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Tricky math problem |
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| May22-09, 02:53 PM | #1 |
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Tricky math problem
A friend of mine showed this to me a few days ago:
x+y+z=0 x3+y3+z3=3 x5+y5+z5=15 x2+y2+z2=? An answer must be supported with justification (so that I know that you didn't guess and get lucky). |
| May22-09, 11:17 PM | #2 |
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I'm going to ask my math teacher :)
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| May26-09, 09:47 AM | #3 |
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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. |
| May28-09, 03:30 PM | #4 |
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Tricky math problem |
| May29-09, 03:20 AM | #5 |
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How do they solve it ?
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| Jun9-09, 01:18 PM | #6 |
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I got the answer but i cantīt get the values os x, y and z, but i got that x.y.z =
Spoiler
x.y.z=1
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| Jun10-09, 11:20 AM | #7 |
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Ok...that was prety hard to solve...it took many hours. I hope you have an easier way to get the answer (tricky):
Spoiler
x= 2 cos (20)
y= 2 cos (140) z= 2 cos (260) the answer to the question is 6 |
| Jun10-09, 11:24 AM | #8 |
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At a glance:
Spoiler
x = -y-z
x<2 and positive y not equal z y and z are negative x.y.z = 1 |
| Jun10-09, 03:08 PM | #9 |
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OK, I got it...itīs really simple to get the answer to x2+y2+z2=?....
Spoiler
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 Thatīs the tricky way... |
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