Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "Northeastern WOOTers Mock AIME I Problems/Problem 14"

(Created page with "== Problem 14 == Consider three infinite sequences of real numbers: <cmath> \begin{eqnarray*} X &=& \left( x_1, x_2, \cdots \right), \\ Y &=& \left( y_1, y_2, \cdots \right),...")
 
m (→‎Solution)
 
(4 intermediate revisions by the same user not shown)
Line 13: Line 13:
  
 
<cmath>\begin{align*} \left( \left( \log_2 x_n \right)^2 + \left( \log_2 y_n \right)^2 \right) \cdot \left( \left( \log_2 y_n \right)^2 + \left( \log_2 z_n \right)^2 \right) &= \left( \log_2 x_n \log_2 y_n + \log_2 y_n \log_2 z_n \right)^2 \\
 
<cmath>\begin{align*} \left( \left( \log_2 x_n \right)^2 + \left( \log_2 y_n \right)^2 \right) \cdot \left( \left( \log_2 y_n \right)^2 + \left( \log_2 z_n \right)^2 \right) &= \left( \log_2 x_n \log_2 y_n + \log_2 y_n \log_2 z_n \right)^2 \\
−
\left( \log_2 y_n \right)^4 + \left( \log_2 x_n \log_2 z_n \right)^2 &= 2 \left( \log_2 y_n \right)^2 \left( \log_2 x_n \log_2 z_n \right) \end{align*} </cmath>
+
\left( \log_2 y_n \right)^4 + \left( \log_2 x_n \log_2 z_n \right)^2 &= 2 \left( \log_2 y_n \right)^2 \left( \log_2 x_n \log_2 z_n \right) \\
 +
\left( \left( \log_2 y_n \right)^2 - \log_2 x_n \log_2 z_n \right)^2 &= 0 \\
 +
\left( \log_2 y_n \right)^2 &= \log_2 x_n \log_2 z_n \end{align*} </cmath>
 +
 
 +
Then the sum becomes:
 +
 
 +
<cmath>\begin{align*} S &= \sum_{n=1}^{\infty} \log_2 x_n \log_2 y_n \log_2 z_n \\
 +
S &= \sum_{n=1}^{\infty} \frac{n^3}{8^n} \end{align*} </cmath>
 +
 
 +
The final step is to take iterative differences, like so:
 +
 
 +
<cmath>\begin{align*} S &= \frac{1}{8} + \frac{8}{64} + \frac{27}{512} + \frac{64}{4096} + \cdots \\
 +
\frac{1}{8}S &= \quad \; \; \; \, \frac{1}{64} + \frac{8}{512} + \frac{27}{4096} + \cdots \\
 +
\Rightarrow \frac{7}{8}S &= \frac{1}{8} + \frac{7}{64} + \frac{19}{512} + \frac{37}{4096} + \cdots \\
 +
\frac{7}{64}S &= \quad \; \; \; \, \frac{1}{64} + \frac{7}{512} + \frac{19}{4096} + \cdots \\
 +
\Rightarrow \frac{49}{64}S &= \frac{1}{8} + \frac{6}{64} + \frac{12}{512} + \frac{18}{4096} + \cdots \\
 +
\frac{49}{512}S &= \quad \; \; \; \, \frac{1}{64} + \frac{6}{512} + \frac{12}{4096} + \cdots \\
 +
\Rightarrow \frac{343}{512}S &= \frac{1}{8} + \frac{5}{64} + \frac{6}{512} + \frac{6}{4096} + \cdots \end{align*} </cmath>
 +
 
 +
Almost done. Now that the numerators are constants <math>(6)</math> instead of cubics <math>(n^3)</math>, we can apply the formula for the sum of an infinite geometric series to get:
 +
 
 +
<cmath>\begin{align*} \frac{343}{512}S &= \frac{1}{8} + \frac{5}{64} + \frac{6}{64 \cdot 7} \\
 +
\frac{343}{512}S &= \frac{97}{64 \cdot 7} \\
 +
S &= \frac{776}{2401} \end{align*} </cmath>
 +
 
 +
Then our answer is <math>\boxed{776}</math>.

