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101 from 3, 4, 8 and 9

  1. Dec 19, 2004 #1
    Using each time all of the four digits: 3, 4, 8 and 9 construct all the integers up to and including 101. You may only use addition, subtraction, multiplication, division, the decimal point, raising to a power, and (if there is no other way) a recurring decimal. Decimal fractions without a leading integer are allowed. You can also run digits together to make numbers like 9 = 48 - 39

    Factorials are not allowed, nor is the square root operation, and you can't use constructions like (3 + 4)8 to mean 78.

    OK, I'll start you off:

    [tex]1 = \frac{4 - 3}{9 - 8}[/tex]

    [tex]2 = 4 + 9 - 3 - 8[/tex]

    [tex]3 = \frac{4}{9.3(recurring) - 8}[/tex]

    I know there are easier ways to make 3, but I wanted to show what I meant by recurring.

    Have fun.
     
  2. jcsd
  3. Dec 19, 2004 #2

    Galileo

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    [tex]4=9-3-\frac{8}{4}[/tex]

    [tex]5=\frac{8}{4}+\frac{9}{3}[/tex]
    [tex]6=3+4+8-9[/tex]

    [tex]7=(4+3)(9-8)[/tex]

    [tex]8=3+4+9-8[/tex]

    [tex]9=9(8-4-3)[/tex]

    [tex]10=9+8-3-4[/tex]

    Go me!
     
  4. Dec 19, 2004 #3

    Galileo

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    [tex]11 = (3)(4)+8-9[/tex]

    [tex]12 = (3)(4)(9-8)[/tex]

    [tex]13 = (3)(4)+9-8[/tex]

    [tex]14 = 9+3+\frac{8}{4}[/tex]

    [tex]15 = \frac{(8)(3)}{4}+9[/tex]

    [tex]16 = 9+8+3-4[/tex]

    [tex]17 = (9+8)(4-3)[/tex]

    [tex]18 = 9+8+4-3[/tex]

    [tex]19 = (3)(8)+4-9[/tex]

    [tex]20 = \frac{9}{3}(4)+8[/tex]
     
  5. Dec 19, 2004 #4

    Gokul43201

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    [tex]11=49-38[/tex]


    [tex]12=3*4*(9-8)[/tex]


    [tex]13=(3*4)+9-8[/tex]


    [tex]14=9+3+\frac{8}{4}[/tex]

    [tex]15=8+4+\frac{9}{3}[/tex]


    [tex]16=9+8-4+3[/tex]


    [tex]17=34-9-8[/tex]


    [tex]18=8+9+4-3[/tex]

    [tex]19=4!-8+\frac{9}{3}[/tex] (cheater !)

    [tex]20=(\frac{4}{3}*9)+8[/tex]


    Someone please fix 19 for me. :redface:
     
    Last edited: Dec 19, 2004
  6. Dec 19, 2004 #5
    [tex]19=(7\times3)-\frac{8}{4}[/tex]

    [tex]19=34-(8+7)[/tex]

    [tex]19=3^{7-4}-8[/tex]

    ------------------------------------------

    [tex]21=(8\times3)+4-7[/tex]

    [tex]22=(8\times4)-(7+3)[/tex]

    [tex]23=\frac{8}{4}+(7\times3)[/tex]

    [tex]24=7^3\bmod{4}*8[/tex]

    [tex]25=(7\times3)+8-4[/tex]

    [tex]26=\frac{38}{4}+7[/tex]

    [tex]27=38-(4+7)[/tex]

    [tex]28=43-(8+7)[/tex]

    [tex]29=(8+4)\times3-7[/tex]

    [tex]30=(7\times4)+8\bmod{3}[/tex]
     
    Last edited by a moderator: Dec 19, 2004
  7. Dec 19, 2004 #6
    Nice work guys. :smile: But that mod function is cheating. :grumpy:
     
  8. Dec 19, 2004 #7
    11 =49-38 =3*8 - 9 - 4 =3*4 +8 -9
    12= (9-3)(8/4)
    13= 3*4 +9 -8
    14= 9+3+8/4
    15= (9)(8/4)-3 = (3/4)*8+9 =9/3+4+8= 3*9 -4 -8
    16= 3+8+9-4
    17= 34-9-8=(4^3)/8 +9
    18= +4 +8 +9 -3 =(9-3)*4-8
    19= 3*8 +4 -9
    20 = (4/3)*9 +8
     
