Solving for $3x+y+2z$ Given Integer Constraints

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Discussion Overview

The discussion revolves around solving for the expression $3x+y+2z$ under specific integer constraints. Participants explore the implications of the equations provided, particularly focusing on the conditions that $x, y, z$ must satisfy, including their relationships and the sum of their values.

Discussion Character

  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant presents the equations $x+y+z=-3$ and $x^3+y^3+z^3-20(x+3)(y+3)(z+3)=2013$ with the condition that $x, y, z \in Z$ and $z \ge y \ge x$.
  • Another participant questions the validity of the equation $x+y+z=-3$ when assuming $x, y, z \in N$, suggesting that it is contradictory since natural numbers cannot sum to a negative value.
  • A later reply acknowledges the mistake regarding the assumption of $x, y, z \in N$ and confirms that they are indeed integers, thanking the previous participant for the correction.

Areas of Agreement / Disagreement

There is a disagreement regarding the initial assumption of the variables being natural numbers, as one participant corrects this to integers. The discussion remains unresolved regarding the implications of the equations under the corrected assumptions.

Contextual Notes

The discussion highlights the importance of the definitions of the sets to which $x, y, z$ belong, as well as the implications of their relationships and the equations provided. The initial assumption of natural numbers leading to a negative sum is a critical point of contention.

anemone
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If $x,\,y,\,z \in Z$ and that $z\ge y \ge x$ and also,

$x+y+z=-3$

$x^3+y^3+z^3-20(x+3)(y+3)(z+3)=2013$

evaluate the value for $3x+y+2z$.
 
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anemone said:
If $x,\,y,\,z \in N$ and that $z\ge y \ge x$ and also,

$x+y+z=-3---(1)$

$x^3+y^3+z^3-20(x+3)(y+3)(z+3)=2013$

evaluate the value for $3x+y+2z$.
$x,y,z\in N,$
why $x+y+z=-3 ? $ in (1)
 
Albert said:
$x,y,z\in N,$
why $x+y+z=-3 ? $ in (1)

It was a mistake, sorry, Albert, you are right, the three of them are integers, and thanks for catching!:o
 
We have
$x+y + z = - 3 \cdots 1$
and $x^3+y^3 + z^3- 20(x+3)(y+3)(z+3)= 2013\cdots(2)$ from (1)
$x+y = -(3+z)\cdots (3)$
$y + z = -(3+x)\cdots (4)$
$z+x = -(3+y)\cdots(5) $

Now
$(x+y+z)^3 = x^3+y^3+z^+3(x+y)(y+z)(z+x)$

or $-27 = 2013 + 20(x+3)(y+3)(z+3) - 3(z+3)(x+3)(y+3)$ (LHS from (1) and RHS from (2), (3),(4),(5)

so $ 17(x+3)(y+3)(z+3) = -(2013+27)=-2040$

or $(x+3)(y+3)(z+3)= - 120$
further $x + 3 + y +3 + z + 3 = 6$
so we need 3 integers product is -120 and sum 6 and the numbers are 10,2,-6
so z = 7, y = -1, x = - 9
$3x + y + 2z = - 27- 1 + 14 = - 14$
 
kaliprasad said:
We have
$x+y + z = - 3 \cdots 1$
and $x^3+y^3 + z^3- 20(x+3)(y+3)(z+3)= 2013\cdots(2)$ from (1)
$x+y = -(3+z)\cdots (3)$
$y + z = -(3+x)\cdots (4)$
$z+x = -(3+y)\cdots(5) $

Now
$(x+y+z)^3 = x^3+y^3+z^+3(x+y)(y+z)(z+x)$

or $-27 = 2013 + 20(x+3)(y+3)(z+3) - 3(z+3)(x+3)(y+3)$ (LHS from (1) and RHS from (2), (3),(4),(5)

so $ 17(x+3)(y+3)(z+3) = -(2013+27)=-2040$

or $(x+3)(y+3)(z+3)= - 120$
further $x + 3 + y +3 + z + 3 = 6$
so we need 3 integers product is -120 and sum 6 and the numbers are 10,2,-6
so z = 7, y = -1, x = - 9
$3x + y + 2z = - 27- 1 + 14 = - 14$

Very well done, kaliprasad!:cool:
 

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