2019 is coming!












4












$begingroup$


Use $2$ $0$ $1 $ and $9$ to make the numbers $66, 74, 76, and$ $77$



RULES




  1. You must use all 4 digits.
    Only the digits 2, 0, 1, and 9 can be used.
    You can make multi-digit numbers out of the numbers. Examples: 20, 92, 2.9


  2. The square function may NOT be used. Nor may the cube, raise to a fourth power, or any other function that raises a number to a specific power.
    You may use the ^ operation if you use a digit, for example, [(9 + 1)^2 - 0!] is acceptable (if you're trying to get 99), because 2, 0, 1, and 9 is used.
    However, [20 ^ 2 / 9 + 1] can't be used to get 40 because it uses an extra 2.


  3. Sorry, but the integer function may NOT be used. Nor may the round, floor, ceiling, more than 2 factorials in a row, repeating symbol, or truncate functions.


  4. +, -, *, /, (), !, sqrt, ^, and !! may be used for functions.


  5. I picked these because these are the hardest for 2019. From the numbers 1-100, these are the only ones I didn't get.



Again, no brute-force methods. Good luck.






mathematics







share|improve this question









$endgroup$












  • $begingroup$
    Are these solutions possible?
    $endgroup$
    – QuantumTwinkie
    Sep 12 '18 at 21:26






  • 1




    $begingroup$
    @QuantumTwinkie let's find out! :D
    $endgroup$
    – user477343
    Sep 12 '18 at 21:47






  • 1




    $begingroup$
    @Alto Doesn't using sqrt contradict your requirements? It's an explicit square root, i.e. it uses 2, or, even worse, it could be represented as x^1/2 which uses 1 and 2.
    $endgroup$
    – rhsquared
    Sep 13 '18 at 2:36






  • 1




    $begingroup$
    @user477343 You are quite right. Maybe I won't be allowed to answer because of the implicit two in my name.
    $endgroup$
    – rhsquared
    Sep 13 '18 at 6:01






  • 1




    $begingroup$
    Are bare decimal points allowed, like $.1$ and $.2$?
    $endgroup$
    – nickgard
    Sep 13 '18 at 8:01
















4












$begingroup$


Use $2$ $0$ $1 $ and $9$ to make the numbers $66, 74, 76, and$ $77$



RULES




  1. You must use all 4 digits.
    Only the digits 2, 0, 1, and 9 can be used.
    You can make multi-digit numbers out of the numbers. Examples: 20, 92, 2.9


  2. The square function may NOT be used. Nor may the cube, raise to a fourth power, or any other function that raises a number to a specific power.
    You may use the ^ operation if you use a digit, for example, [(9 + 1)^2 - 0!] is acceptable (if you're trying to get 99), because 2, 0, 1, and 9 is used.
    However, [20 ^ 2 / 9 + 1] can't be used to get 40 because it uses an extra 2.


  3. Sorry, but the integer function may NOT be used. Nor may the round, floor, ceiling, more than 2 factorials in a row, repeating symbol, or truncate functions.


  4. +, -, *, /, (), !, sqrt, ^, and !! may be used for functions.


  5. I picked these because these are the hardest for 2019. From the numbers 1-100, these are the only ones I didn't get.



Again, no brute-force methods. Good luck.






mathematics







share|improve this question









$endgroup$












  • $begingroup$
    Are these solutions possible?
    $endgroup$
    – QuantumTwinkie
    Sep 12 '18 at 21:26






  • 1




    $begingroup$
    @QuantumTwinkie let's find out! :D
    $endgroup$
    – user477343
    Sep 12 '18 at 21:47






  • 1




    $begingroup$
    @Alto Doesn't using sqrt contradict your requirements? It's an explicit square root, i.e. it uses 2, or, even worse, it could be represented as x^1/2 which uses 1 and 2.
    $endgroup$
    – rhsquared
    Sep 13 '18 at 2:36






  • 1




    $begingroup$
    @user477343 You are quite right. Maybe I won't be allowed to answer because of the implicit two in my name.
    $endgroup$
    – rhsquared
    Sep 13 '18 at 6:01






  • 1




    $begingroup$
    Are bare decimal points allowed, like $.1$ and $.2$?
    $endgroup$
    – nickgard
    Sep 13 '18 at 8:01














