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    Maths problem

    Whats is the greatest number that can be achieved using these three numbers

    1, 2, 3

    I am almost certain a number greater than 6 can be achieved, I'm thinking the square root divided by one of the numbers, but cant work it out

    54 Comments

    3+1 x 2

    1+3x2=8

    3/(1/(2^2)) = 12

    2/(1/(3^2)) = 18

    321 Lol!

    kungfu;8159980

    3+1 x 2



    = 8

    (2+1) x 3 = 9.

    No idea if anything over that though.

    kungfu;8159980

    3+1 x 2



    +1 :thumbsup:

    2+1x3=9

    321 add up:-D

    Original Poster

    r3tract;8160005

    3/(sqrt(1/2)) = 12



    Is that 3 divided by (1 divided by 2 sqrt)

    2 divided by the square root of 1/3 =18?

    3 to the power of 2 +1?

    Original Poster

    choc1969;8160011

    2+1x3=9



    Thanks babes x

    Original Poster

    chesso;8160023

    2 divided by the square root of 1/3 =18?



    Thats 2 sums on one number, if you get what I mean.

    You can have 1 divided by 3 to get 1/3 and then squareroot the result, you are only allowed one sum per number, else I could just keep adding 3 to 3 until I feel like I want to stop

    Original Poster

    awoodhall2003;8160027

    3 to the power of 2 +1?



    thats 10

    loll1es;8160052

    thats 10



    thus a greater number then 6.

    3 ^ (2+1) = 27

    Original Poster

    awoodhall2003;8160062

    thus a greater number then 6.



    yeah but there are already results up to 18 (might not be correct though)

    linuxlinks;8160068

    3 ^ (2+1) = 27


    Yep, that's the best you can get I think

    2+1 = 3
    3^3
    = 27

    linuxlinks;8160068

    3 ^ (2+1) = 27



    +1

    not 27+1 lol but 27 is the highest I believe.

    loll1es;8160072

    yeah but there are already results up to 18 (might not be correct though)


    Sorry couldnt see the others when i typed it.

    linuxlinks;8160068

    3 ^ (2+1) = 27



    baffledsalmon;8160085

    2+1 = 33^3= 27


    as above...

    But...2 to the power of (1+3)

    awoodhall2003;8160098

    But...2 to the power of (1+3)



    That's a mere 16 though

    Original Poster

    baffledsalmon;8160085

    2+1 = 33^3= 27



    what is ^ mean? :?

    loll1es;8160155

    what is ^ mean? :?


    to the power of

    linuxlinks;8160117

    That's a mere 16 though


    throwing alternatives, the highest I agree with, i cant think of anything better.
    loll1es;8160155

    what is ^ mean? :?


    to the power of.

    linuxlinks;8160117

    That's a mere 16 though



    yep.. and 27 is greater than 16 last time I checked too. Unless its a trick question... 27 is correct.

    Original Poster

    I think 27 is the highest

    5nowman;8160165

    yep.. and 27 is greater than 16 last time I checked too. Unless its a … yep.. and 27 is greater than 16 last time I checked too. Unless its a trick question... 27 is correct.



    Yeah, such as if we were allowed to use ! (factorial

    31 to the power of 2 = 961

    Does this count? :thinking:

    ScreechMF;8160185

    31 to the power of 2 = 961Does this count? :thinking:



    No

    If that's allowed then 21 to the power of 3=9261

    Original Poster

    ScreechMF;8160185

    31 to the power of 2 = 961Does this count? :thinking:



    Nope

    (3^(2+1))! = 10888869450418352160768000000

    Quite a resounding victory I think!

    Original Poster

    InfernoZeus;8160225

    (3^(2+1))! = 10888869450418352160768000000Quite a resounding victory I … (3^(2+1))! = 10888869450418352160768000000Quite a resounding victory I think!



    Think your calculator is broke, cos mine makes that 27 :lol:

    loll1es;8160241

    Think your calculator is broke, cos mine makes that 27 :lol:


    Then you forgot to use the factorial button :w00t:

    (3^(2+1))! = 27! = 10888869450418352160768000000

    ! isn't permitted

    linuxlinks;8160274

    ! isn't permitted



    Why not? :?

    Original Poster

    InfernoZeus;8160261

    Then you forgot to use the factorial button :w00t:(3^(2+1))! = 27! = … Then you forgot to use the factorial button :w00t:(3^(2+1))! = 27! = 10888869450418352160768000000



    thats using 2 sums on one number, not permitted, sorry

    Original Poster

    InfernoZeus;8160279

    Why not? :?



    cos I said so, and I make the rules.....ha ha!!

    loll1es;8160306

    thats using 2 sums on one number, not permitted, sorry


    Not really, I make one sumation of 1 and 2, then the product of the result with 3, and then the factorial of the product.

    No different to summing 1 and 2, and then raising the result to the power 3.
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