Difference between revisions of "Mock AIME 6 2006-2007 Problems/Problem 8"

Revision as of 16:52, 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.

~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.

@@ Line 10: / Line 10: @@
 <math>a_2=\frac{a_1}{a_0}=\frac{y}{x}</math>
-<math>a_3=\frac{a_2}{a_1}=\frac{\frac{y}{x}}{y}=\frac{y}{x}</math>
+<math>a_3=\frac{a_2}{a_1}=\frac{\frac{y}{x}}{y}=\frac{1}{x}</math>
+<math>a_4=\frac{a_3}{a_2}=\frac{\frac{1}{x}}{\frac{y}{x}}=\frac{1}{y}</math>
+<math>a_5=\frac{a_4}{a_3}=\frac{\frac{1}{y}}{\frac{1}{x}}=\frac{x}{y}</math>
+<math>a_6=\frac{a_5}{a_4}=\frac{\frac{x}{y}}{\frac{1}{y}}=x=a_0</math>
+<math>a_7=\frac{a_6}{a_5}=\frac{x}{\frac{x}{y}}=y=a_1</math>
+And the sequence repeats every 6 steps.

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Difference between revisions of "Mock AIME 6 2006-2007 Problems/Problem 8"

Revision as of 16:52, 26 November 2023

Problem

Solution