4












4








4


1



$begingroup$


Use $2$ $0$ $1 $ and $9$ to make the numbers $66, 74, 76, and$ $77$



RULES




  1. You must use all 4 digits.
    Only the digits 2, 0, 1, and 9 can be used.
    You can make multi-digit numbers out of the numbers. Examples: 20, 92, 2.9


  2. The square function may NOT be used. Nor may the cube, raise to a fourth power, or any other function that raises a number to a specific power.
    You may use the ^ operation if you use a digit, for example, [(9 + 1)^2 - 0!] is acceptable (if you're trying to get 99), because 2, 0, 1, and 9 is used.
    However, [20 ^ 2 / 9 + 1] can't be used to get 40 because it uses an extra 2.


  3. Sorry, but the integer function may NOT be used. Nor may the round, floor, ceiling, more than 2 factorials in a row, repeating symbol, or truncate functions.


  4. +, -, *, /, (), !, sqrt, ^, and !! may be used for functions.


  5. I picked these because these are the hardest for 2019. From the numbers 1-100, these are the only ones I didn't get.



Again, no brute-force methods. Good luck.






mathematics







share|improve this question









$endgroup$




Use $2$ $0$ $1 $ and $9$ to make the numbers $66, 74, 76, and$ $77$



RULES




  1. You must use all 4 digits.
    Only the digits 2, 0, 1, and 9 can be used.
    You can make multi-digit numbers out of the numbers. Examples: 20, 92, 2.9


  2. The square function may NOT be used. Nor may the cube, raise to a fourth power, or any other function that raises a number to a specific power.
    You may use the ^ operation if you use a digit, for example, [(9 + 1)^2 - 0!] is acceptable (if you're trying to get 99), because 2, 0, 1, and 9 is used.
    However, [20 ^ 2 / 9 + 1] can't be used to get 40 because it uses an extra 2.


  3. Sorry, but the integer function may NOT be used. Nor may the round, floor, ceiling, more than 2 factorials in a row, repeating symbol, or truncate functions.


  4. +, -, *, /, (), !, sqrt, ^, and !! may be used for functions.


  5. I picked these because these are the hardest for 2019. From the numbers 1-100, these are the only ones I didn't get.



Again, no brute-force methods. Good luck.






mathematics




mathematics






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked Sep 12 '18 at 20:30









AltoAlto

1,094119




1,094119












  • $begingroup$
    Are these solutions possible?
    $endgroup$
    – QuantumTwinkie
    Sep 12 '18 at 21:26






  • 1




    $begingroup$
    @QuantumTwinkie let's find out! :D
    $endgroup$
    – user477343
    Sep 12 '18 at 21:47






  • 1




    $begingroup$
    @Alto Doesn't using sqrt contradict your requirements? It's an explicit square root, i.e. it uses 2, or, even worse, it could be represented as x^1/2 which uses 1 and 2.
    $endgroup$
    – rhsquared
    Sep 13 '18 at 2:36






  • 1




    $begingroup$
    @user477343 You are quite right. Maybe I won't be allowed to answer because of the implicit two in my name.
    $endgroup$
    – rhsquared
    Sep 13 '18 at 6:01






  • 1




    $begingroup$
    Are bare decimal points allowed, like $.1$ and $.2$?
    $endgroup$
    – nickgard
    Sep 13 '18 at 8:01


















  • $begingroup$
    Are these solutions possible?
    $endgroup$
    – QuantumTwinkie
    Sep 12 '18 at 21:26






  • 1




    $begingroup$
    @QuantumTwinkie let's find out! :D
    $endgroup$
    – user477343
    Sep 12 '18 at 21:47






  • 1




    $begingroup$
    @Alto Doesn't using sqrt contradict your requirements? It's an explicit square root, i.e. it uses 2, or, even worse, it could be represented as x^1/2 which uses 1 and 2.
    $endgroup$
    – rhsquared
    Sep 13 '18 at 2:36






  • 1




    $begingroup$
    @user477343 You are quite right. Maybe I won't be allowed to answer because of the implicit two in my name.
    $endgroup$
    – rhsquared
    Sep 13 '18 at 6:01






  • 1




    $begingroup$
    Are bare decimal points allowed, like $.1$ and $.2$?
    $endgroup$
    – nickgard
    Sep 13 '18 at 8:01
















$begingroup$
Are these solutions possible?
$endgroup$
– QuantumTwinkie
Sep 12 '18 at 21:26




