 = 1 + 9 * 6 + 8

          67 = 19 + 6 * 8

          68 = (1 + 9) * 6 + 8

          76 = -1 + 9 + 68

          77 = 1 * 9 + 68

          78 = 1 + 9 + 68

          80 = -1 - 9 + 6! / 8

          81 = -1 * 9 + 6! / 8

          82 = 1 - 9 + 6! / 8

          87 = -1 + 96 - 8

          88 = 1 * 96 - 8

          89 = 1 + 96 - 8

          98 = -1 + 9 + 6! / 8

          99 = 1 * 9 + 6! / 8

          100 = 1 + 9 + 6! / 8








          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered 20 hours ago









          trolley813trolley813

          1,14638




          1,14638























              1












              $begingroup$

              Partial Answer (too lazy to type them out :P)




              enter image description here







              share|improve this answer









              $endgroup$


















                1












                $begingroup$

                Partial Answer (too lazy to type them out :P)




                enter image description here







                share|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  Partial Answer (too lazy to type them out :P)




                  enter image description here







                  share|improve this answer









                  $endgroup$



                  Partial Answer (too lazy to type them out :P)




                  enter image description here








                  share|improve this answer












                  share|improve this answer



                  share|improve this answer










                  answered 16 hours ago









                  Omega KryptonOmega Krypton

                  4,2251339




                  4,2251339























                      1












                      $begingroup$

                      Here are all the numbers which have not been obtained already by trolley813, Omega Krypton and ppgdev




                      $28 = 1 times ((sqrt{9})! times 6) - 8$
                      $30 =-(1times(sqrt{9})!) + sqrt{sqrt{6^8}} $
                      $33 = 19 + 6 + 8$
                      $34 = ((1+(sqrt{9})!) times 6) - 8 $
                      $35 = (1 times sqrt{(sqrt{9})^6}) + 8$
                      $36 = (1 + sqrt{(sqrt{9})^6}) + 8$
                      $37 = 1^9 + sqrt{sqrt{6^8}}$
                      $38 = (1+sqrt{9})! + 6 + 8$
                      $41 = -(1 + (sqrt{9})!) + (6 times 8)$
                      $42 = -(1 times (sqrt{9})!) + (6 times 8)$
                      $43 = -1 + ((sqrt{9})! times 6) + 8$
                      $44 = 1 times ((sqrt{9})! times 6) + 8$
                      $45 = 1 + ((sqrt{9})! times 6) + 8$
                      $ 55 = 1 + (sqrt{9})! + (6 times 8)$
                      $56 = (1^9 + 6) times 8 $
                      $ 64 = ((-1 + sqrt{9}) + 6) times 8$
                      $65 = -(1 times sqrt{9}) + 68 $
                      $66 = 1 - sqrt{9} + 68 $
                      $69 = 1^9 + 68$
                      $70 = -1 + sqrt{9} + 68 $
                      $71 = -1 + ((sqrt{9} + 6)times 8)$
                      $72 = ((1 times sqrt{9}) + 6) times 8$
                      $73 = 1 + ((sqrt{9} + 6)times 8) $
                      $74 = (1 times (sqrt{9})!) + 68$
                      $75 = 1 + (sqrt{9})!) + 68$
                      $ 79 = -1 + ((9 + sqrt{sqrt{ldots sqrt{6}}}) times 8)$
                      $ 86 = -1 - sqrt{9} + (6!/8)$
                      $90 = (1^9 times 6!)/8 $
                      $91 = 1^9 + (6!/8)$







                      share|improve this answer











                      $endgroup$













                      • $begingroup$
                        @Oray which numbers are impossible?
                        $endgroup$
                        – Auroxa
                        7 hours ago










                      • $begingroup$
                        nice one @Auroxa
                        $endgroup$
                        – Omega Krypton
                        7 hours ago
















                      1












                      $begingroup$

                      Here are all the numbers which have not been obtained already by trolley813, Omega Krypton and ppgdev