Latest revision as of 17:23, 8 August 2021

Problem 14

Consider three infinite sequences of real numbers: \begin{eqnarray*} X &=& \left( x_1, x_2, \cdots \right), \\ Y &=& \left( y_1, y_2, \cdots \right), \\ Z &=& \left( z_1, z_2, \cdots \right). \end{eqnarray*} It is known that, for all integers $n$, the following statement holds: \begin{align*} \left( \left( \log_2 x_n \right)^2 + \left( \log_2 y_n \right)^2 \right) \cdot \left( \left( \log_2 y_n \right)^2 + \left( \log_2 z_n \right)^2 \right) \\ &= \left( \log_2 x_n \log_2 y_n + \log_2 y_n \log_2 z_n \right)^2.\end{align*}The elements of $Y$ are defined by the relation $y_n=2^{\frac{n}{2^n}}$. Let \[S =\sum_{n=1}^{\infty} \log_2 x_n \log_2 y_n \log_2 z_n.\]Then, $S$ can be represented as a fraction $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m$.



Solution

From the given condition, we have:

\begin{align*} \left( \left( \log_2 x_n \right)^2 + \left( \log_2 y_n \right)^2 \right) \cdot \left( \left( \log_2 y_n \right)^2 + \left( \log_2 z_n \right)^2 \right) &= \left( \log_2 x_n \log_2 y_n + \log_2 y_n \log_2 z_n \right)^2 \\ \left( \log_2 y_n \right)^4 + \left( \log_2 x_n \log_2 z_n \right)^2 &= 2 \left( \log_2 y_n \right)^2 \left( \log_2 x_n \log_2 z_n \right) \\ \left( \left( \log_2 y_n \right)^2 - \log_2 x_n \log_2 z_n \right)^2 &= 0 \\ \left( \log_2 y_n \right)^2 &= \log_2 x_n \log_2 z_n \end{align*}

Then the sum becomes:

\begin{align*} S &= \sum_{n=1}^{\infty} \log_2 x_n \log_2 y_n \log_2 z_n \\ S &= \sum_{n=1}^{\infty} \frac{n^3}{8^n} \end{align*}

The final step is to take iterative differences, like so:

\begin{align*} S &= \frac{1}{8} + \frac{8}{64} + \frac{27}{512} + \frac{64}{4096} + \cdots \\ \frac{1}{8}S &= \quad \; \; \; \, \frac{1}{64} + \frac{8}{512} + \frac{27}{4096} + \cdots \\ \Rightarrow \frac{7}{8}S &= \frac{1}{8} + \frac{7}{64} + \frac{19}{512} + \frac{37}{4096} + \cdots \\ \frac{7}{64}S &= \quad \; \; \; \, \frac{1}{64} + \frac{7}{512} + \frac{19}{4096} + \cdots \\ \Rightarrow \frac{49}{64}S &= \frac{1}{8} + \frac{6}{64} + \frac{12}{512} + \frac{18}{4096} + \cdots \\ \frac{49}{512}S &= \quad \; \; \; \, \frac{1}{64} + \frac{6}{512} + \frac{12}{4096} + \cdots \\ \Rightarrow \frac{343}{512}S &= \frac{1}{8} + \frac{5}{64} + \frac{6}{512} + \frac{6}{4096} + \cdots \end{align*}

Almost done. Now that the numerators are constants $(6)$ instead of cubics $(n^3)$, we can apply the formula for the sum of an infinite geometric series to get:

\begin{align*} \frac{343}{512}S &= \frac{1}{8} + \frac{5}{64} + \frac{6}{64 \cdot 7} \\ \frac{343}{512}S &= \frac{97}{64 \cdot 7} \\ S &= \frac{776}{2401} \end{align*}

Then our answer is $\boxed{776}$.

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=Northeastern_WOOTers_Mock_AIME_I_Problems/Problem_14&oldid=159806"