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Simple maths quiz

  1. Dec 18, 2004 #1
    simple I think not - my son has homework/quiz for fun - try to make numbers from 1 to 50 using the numbers 2,5,1,2 (xmas date)- I think all of them must be used.

    for example

    25+21=46

    (5+1)2-2=34 five plus one squared minus two equals 34

    15x2+2=32


    weve got some we cant do even usng factorials(dbl/trpl) - decimals and fractions

    the ones we cant do are 41/42/43

    any help with these appreciated
     
  2. jcsd
  3. Dec 19, 2004 #2
    5-(2)(2)(1)=1
    5-[2+(2-1)]=2
    5-[(2/2)+1]=3
    (5-1)+2-2=4
    12-(5+2)=5
    5+(2/2)(1)=6
    5+(2/2)+1=7
    5+2+(2-1)=8
    5+(2+2)1=9
    5+2+2+1=10
    5(2)+(2-1)=11
    5(2)+2(1)=12
     
  4. Dec 19, 2004 #3
    are you allowed to use trig functions? not that it would necessarily help
    in fact i doubt it would
     
  5. Dec 19, 2004 #4
    im assuming they cannot be used more than once because you could just say:
    5*2*2*2+1
     
  6. Dec 19, 2004 #5
    hahaha [5!/(2+1)]+2=42
     
  7. Dec 19, 2004 #6
    lol if it could be used more than once u can just do 1^5-1^2+1.... all the way to whatever and making all numbers you could posibbly want.
     
  8. Dec 19, 2004 #7
    No lol they cant be used more than once - posted this on other sites - maths goodies did best we just need number 41 now and we have one for number 36 that we dont understand but we hope the tutor will .

    (((1+5/2)!)^2 = 36
     
  9. Dec 19, 2004 #8
    [tex]36 = (12 \times .5)^2[/tex]

    [tex]41 = \left(\frac{2}{.1} + .5\right) \times 2[/tex]
     
  10. Dec 19, 2004 #9
    [tex]51 = 52 + 1 - 2[/tex]

    [tex]52 = (25 + 1) \times 2[/tex]

    [tex]53 = 52 + 2 -1[/tex]

    [tex]54 = 52 + 2 \times 1[/tex]

    [tex]55 = 52 + 2 + 1[/tex]

    [tex]56 = \frac{\frac{2}{.5(recurring)} + 2}{.1}[/tex]

    [tex]57 = 52 + \frac{1}{.2}[/tex]

    [tex]58 = 5 \times 12 - 2[/tex]

    [tex]59 = 5 \times (12 - .2)[/tex]

    [tex]60 = 15 \times 2 \times 2[/tex]

    :tongue2:
     
    Last edited: Dec 19, 2004
  11. Dec 19, 2004 #10

    dextercioby

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    Good one,ceptimus...It's cheating............... :cry: It sould have been "0" before the dot. :tongue2: There's my share of cheating:
    [tex]36=[(5-2)!]^{2\cdot1} [/tex]
    [tex]41=[5-(2-2)!]1 [/tex]

    Daniel.

    PS.In the second line,there is no multiplication between 4 and 1. :tongue2:
    EDIT:[tex] 41=\frac{5}{.(1)}-(2\cdot2) [/tex]
     
    Last edited: Dec 19, 2004
  12. Dec 19, 2004 #11
    I couldn't find answers for 69, 79, 83 and 87 but otherwise I have all the numbers up to 113.

    Only functions I've used are +, -, *, /, ^, the decimal point, and (recurring)
     
  13. Dec 19, 2004 #12
    OK then. Without the 0. 'cheat'

    [tex]36 = 12 \times (5 - 2)[/tex] D'Oh!
     
  14. Dec 19, 2004 #13

    dextercioby

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    [tex] 69=(5+2)1-2 [/tex]

    Daniel.
     
  15. Dec 19, 2004 #14
    You're not seriously expecting anyone to accept that [tex]71 = (5+2)1[/tex] is an acceptable notation are you? :yuck: :rofl:
     
    Last edited: Dec 19, 2004
  16. Dec 19, 2004 #15
    These are the answers we came up with mostly by ourselves and with help from this forum - while we have checked these they may not all be correct - tutor will mark this week. Number 44 was a double factorial and we hope this is legal (as well as the .5 in number 41 lol). The alternatives where afterthoughts while checking when typing.

    1- 5-(2+2)x1
    2- 5-(2+2)+1
    3- 5-2x1^2
    4- 5-(1+2)+2
    5- (5x1)+2-2
    6- 5+1+2-2
    7- 5+2x1^2
    8- 5+1^2+2
    9- (5+2^2)x1
    10- 5+2+2+1
    11- 5+(2x(2+1))
    12- (2x5)+2x1
    13- (2x5)+2+1
    14- (5+1)x2+2
    15- (5+2)x2+1
    16- (5+2+1)x2
    17- (2+1)x5x2
    18- ((2x2)!)-5-1
    19- 5x(2+2)-1
    20- 5x(2+2)x1
    21- 5x(2+2)+1
    22- ((5x2)+1)x2
    23- 5^2-2x1
    24- (2x2)x(5+1)
    25- 5x(2+2+1)
    26- 5^2-2+1
    27- 25+2x1
    28- 25+2+1
    29- (((2+2)!)+5)x1
    30- (5x(2+1))x2
    31- ((2+5)!!!)+2+1 alternative ((5!)/(2+2))+1
    32- 15x2+2 alternative((5-1)^2)x2
    33- 2^5+2-1
    34- (5+1)^2-2
    35- 2^5+2+1
    36- (((1+5)/2)!)^2
    37- 21x2-5
    38- (5+1)^2+2
    39- 5!/2-21
    40- 2^(2+1)x5
    41- (21-.5)x2
    42- (5+2)x((2+1)!)
    43- 2x21.5
    44- ((5+1)!!)-2-2
    45- (2+1)^2x5
    46- ((5-1)!)*2-2
    47- 21x2+5
    48- (5+2)^2-1
    49- 25x2-1
    50 25x2x1

    alternative to 41/44 welcome just in case
    Many thanks to all who helped and a Happy Xmas and New Year to all from here in the UK.
     
  17. Dec 19, 2004 #16
    44 = 2^5 + 12

    see other answers above.
     
  18. Dec 19, 2004 #17

    dextercioby

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    I would call that an identity.At least in the decimal basis... :approve:
    Read and weep:
    [tex]41=[(2+1)!]^{2} +5 [/tex] :tongue2:

    Daniel.
     
  19. Dec 19, 2004 #18
    beautifully done dexter
     
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