                      $28 = 1 times ((sqrt{9})! times 6) - 8$
                      $30 =-(1times(sqrt{9})!) + sqrt{sqrt{6^8}} $
                      $33 = 19 + 6 + 8$
                      $34 = ((1+(sqrt{9})!) times 6) - 8 $
                      $35 = (1 times sqrt{(sqrt{9})^6}) + 8$
                      $36 = (1 + sqrt{(sqrt{9})^6}) + 8$
                      $37 = 1^9 + sqrt{sqrt{6^8}}$
                      $38 = (1+sqrt{9})! + 6 + 8$
                      $41 = -(1 + (sqrt{9})!) + (6 times 8)$
                      $42 = -(1 times (sqrt{9})!) + (6 times 8)$
                      $43 = -1 + ((sqrt{9})! times 6) + 8$
                      $44 = 1 times ((sqrt{9})! times 6) + 8$
                      $45 = 1 + ((sqrt{9})! times 6) + 8$
                      $ 55 = 1 + (sqrt{9})! + (6 times 8)$
                      $56 = (1^9 + 6) times 8 $
                      $ 64 = ((-1 + sqrt{9}) + 6) times 8$
                      $65 = -(1 times sqrt{9}) + 68 $
                      $66 = 1 - sqrt{9} + 68 $
                      $69 = 1^9 + 68$
                      $70 = -1 + sqrt{9} + 68 $
                      $71 = -1 + ((sqrt{9} + 6)times 8)$
                      $72 = ((1 times sqrt{9}) + 6) times 8$
                      $73 = 1 + ((sqrt{9} + 6)times 8) $
                      $74 = (1 times (sqrt{9})!) + 68$
                      $75 = 1 + (sqrt{9})!) + 68$
                      $ 79 = -1 + ((9 + sqrt{sqrt{ldots sqrt{6}}}) times 8)$
                      $ 86 = -1 - sqrt{9} + (6!/8)$
                      $90 = (1^9 times 6!)/8 $
                      $91 = 1^9 + (6!/8)$







                      share|improve this answer











                      $endgroup$













                      • $begingroup$
                        @Oray which numbers are impossible?
                        $endgroup$
                        – Auroxa
                        7 hours ago










                      • $begingroup$
                        nice one @Auroxa
                        $endgroup$
                        – Omega Krypton
                        7 hours ago














                      1












                      1








                      1





                      $begingroup$

                      Here are all the numbers which have not been obtained already by trolley813, Omega Krypton and ppgdev




                      $28 = 1 times ((sqrt{9})! times 6) - 8$
                      $30 =-(1times(sqrt{9})!) + sqrt{sqrt{6^8}} $
                      $33 = 19 + 6 + 8$
                      $34 = ((1+(sqrt{9})!) times 6) - 8 $
                      $35 = (1 times sqrt{(sqrt{9})^6}) + 8$
                      $36 = (1 + sqrt{(sqrt{9})^6}) + 8$
                      $37 = 1^9 + sqrt{sqrt{6^8}}$
                      $38 = (1+sqrt{9})! + 6 + 8$
                      $41 = -(1 + (sqrt{9})!) + (6 times 8)$
                      $42 = -(1 times (sqrt{9})!) + (6 times 8)$
                      $43 = -1 + ((sqrt{9})! times 6) + 8$
                      $44 = 1 times ((sqrt{9})! times 6) + 8$
                      $45 = 1 + ((sqrt{9})! times 6) + 8$
                      $ 55 = 1 + (sqrt{9})! + (6 times 8)$
                      $56 = (1^9 + 6) times 8 $
                      $ 64 = ((-1 + sqrt{9}) + 6) times 8$
                      $65 = -(1 times sqrt{9}) + 68 $
                      $66 = 1 - sqrt{9} + 68 $
                      $69 = 1^9 + 68$
                      $70 = -1 + sqrt{9} + 68 $
                      $71 = -1 + ((sqrt{9} + 6)times 8)$
                      $72 = ((1 times sqrt{9}) + 6) times 8$
                      $73 = 1 + ((sqrt{9} + 6)times 8) $
                      $74 = (1 times (sqrt{9})!) + 68$
                      $75 = 1 + (sqrt{9})!) + 68$
                      $ 79 = -1 + ((9 + sqrt{sqrt{ldots sqrt{6}}}) times 8)$
                      $ 86 = -1 - sqrt{9} + (6!/8)$
                      $90 = (1^9 times 6!)/8 $
                      $91 = 1^9 + (6!/8)$







                      share|improve this answer











                      $endgroup$



                      Here are all the numbers which have not been obtained already by trolley813, Omega Krypton and ppgdev




                      $28 = 1 times ((sqrt{9})! times 6) - 8$
                      $30 =-(1times(sqrt{9})!) + sqrt{sqrt{6^8}} $
                      $33 = 19 + 6 + 8$
                      $34 = ((1+(sqrt{9})!) times 6) - 8 $
                      $35 = (1 times sqrt{(sqrt{9})^6}) + 8$
                      $36 = (1 + sqrt{(sqrt{9})^6}) + 8$
                      $37 = 1^9 + sqrt{sqrt{6^8}}$
                      $38 = (1+sqrt{9})! + 6 + 8$
                      $41 = -(1 + (sqrt{9})!) + (6 times 8)$
                      $42 = -(1 times (sqrt{9})!) + (6 times 8)$
                      $43 = -1 + ((sqrt{9})! times 6) + 8$
                      $44 = 1 times ((sqrt{9})! times 6) + 8$
                      $45 = 1 + ((sqrt{9})! times 6) + 8$
                      $ 55 = 1 + (sqrt{9})! + (6 times 8)$
                      $56 = (1^9 + 6) times 8 $
                      $ 64 = ((-1 + sqrt{9}) + 6) times 8$
                      $65 = -(1 times sqrt{9}) + 68 $
                      $66 = 1 - sqrt{9} + 68 $
                      $69 = 1^9 + 68$
                      $70 = -1 + sqrt{9} + 68 $
                      $71 = -1 + ((sqrt{9} + 6)times 8)$
                      $72 = ((1 times sqrt{9}) + 6) times 8$
                      $73 = 1 + ((sqrt{9} + 6)times 8) $
                      $74 = (1 times (sqrt{9})!) + 68$
                      $75 = 1 + (sqrt{9})!) + 68$
                      $ 79 = -1 + ((9 + sqrt{sqrt{ldots sqrt{6}}}) times 8)$
                      $ 86 = -1 - sqrt{9} + (6!/8)$
                      $90 = (1^9 times 6!)/8 $
                      $91 = 1^9 + (6!/8)$








