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Difference between revisions of "Mock AIME 6 2006-2007 Problems/Problem 8"
 
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Then, <math>(x+y)^2=81</math> which gives <math>x^2+2xy+y^2=81</math> which gives <math>x^2+y^2=81-2xy=81-(2)(3)=75</math>
 
Then, <math>(x+y)^2=81</math> which gives <math>x^2+2xy+y^2=81</math> which gives <math>x^2+y^2=81-2xy=81-(2)(3)=75</math>
  
 +
Finally, <math>x^3+y^3=(x+y)(x^2-xy+y^2)=(9)(75-3)=\boxed{648}</math>
  
 
~Tomas Diaz. orders@tomasdiaz.com
 
~Tomas Diaz. orders@tomasdiaz.com
 +
 
{{alternate solutions}}
 
{{alternate solutions}}

Latest revision as of 17:06, 26 November 2023

Problem

A sequence of positive reals defined by $a_0=x$, $a_1=y$, and $a_n\cdot a_{n+2}=a_{n+1}$ for all integers $n\ge 0$. Given that $a_{2007}+a_{2008}=3$ and $a_{2007}\cdot a_{2008}=\frac 13$, find $x^3+y^3$.

Solution

$a_0=x$

$a_1=y$

$a_2=\frac{a_1}{a_0}=\frac{y}{x}$

$a_3=\frac{a_2}{a_1}=\frac{\frac{y}{x}}{y}=\frac{1}{x}$

$a_4=\frac{a_3}{a_2}=\frac{\frac{1}{x}}{\frac{y}{x}}=\frac{1}{y}$

$a_5=\frac{a_4}{a_3}=\frac{\frac{1}{y}}{\frac{1}{x}}=\frac{x}{y}$

$a_6=\frac{a_5}{a_4}=\frac{\frac{x}{y}}{\frac{1}{y}}=x=a_0$

$a_7=\frac{a_6}{a_5}=\frac{x}{\frac{x}{y}}=y=a_1$

And the sequence repeats every 6 steps.

Therefore,

$a_n=a_{n\;mod\;6}$

Since, $2007 \equiv 3\;(mod\;6)$ and $2008 \equiv 4\;(mod\;6)$, then $a_{2007}=a_3=\frac{1}{x}$, and $a_{2008}=a_4=\frac{1}{y}$

From $a_{2007}\cdot a_{2008}=\frac 13$, we get $\frac{1}{xy}=\frac{1}{3}$, thus $xy=3$

and from $a_{2007}+a_{2008}=3$, we get $\frac{1}{x}+\frac{1}{y}=3$.

Therefore, $\frac{x+y}{xy}=3$ which gives $x+y=3xy=(3)(3)=9$

Then, $(x+y)^2=81$ which gives $x^2+2xy+y^2=81$ which gives $x^2+y^2=81-2xy=81-(2)(3)=75$

Finally, $x^3+y^3=(x+y)(x^2-xy+y^2)=(9)(75-3)=\boxed{648}$

~Tomas Diaz. orders@tomasdiaz.com

Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

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