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Replies
  • raj87verma88

    MemberMay 2, 2008

    Rather difficult
    But let me give it a shot

    cos-1 (inverse) 0 = 90
    √90 = 9.5
    and 3√90 = 4.5
    √cos (inverse) 0 + √cos (inverse) 0 + √cos (inverse) 0 -3√cos (inverse) 0
    = √90 + √90 + √90 - 3√90
    = 9.48 +9.48 + 9.48 – 4.48
    = 28.44 – 4.48
    = 23.9

    after taking square root and cube root if we approximate

    that is

    9.5 + 9.5 + 9.5 – 4.5
    then answer is exactly 24

    How about it Biggie
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  • raj87verma88

    MemberMay 2, 2008

    Also instead of cube root of 90 we can do
    ln90 = 4.49
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  • Kaustubh Katdare

    AdministratorMay 2, 2008

    Hint -

    K.I.S.S 😁

    ...and since when did we start relying on 'approximate' values?

    On the same lines, I've another question -> is 0.999999...(infinite times) equal to 1 ?

    If yes, how? If not, why? 😉
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  • devesh

    MemberMay 4, 2008

    do we only have to use 0's?
    I mean nothing else(no other numerics),only operators and zeroes!!!
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  • Kaustubh Katdare

    AdministratorMay 4, 2008

    Yes, that's right. Only four zeros! (and of course, mathematical operators).

    Is it really 'that' difficult?
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  • gohm

    MemberMay 4, 2008

    Are we allowed to rotate the operators?
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  • gohm

    MemberMay 4, 2008

    Nope, it is never equal to 1. Only 1 is equal to 1. In most practical uses it is close enough mathematically to round the .99999 to 1 however it is never actually equal to 1.

    The_Big_K
    Hint -

    K.I.S.S 😁

    ...and since when did we start relying on 'approximate' values?

    On the same lines, I've another question -> is 0.999999...(infinite times) equal to 1 ?

    If yes, how? If not, why? 😉
    Are you sure? This action cannot be undone.
    Cancel
  • Kaustubh Katdare

    AdministratorMay 4, 2008

    1. No, you cannot 'rotate' operators. Come on, its just basic mathematics!

    2. If 0.99999...(infinite) is not equal to one, what's wrong with the following? -

    x = 0.99999...(infinite times)
    10x = 9.9999.......

    => 9x = 9

    => x = 1

    😁

    What say? 😉
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  • gohm

    MemberMay 4, 2008

    Here is your problem, the tricky business between the 10x=9.999999... and then 9x=9. Did you change the value of x?

    The_Big_K
    1. No, you cannot 'rotate' operators. Come on, its just basic mathematics!

    2. If 0.99999...(infinite) is not equal to one, what's wrong with the following? -

    x = 0.99999...(infinite times)
    10x = 9.9999.......

    => 9x = 9

    => x = 1

    😁

    What say? 😉
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  • Kaustubh Katdare

    AdministratorMay 4, 2008

    gohm
    Here is your problem, the tricky business between the 10x=9.999999... and then 9x=9. Did you change the value of x?

    Umm, okay, here we go -

    x = 0.99999...(infinite times) - Equation (A)
    10x = 9.9999....... - Equation (B)

    Equation (B) minus (A)

    => 9x = 9

    => x = 1

    👍
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  • mahul

    MemberMay 4, 2008

    Four zeroes to score 24, that's easy-->>

    (0!+0!+0!+0!)!=24

    i hope that would do, unless _k has a problem with my repeated use of factorials 😀
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  • Kaustubh Katdare

    AdministratorMay 4, 2008

    mahul
    Four zeroes to score 24, that's easy-->>

    (0!+0!+0!+0!)!=24

    i hope that would do, unless _k has a problem with my repeated use of factorials 😀
    Hah, no problem with factorials! That's the right way to get 24. Pretty easy, eh? 😀
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