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Topic: solve (Read 494 times) |
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tony123
Junior Member
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x+y+z=1
x^2+y^2+z^2=21
x^3+y^3+z^3=55
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Sameer
Uberpuzzler
Pie = pi * e
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Re: solve
« Reply #1 on: Jun 27th, 2007, 10:31pm » |
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er... are we doing your homework?
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"Obvious" is the most dangerous word in mathematics.
--Bell, Eric Temple
Proof is an idol before which the mathematician tortures himself.
Sir Arthur Eddington, quoted in Bridges to Infinity
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Grimbal
wu::riddles Moderator
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Re: solve
« Reply #2 on: Jun 28th, 2007, 4:51am » |
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With the simple assumption that x, y, z are integers, it becomes quite easy.
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K Sengupta
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Re: solve
« Reply #3 on: Jun 28th, 2007, 6:34pm » |
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(x, y, z) = (-1, 4, -2); (-1, 2, -4); (-2, -1, 4); (-2, 4, -1); (4, -1, -2), (4, -2, -1) gives all possible solutions to the problem under reference
EXPLANATION:
For the sake of simplicity, let us take (m, n) = (xy+yz+ zx, xyz)
Then, m = (1-21)/2 = -10
Also, 55- 3n = (1)*(21+10), giving: n = 8
But, xy+yz+zx = -10
Or, 8/x + x(1-x) = -10
Or, 8+ x^2 – x^3 = -10x
Or, x^3-x^2-10x – 8 = 0
or, (x+1)(x+2)(x-4) = 0
Or, x = -1, -2, 4
Substituting x=-1, we obtain (y+z, yz) = (2, -8) giving: (y, z) = (4, -2); (2, -4)
Similarly, for x=-2, we get (y, z) = (-1, 4); (4, -1)
For, x = 4, we obtain: (y, z) = (-1, -2); (-2, -1)
Thus, (x, y, z) = (-1, 4, -2); (-1, 2, -4); (-2, -1, 4); (-2, 4, -1); (4, -1, -2); (4, -2, -1) gives all possible solutions to the problem under reference.
** In my opinion the fundamental tenets inclusive of the puzzle under reference are rather simple, and accordingly, this problem should have been posted in the "Easy" Forum.
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| « Last Edit: Jun 28th, 2007, 6:38pm by K Sengupta » |
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K Sengupta
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Re: solve
« Reply #4 on: Jun 28th, 2007, 6:46pm » |
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on Jun 28th, 2007, 4:51am, Grimbal wrote:| With the simple assumption that x, y, z are integers, it becomes quite easy. |
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True.
In conformity with the previous post:
xyz = 8 and by conditions of the problem x+y+z =1; x^2+ y^2+ z^2 = 21
Since we know that the factors of 8 are -1, -2, -4, -8, 1, 2, 4, 8; the problem reduces to determining just three of the factors that sum to 1 and whose sum of squares is 21.
So, by a little trial and error, | hidden: | | (x, y, z) = (-1, 4, -2); (-1, 2, -4); (-2, -1, 4); (-2, 4, -1); (4, -1, -2); (4, -2, -1) |
corresponds to all possible solutions to the given problem.
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| « Last Edit: Jun 28th, 2007, 7:05pm by K Sengupta » |
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