$begingroup$
Are these solutions possible?
$endgroup$
– QuantumTwinkie
Sep 12 '18 at 21:26




1




1




$begingroup$
@QuantumTwinkie let's find out! :D
$endgroup$
– user477343
Sep 12 '18 at 21:47




$begingroup$
@QuantumTwinkie let's find out! :D
$endgroup$
– user477343
Sep 12 '18 at 21:47




1




1




$begingroup$
@Alto Doesn't using sqrt contradict your requirements? It's an explicit square root, i.e. it uses 2, or, even worse, it could be represented as x^1/2 which uses 1 and 2.
$endgroup$
– rhsquared
Sep 13 '18 at 2:36




$begingroup$
@Alto Doesn't using sqrt contradict your requirements? It's an explicit square root, i.e. it uses 2, or, even worse, it could be represented as x^1/2 which uses 1 and 2.
$endgroup$
– rhsquared
Sep 13 '18 at 2:36




1




1




$begingroup$
@user477343 You are quite right. Maybe I won't be allowed to answer because of the implicit two in my name.
$endgroup$
– rhsquared
Sep 13 '18 at 6:01




$begingroup$
@user477343 You are quite right. Maybe I won't be allowed to answer because of the implicit two in my name.
$endgroup$
– rhsquared
Sep 13 '18 at 6:01




1




1




$begingroup$
Are bare decimal points allowed, like $.1$ and $.2$?
$endgroup$
– nickgard
Sep 13 '18 at 8:01




$begingroup$
Are bare decimal points allowed, like $.1$ and $.2$?
$endgroup$
– nickgard
Sep 13 '18 at 8:01










3 Answers
3






active

oldest

votes


















6












$begingroup$

66:




$(sqrt{9})!times(12-0!)$




74:




$9/.12-0!$




76:




$9/.12+0!$




77:




$((sqrt{9})!)!times.1+(0!/.2)$







share|improve this answer









$endgroup$













  • $begingroup$
    I think the square function cannot be used
    $endgroup$
    – Ian Fako
    Sep 13 '18 at 9:18










  • $begingroup$
    No it can't even I tried using it.
    $endgroup$
    – Quark-epoch
    Sep 13 '18 at 9:38










  • $begingroup$
    And you can't concatenate to make $12$, either, but $(+1)$ in at least $13$ hours for the effort (DVL) :D
    $endgroup$
    – user477343
    Sep 13 '18 at 10:44












  • $begingroup$
    @user477343 you can concatenate 1 and 2. See examples above 20, 92, 2.9
    $endgroup$
    – Ian Fako
    Sep 13 '18 at 10:52










  • $begingroup$
    @IanFako Wait... I thought it said "can't" .... dammit, I would have had more of a chance finding a solution otherwise. Eh well, everything happens for a reason... at least in my philosophy.
    $endgroup$
    – user477343
    Sep 13 '18 at 10:54





















0












$begingroup$

Also here is another solution to 77 that might be easier- (√9!)!*.1+(0!/.2)



The equation doesn't require the use of a double factorial.



KG






share|improve this answer









$endgroup$













  • $begingroup$
    You should check the other answers before adding a new one. This is the same as the accepted answer for 77.
    $endgroup$
    – Herb Wolfe
    Jan 7 at 21:52



















0












$begingroup$

How do you get 83 and 86 using these numbers?





share








New contributor




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






$endgroup$













    Your Answer





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    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    6












    $begingroup$

    66:




    $(sqrt{9})!times(12-0!)$




    74:




    $9/.12-0!$




    76:




    $9/.12+0!$




    77:




    $((sqrt{9})!)!times.1+(0!/.2)$







    share|improve this answer









    $endgroup$













    • $begingroup$
      I think the square function cannot be used
      $endgroup$
      – Ian Fako
      Sep 13 '18 at 9:18










    • $begingroup$
      No it can't even I tried using it.
      $endgroup$
      – Quark-epoch
      Sep 13 '18 at 9:38










    • $begingroup$
      And you can't concatenate to make $12$, either, but $(+1)$ in at least $13$ hours for the effort (DVL) :D
      $endgroup$
      – user477343
      Sep 13 '18 at 10:44












    • $begingroup$
      @user477343 you can concatenate 1 and 2. See examples above 20, 92, 2.9
      $endgroup$
      – Ian Fako
      Sep 13 '18 at 10:52