  9. Dec 19, 2004 #8
    He want's 9's not 7's
     
  10. Dec 19, 2004 #9

    dextercioby

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    [tex]41=9\cdot 4+8-3[/tex]
    [tex]42=43-9+8[/tex]
    [tex]44=43-8+9=8\cdot 4+9+3=(8+\frac{9}{3})\cdot 4[/tex]
    [tex]46=89-43[/tex]
    [tex]47=9\cdot 4+8+3[/tex]
    [tex]51=34+8+9[/tex]
    [tex]54=\frac{9\cdot 8\cdot 3}{4}=\frac{9\cdot 8}{.(4)\cdot 3}[/tex]
    [tex]55=89-34=98-43[/tex]
    [tex]56=4^{\frac{9}{3}}-8[/tex]
    [tex]59=8\cdot 4+9\cdot 3[/tex]
    [tex]60=43+8+9=8\cdot 9-3\cdot 4=8\cdot 3+9\cdot 4[/tex]
    [tex]65=8\cdot 9-4-3=98-34[/tex]
    [tex]71=8\cdot 9-4+3=8.(3)\cdot 9-4[/tex]
    [tex]72=\frac{4^{3}}{8}\cdot 9=4^{\frac{9}{3}}+8[/tex]
    [tex]73=8\cdot 9-3+4[/tex]
    [tex]76=(9\cdot 3-8)\cdot 4[/tex]
    [tex]79=8\cdot 9+3+4[/tex]
    [tex]81=9\cdot 3^{\frac{8}{4}}[/tex]
    [tex]82=89-3-4[/tex]
    [tex]84=8\cdot 9+3\dot 4[/tex]
    [tex]87=(\frac{3}{.(4)}\cdot 8)-9[/tex]
    [tex]88=89-4+3[/tex]
    [tex]90=89-3+4[/tex]
    [tex]96=89+3+4=\frac{9\cdot 8\cdot 4}{3}=.(4)\cdot 3\cdot 9\cdot 8=.(3)\cdot 4\cdot 8\cdot 9[/tex]
    [tex]97=98-4+3[/tex]
    [tex]99=98+4-3[/tex]
    [tex]100=\frac{3}{.(4)}\cdot 9-8[/tex]

    Daniel.

    Fill in the gaps
     
    Last edited: Dec 19, 2004
  11. Dec 19, 2004 #10
    I think we need to put in the (recurring) or some symbol that means that. I think Daniel's solution for 30, as written, is actually a solution for 32.

    No doubt one of you LaTeX gurus will know the correct formatting for a recurring decimal. :smile:
     
  12. Dec 19, 2004 #11

    dextercioby

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    I feel like a jackass. :mad: I erased the message by mistake. :cry:
    [tex]19=\frac{8}{.(3)}+4-9[/tex]
    [tex]20=\frac{9\cdot 8}{3}-4=8\cdot 4-9-3[/tex]
    [tex]21=\frac{8}{.(4)}+\frac{9}{3}[/tex]
    [tex]23=9\cdot 3-8+4=\frac{8}{.4}+\frac{9}{3}[/tex]
    [tex]24=\frac{8}{.(4)}+9-3[/tex]
    [tex]25=9\cdot 3-\frac{8}{4}[/tex]
    [tex]26=\frac{8}{.4}+9-3=8\cdot 4-9+3[/tex]
    [tex]28=\frac{9\cdot 8}{3}+4=\frac{9}{.3}-\frac{8}{4}[/tex]
    [tex]29=8\cdot 3+9-4[/tex]
    [tex]30=\frac{8}{.(4)}+9+3[/tex]
    [tex]31=9\cdot 3+8-4[/tex]
    [tex]32=\frac{8}{.4}+9+3=\frac{9}{.3}+\frac{8}{4}[/tex]

    [tex]33=34-9+8[/tex]
    [tex]35=34-8+9=8\cdot 4+\frac{9}{3}[/tex]
    [tex]37=8\cdot 3+9+4[/tex]
    [tex]38=8\cdot 4+9-3[/tex]
    [tex]39=9\cdot 3+8+4[/tex]

    Daniel.
    Fill in the gaps.
    PS.I may have not transcripted everything i had before. :grumpy: I'll be a guru next year.Guru of physics... :tongue2:
     
    Last edited: Dec 19, 2004
  13. Dec 19, 2004 #12
    oops? :frown:

    Btw, is this notation any good?