                      share|improve this answer














                      share|improve this answer



                      share|improve this answer








                      edited 12 hours ago

























                      answered 12 hours ago









                      hexominohexomino

                      41k3122191




                      41k3122191












                      • $begingroup$
                        @Oray which numbers are impossible?
                        $endgroup$
                        – Auroxa
                        7 hours ago










                      • $begingroup$
                        nice one @Auroxa
                        $endgroup$
                        – Omega Krypton
                        7 hours ago


















                      • $begingroup$
                        @Oray which numbers are impossible?
                        $endgroup$
                        – Auroxa
                        7 hours ago










                      • $begingroup$
                        nice one @Auroxa
                        $endgroup$
                        – Omega Krypton
                        7 hours ago
















                      $begingroup$
                      @Oray which numbers are impossible?
                      $endgroup$
                      – Auroxa
                      7 hours ago




                      $begingroup$
                      @Oray which numbers are impossible?
                      $endgroup$
                      – Auroxa
                      7 hours ago












                      $begingroup$
                      nice one @Auroxa
                      $endgroup$
                      – Omega Krypton
                      7 hours ago




                      $begingroup$
                      nice one @Auroxa
                      $endgroup$
                      – Omega Krypton
                      7 hours ago











                      1












                      $begingroup$

                      Here is an answer for 39 that was missing so far:




                      39 = -(1*9) + (6*8)




                      I also got 79




                      79 = ((-1 + (sqrt(9)!))!! + (8!!)/6







                      share|improve this answer










                      New contributor




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






                      $endgroup$













                      • $begingroup$
                        @Auroxa for my 79 answer I assumed double factorial (!!) as defined on Wolfram Mathworld is allowed.
                        $endgroup$
                        – ppgdev
                        7 hours ago
















                      1












                      $begingroup$

                      Here is an answer for 39 that was missing so far:




                      39 = -(1*9) + (6*8)




                      I also got 79




                      79 = ((-1 + (sqrt(9)!))!! + (8!!)/6







                      share|improve this answer










                      New contributor




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






                      $endgroup$













                      • $begingroup$
                        @Auroxa for my 79 answer I assumed double factorial (!!) as defined on Wolfram Mathworld is allowed.
                        $endgroup$
                        – ppgdev
                        7 hours ago














                      1












                      1








                      1





                      $begingroup$

                      Here is an answer for 39 that was missing so far:




                      39 = -(1*9) + (6*8)




                      I also got 79




                      79 = ((-1 + (sqrt(9)!))!! + (8!!)/6







                      share|improve this answer










                      New contributor




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






                      $endgroup$



                      Here is an answer for 39 that was missing so far:




                      39 = -(1*9) + (6*8)




                      I also got 79




                      79 = ((-1 + (sqrt(9)!))!! + (8!!)/6








                      share|improve this answer










                      New contributor




                      ppgdev 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 9 hours ago





















                      New contributor




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









                      answered 12 hours ago









                      ppgdevppgdev

                      314




                      314




                      New contributor




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





                      New contributor





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






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












                      • $begingroup$
                        @Auroxa for my 79 answer I assumed double factorial (!!) as defined on Wolfram Mathworld is allowed.
                        $endgroup$
                        – ppgdev
                        7 hours ago


















                      • $begingroup$
                        @Auroxa for my 79 answer I assumed double factorial (!!) as defined on Wolfram Mathworld is allowed.
                        $endgroup$
                        – ppgdev
                        7 hours ago
















                      $begingroup$
                      @Auroxa for my 79 answer I assumed double factorial (!!) as defined on Wolfram Mathworld is allowed.
                      $endgroup$
                      – ppgdev
                      7 hours ago




                      $begingroup$
                      @Auroxa for my 79 answer I assumed double factorial (!!) as defined on Wolfram Mathworld is allowed.
                      $endgroup$
                      – ppgdev
                      7 hours ago










                      Auroxa is a new contributor. Be nice, and check out our Code of Conduct.










                      draft saved

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                      Auroxa is a new contributor. Be nice, and check out our Code of Conduct.













                      Auroxa is a new contributor. Be nice, and check out our Code of Conduct.












                      Auroxa is a new contributor. Be nice, and check out our Code of Conduct.
















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