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Suppose that \(x, y, z\) are positive integers satisfying \(x \le y \le z\), and such that the product of all three numbers is twice their sum. What is the sum of all possible values of \(z\)?

 Sep 24, 2018
 #1
avatar+5224 
+2

I'm only seeing 3 triplets of integers that meet the specifications

 

(1, 3, 8), (1, 4, 5), (2, 2, 4)

 

Thus the sum asked for is

 

8 + 5 + 4 = 17

 Sep 24, 2018
 #5
avatar
+1

How  did you see the three triplets without shaking and baking like heureka did?

Guest Sep 26, 2018
 #2
avatar+22489 
+10

Suppose that \(x,y,z\) are positive integers satisfying \(x \le y \le z\) ,
and such that the product of all three numbers is twice their sum.
What is the sum of all possible values of \(z\)?

 

\(\mathbf{z=\ ?}\)

\(\begin{array}{|rcll|} \hline 2(x+y+z) &=& xyz \\ 2(x+y)+2z &=& xyz \\ xyz-2z &=& 2(x+y) \\ z(xy-2) &=& 2(x+y) \\ \mathbf{z} &\mathbf{=} & \mathbf{ \dfrac{2(x+y)}{xy-2} }\\ \hline \end{array}\)

 

\(\mathbf{x \le y \le z}\)

\(\begin{array}{|ll|} \hline x \le y \le \dfrac{2(x+y)}{xy-2} \quad & | \quad x,y,z \gt 0 \\ & \small{\text{The denominator must be greather than zero:}} \\ & \begin{array}{|rcl|} \hline xy-2 &>& 0 \\ xy &>& 2 \\ \mathbf{y} & \mathbf{>} & \mathbf{\dfrac{2}{x}} \quad\text{ or } \quad \mathbf{\dfrac{2}{x} < y }\\ \hline \end{array}\\ \dfrac{2}{x} \lt y \le \dfrac{2(x+y)}{xy-2} \\\\ & \begin{array}{|rcll|} \hline y &\le& \dfrac{2(x+y)}{xy-2} \quad & | \quad \cdot (xy-2) \\ y(xy-2) &\le& 2(x+y) \\ xy^2 -2y &\le& 2x+2y \\ xy^2 -4y -2x &\le& 0 \\ xy^2 -4y -2x &=& 0 \\\\ y &=& \dfrac{4\pm \sqrt{16-4x(-2x)} }{2x} \\ y &=& \dfrac{4\pm \sqrt{16+8x^2} }{2x} \\ y &=& \dfrac{4\pm \sqrt{4(4+2x^2)} }{2x} \\ y &=& \dfrac{4\pm 2\sqrt{ 4+2x^2 } }{2x} \\ y &=& \dfrac{4}{2x} + \dfrac{2}{2x} \sqrt{ 4+2x^2 } \quad & | \quad y > 0 ! \\ \mathbf{ y } & \mathbf{=} & \mathbf{\dfrac{2}{x} + \dfrac{\sqrt{ 4+2x^2 }}{x} } \\ \hline \end{array}\\ \dfrac{2}{x} \lt y \le \dfrac{2}{x} + \dfrac{\sqrt{ 4+2x^2 }}{x} \\ \hline \end{array} \)

 

\(\mathbf{x \text{ and } y =\ ?}\)

\(\begin{array}{|lrcll|} \hline \mathbf{x=1}: & \dfrac{2}{1} \lt &y& \le \dfrac{2}{1} + \dfrac{\sqrt{ 4+2\cdot 1^2 }}{1} \\\\ & 2 \lt &y& \le 2 + \sqrt{ 6 } \\ & 2 \lt &y& \le 2 + 2.4\ldots \\ & 2 \lt &y& \le 4 \\ & && \Rightarrow 2 \lt 3 \le 4 & \mathbf{y = 3} \\ & && \Rightarrow 2 \lt 4 \le 4 & \mathbf{y = 4} \\\\ \mathbf{x=2}: & \dfrac{2}{2} \lt &y& \le \dfrac{2}{2} + \dfrac{\sqrt{ 4+2\cdot 2^2 }}{2} \\\\ & 1 \lt &y& \le 1 + \dfrac{ \sqrt{ 12 } }{2} \\ & 1 \lt &y& \le 1 + 1.7\ldots \\ & 1 \lt &y& \le 2 \\ & && \Rightarrow 1 \lt 2 \le 2 & \mathbf{y = 2} \\\\ \mathbf{x=3}: & \dfrac{2}{3} \lt &y& \le \dfrac{2}{3} + \dfrac{\sqrt{ 4+2\cdot 3^2 }}{3} \\\\ & \dfrac{2}{3} \lt &y& \le \dfrac{2}{3} + \dfrac{ \sqrt{ 22 }}{3} \\ & \dfrac{2}{3} \lt &y& \le \dfrac{2}{3} + 1.5\ldots \\ & \dfrac{2}{3} \lt &y& \le 2.23013858661 \\ & \dfrac{2}{3} \lt &y& \le 2 \\ & && \Rightarrow \dfrac{2}{3} \lt 1 \le 3 & \mathbf{y = 1} \\ & && \Rightarrow \dfrac{2}{3} \lt 2 \le 3 & \mathbf{y = 2} \\\\ \mathbf{x=4}: & \dfrac{2}{4} \lt &y& \le \dfrac{2}{4} + \dfrac{\sqrt{ 4+2\cdot 4^2 }}{4} \\\\ & \dfrac{1}{2} \lt &y& \le \dfrac{1}{2} + \dfrac{ \sqrt{ 36 }}{4} \\ & \dfrac{1}{2} \lt &y& \le \dfrac{1}{2} + 1.5 \\ & \dfrac{1}{2} \lt &y& \le 2 \\ & && \Rightarrow \dfrac{1}{2} \lt 1 \le 2 & \mathbf{y = 1} \\ & && \Rightarrow \dfrac{1}{2} \lt 2 \le 2 & \mathbf{y = 2} \\\\ \mathbf{x\gt 4}:\\ \lim \limits_{x\to \infty} \frac{2}{x} + \dfrac{\sqrt{ 4+2x^2 }}{x} \\\\ = \lim \limits_{x\to \infty} \frac{2}{x} + \sqrt{ \frac{4}{x^2}+2 } = \sqrt{2} \\ &&& \Rightarrow & \mathbf{y = 2} \\ &&& \Rightarrow & \mathbf{y = 1} \\ \hline \end{array}\)

 

Solution x,y,z:

\(\begin{array}{|r|r|r|c|} \hline x & y\ge x & z=\dfrac{2(x+y)}{xy-2} & \text{solution} \\ \hline 1 & 3 & 8 & \checkmark \\ & 4 & 5 & \checkmark \\ \hline 2 & 2 & 4 & \checkmark \\ \hline & & \mathbf{\text{sum }z = 17} \\ \hline \end{array}\)

 

 

laugh

 Sep 25, 2018
edited by heureka  Sep 26, 2018
 #3
avatar+102441 
+1

Very impressive Heureka :)

Melody  Sep 25, 2018
 #4
avatar+22489 
+9

Thank you, Melody.

 

laugh

heureka  Sep 26, 2018

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