Tricky math problem

  1. A friend of mine showed this to me a few days ago:



    An answer must be supported with justification (so that I know that you didn't guess and get lucky).
  2. jcsd
  3. I'm going to ask my math teacher :)

  4. Unless you can demonstrate that there is a clever shortcut to this problem which would raise it to the level of a brainteaser, this is just a set of simultaneous equations which is easily solvable with, for example, the Wolfram Alpha website.
  5. I had no idea that such a site even existed.
  6. How do they solve it ?
  7. I got the answer but i cant´t get the values os x, y and z, but i got that x.y.z =
  8. Ok...that was prety hard to took many hours. I hope you have an easier way to get the answer (tricky):
    x= 2 cos (20)
    y= 2 cos (140)
    z= 2 cos (260)
    the answer to the question is 6
  9. At a glance:

    x = -y-z
    x<2 and positive
    y not equal z
    y and z are negative
    x.y.z = 1
  10. OK, I got´s really simple to get the answer to x2+y2+z2=?....
    the answer is 6. and you just have to set values to x (positive) y(negative) and z(negative) and calcutale the proportion like: set x=3 y=-2 z=-1, then
    ^2 = 14 --> ?
    ^3 = 18 --> 3
    ^5 = 210 --> 15

    so 14 . 18 . 15 = ? . 3 . 210
    ? = 6

    That´s the tricky way...
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