    [tex]19=\frac{8}{.\overline{3}} + 4 - 9[/tex]
     
  14. Dec 19, 2004 #13

    dextercioby

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    In my fifth grade (11 years ago :wink: ) i learnt that:
    [tex] 4,33333333333...=4.(3)=4\frac{3}{9}=4\frac{1}{3}=\frac{13}{3} [/tex]

    So,for me,it's obvious the notation.And besides why write ".\overline{3}" when u can easily put two brackets:".(3)"...?????? :tongue2:

    Daniel.
     
  15. Dec 19, 2004 #14

    dextercioby

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    Some more

    [tex]64=3^{4}-9-8[/tex]
    [tex]80=3^{4}-9+8[/tex]
    [tex]82=3^{4}-8+9[/tex]
    [tex]98=3^{4}+9+8[/tex]

    [tex]72=\frac{3^{4}}{9}\cdot 8[/tex]
    [tex]69=\frac{3}{.4}\cdot 8+9[/tex]
    [tex]51=\frac{3}{.4}\cdot 8-9[/tex]
    [tex]81=9^{4-\sqrt[3]{8}} [/tex] CHEATING!! :tongue2:

    Daniel.


    PS.Ceptimus,it's not fair!!Let us use at least [tex] \sqrt[3]{8} [/tex] or [tex] \sqrt{9};\sqrt{4} [/tex].Please... :cry:
     
    Last edited: Dec 19, 2004
  16. Dec 19, 2004 #15

    Galileo

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    Just for clarity. I've attempted to fill the gaps for 21-30.
    And I don't really like using the recurring decimal or decimal point. 22 and 30 are still missing non-decimal form.

    [tex]21=3(9-\frac{8}{4})[/tex]

    [tex]22=..? [/tex]

    [tex]23=(9)(3)+4-8[/tex]

    [tex]24=3+4+8+9[/tex]

    [tex]25=(3)(9)-\frac{8}{4}[/tex]

    [tex]26=(8)(4)+3-9[/tex]

    [tex]27=3^{4-(9-8))[/tex]

    [tex]28=\frac{9}{3}8+4[/tex]

    [tex]29=(3)(4)+8+9[/tex]

    [tex]30=..??[/tex]
     
  17. Dec 19, 2004 #16

    dextercioby

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    Since that 22 is really eataing me alive,i decided to take the sword and cut the Gordian knot.
    [tex]87=48+39=49+38[/tex]
    [tex]22=(\sqrt[3]{8}\cdot 9)+4=8\cdot 3-[\frac{9}{4}]=[\frac{9}{.(8)}]+4\cdot 3=(9+\sqrt[3]{8})\cdot\sqrt{4}=...[/tex]


    Daniel.
     
  18. Dec 19, 2004 #17

    Gokul43201

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    Ha ha...using the greatest integer function is one quick way to cut a Gordian Knot all right ! :rofl:
     
  19. Dec 19, 2004 #18
    [tex] 22 = 9\cdot 3 - 4/.8 [/tex]

    [tex] 30 = 4\cdot 8-3+.\overline{9} [/tex]

    [tex] 87 = 3 \cdot 4 \cdot 8 - 9 [/tex]

    [tex] 81 = \frac{(84 - 3)} {.\overline{9}} [/tex]

    [tex]101 = 89+ \frac{3} {.4} [/tex]
     
    Last edited: Dec 19, 2004
  20. Dec 20, 2004 #19
    [tex] 43 = \frac{43} {9-8} [/tex]

    [tex] 45 = 93 - 48 [/tex]

    [tex] 48 = (3+9) \cdot (8-4) [/tex]

    [tex] 49 = (3+4) \cdot (8-.\overline{9}) [/tex]

    [tex] 50 = 48+3-.\overline{9} [/tex]

    [tex] 52 = 48+3+.\overline{9} [/tex]

    [tex] 57 = (3+4) \cdot 8 + .\overline{9} [/tex]

    [tex] 63 = (3+8-4) \cdot 9 [/tex]

    [tex] 70 = \frac{3+4} {.9-.8} [/tex]

    [tex] 77 = 89 - \frac{4}{.\overline{3}} [/tex]

    [tex] 78 = 39 \cdot \frac{8}{4} [/tex]

    [tex] 86 = 98 - 3 \cdot 4 [/tex]

    [tex] 89 = \frac{89}{4-3} [/tex]

    [tex] 91 = 98 - 3 - 4[/tex]
     
    Last edited: Dec 20, 2004
  21. Dec 20, 2004 #20
    Someone required to play with 53 58 61 62 66 67 68 74 75 83 85 92 93 94 :-)
     
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