    • $begingroup$
      @IanFako Wait... I thought it said "can't" .... dammit, I would have had more of a chance finding a solution otherwise. Eh well, everything happens for a reason... at least in my philosophy.
      $endgroup$
      – user477343
      Sep 13 '18 at 10:54


















    6












    $begingroup$

    66:




    $(sqrt{9})!times(12-0!)$




    74:




    $9/.12-0!$




    76:




    $9/.12+0!$




    77:




    $((sqrt{9})!)!times.1+(0!/.2)$







    share|improve this answer









    $endgroup$













    • $begingroup$
      I think the square function cannot be used
      $endgroup$
      – Ian Fako
      Sep 13 '18 at 9:18










    • $begingroup$
      No it can't even I tried using it.
      $endgroup$
      – Quark-epoch
      Sep 13 '18 at 9:38










    • $begingroup$
      And you can't concatenate to make $12$, either, but $(+1)$ in at least $13$ hours for the effort (DVL) :D
      $endgroup$
      – user477343
      Sep 13 '18 at 10:44












    • $begingroup$
      @user477343 you can concatenate 1 and 2. See examples above 20, 92, 2.9
      $endgroup$
      – Ian Fako
      Sep 13 '18 at 10:52










    • $begingroup$
      @IanFako Wait... I thought it said "can't" .... dammit, I would have had more of a chance finding a solution otherwise. Eh well, everything happens for a reason... at least in my philosophy.
      $endgroup$
      – user477343
      Sep 13 '18 at 10:54
















    6












    6








    6





    $begingroup$

    66:




    $(sqrt{9})!times(12-0!)$




    74:




    $9/.12-0!$




    76:




    $9/.12+0!$




    77:




    $((sqrt{9})!)!times.1+(0!/.2)$







    share|improve this answer









    $endgroup$



    66:




    $(sqrt{9})!times(12-0!)$




    74:




    $9/.12-0!$




    76:




    $9/.12+0!$




    77:




    $((sqrt{9})!)!times.1+(0!/.2)$








    share|improve this answer












    share|improve this answer



    share|improve this answer










    answered Sep 13 '18 at 8:35









    nickgardnickgard

    1,7601815




    1,7601815












    • $begingroup$
      I think the square function cannot be used
      $endgroup$
      – Ian Fako
      Sep 13 '18 at 9:18










    • $begingroup$
      No it can't even I tried using it.
      $endgroup$
      – Quark-epoch
      Sep 13 '18 at 9:38










    • $begingroup$
      And you can't concatenate to make $12$, either, but $(+1)$ in at least $13$ hours for the effort (DVL) :D
      $endgroup$
      – user477343
      Sep 13 '18 at 10:44












    • $begingroup$
      @user477343 you can concatenate 1 and 2. See examples above 20, 92, 2.9
      $endgroup$
      – Ian Fako
      Sep 13 '18 at 10:52










    • $begingroup$
      @IanFako Wait... I thought it said "can't" .... dammit, I would have had more of a chance finding a solution otherwise. Eh well, everything happens for a reason... at least in my philosophy.
      $endgroup$
      – user477343
      Sep 13 '18 at 10:54




















    • $begingroup$
      I think the square function cannot be used
      $endgroup$
      – Ian Fako
      Sep 13 '18 at 9:18










    • $begingroup$
      No it can't even I tried using it.
      $endgroup$
      – Quark-epoch
      Sep 13 '18 at 9:38










    • $begingroup$
      And you can't concatenate to make $12$, either, but $(+1)$ in at least $13$ hours for the effort (DVL) :D
      $endgroup$
      – user477343
      Sep 13 '18 at 10:44












    • $begingroup$
      @user477343 you can concatenate 1 and 2. See examples above 20, 92, 2.9
      $endgroup$
      – Ian Fako
      Sep 13 '18 at 10:52










    • $begingroup$
      @IanFako Wait... I thought it said "can't" .... dammit, I would have had more of a chance finding a solution otherwise. Eh well, everything happens for a reason... at least in my philosophy.
      $endgroup$
      – user477343
      Sep 13 '18 at 10:54


















    $begingroup$
    I think the square function cannot be used
    $endgroup$
    – Ian Fako
    Sep 13 '18 at 9:18




    $begingroup$
    I think the square function cannot be used
    $endgroup$
    – Ian Fako
    Sep 13 '18 at 9:18












    $begingroup$
    No it can't even I tried using it.
    $endgroup$
    – Quark-epoch
    Sep 13 '18 at 9:38




    $begingroup$
    No it can't even I tried using it.
    $endgroup$
    – Quark-epoch
    Sep 13 '18 at 9:38












    $begingroup$
    And you can't concatenate to make $12$, either, but $(+1)$ in at least $13$ hours for the effort (DVL) :D
    $endgroup$
    – user477343
    Sep 13 '18 at 10:44






    $begingroup$
    And you can't concatenate to make $12$, either, but $(+1)$ in at least $13$ hours for the effort (DVL) :D
    $endgroup$
    – user477343
    Sep 13 '18 at 10:44














    $begingroup$
    @user477343 you can concatenate 1 and 2. See examples above 20, 92, 2.9
    $endgroup$
    – Ian Fako
    Sep 13 '18 at 10:52




    $begingroup$
    @user477343 you can concatenate 1 and 2. See examples above 20, 92, 2.9
    $endgroup$
    – Ian Fako
    Sep 13 '18 at 10:52












    $begingroup$
    @IanFako Wait... I thought it said "can't" .... dammit, I would have had more of a chance finding a solution otherwise. Eh well, everything happens for a reason... at least in my philosophy.
    $endgroup$
    – user477343
    Sep 13 '18 at 10:54






    $begingroup$
    @IanFako Wait... I thought it said "can't" .... dammit, I would have had more of a chance finding a solution otherwise. Eh well, everything happens for a reason... at least in my philosophy.
    $endgroup$
    – user477343
    Sep 13 '18 at 10:54













    0












    $begingroup$

    Also here is another solution to 77 that might be easier- (√9!)!*.1+(0!/.2)



    The equation doesn't require the use of a double factorial.



    KG






    share|improve this answer









    $endgroup$













    • $begingroup$
      You should check the other answers before adding a new one. This is the same as the accepted answer for 77.
      $endgroup$
      – Herb Wolfe
      Jan 7 at 21:52
















    0












    $begingroup$

    Also here is another solution to 77 that might be easier- (√9!)!*.1+(0!/.2)



    The equation doesn't require the use of a double factorial.



    KG






    share|improve this answer









    $endgroup$













    • $begingroup$
      You should check the other answers before adding a new one. This is the same as the accepted answer for 77.
      $endgroup$
      – Herb Wolfe
      Jan 7 at 21:52














    0












    0








    0





    $begingroup$

    Also here is another solution to 77 that might be easier- (√9!)!*.1+(0!/.2)



    The equation doesn't require the use of a double factorial.



    KG






    share|improve this answer









    $endgroup$



    Also here is another solution to 77 that might be easier- (√9!)!*.1+(0!/.2)



    The equation doesn't require the use of a double factorial.



    KG







    share|improve this answer












    share|improve this answer



    share|improve this answer










    answered Jan 7 at 21:36









    MathMaster920MathMaster920

    12




    12












    • $begingroup$
      You should check the other answers before adding a new one. This is the same as the accepted answer for 77.
      $endgroup$
      – Herb Wolfe
      Jan 7 at 21:52


















    • $begingroup$
      You should check the other answers before adding a new one. This is the same as the accepted answer for 77.
      $endgroup$
      – Herb Wolfe
      Jan 7 at 21:52
















    $begingroup$
    You should check the other answers before adding a new one. This is the same as the accepted answer for 77.
    $endgroup$
    – Herb Wolfe
    Jan 7 at 21:52




    $begingroup$
    You should check the other answers before adding a new one. This is the same as the accepted answer for 77.
    $endgroup$
    – Herb Wolfe
    Jan 7 at 21:52











    0












    $begingroup$

    How do you get 83 and 86 using these numbers?





    share








    New contributor




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






    $endgroup$


















      0












      $begingroup$

      How do you get 83 and 86 using these numbers?





      share








      New contributor




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






      $endgroup$
















        0












        0








        0





        $begingroup$

        How do you get 83 and 86 using these numbers?





        share








        New contributor




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






        $endgroup$



        How do you get 83 and 86 using these numbers?






        share








        New contributor




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








        share


        share






        New contributor




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









        answered 2 mins ago









        AllenAllen

        1




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