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 "2004 AIME I Problems/Problem 12"
(Added Solution)
Line 3: Line 3:
  
 
== Solution ==
 
== Solution ==
−
{{solution}}
+
<math>\left\lfloor\log_2\left(\frac{1}{x}\right)\right\rfloor</math> is even when
 +
 
 +
<math>x \in \left(\frac{1}{2},1\right) \cup \left(\frac{1}{8},\frac{1}{4}\right) \cup \left(\frac{1}{32},\frac{1}{16}\right) \cup...</math>
 +
 
 +
Likewise:
 +
<math>\left\lfloor\log_2\left(\frac{1}{y}\right)\right\rfloor</math> is even when
 +
 
 +
<math>y \in \left(\frac{1}{5},1\right) \cup \left(\frac{1}{125},\frac{1}{25}\right) \cup \left(\frac{1}{3125},\frac{1}{625}\right) \cup...</math>
 +
 
 +
Graphing this creates a series of rectangles which become smaller as you move toward the origin.
 +
 
 +
The <math>x</math> interval of each box is given by the sequence <math>\frac{1}{2} , \frac{1}{8}, \frac{1}{32} ...</math>
 +
 
 +
The <math>y</math> interval is given by <math>\frac{4}{5} , \frac{4}{125}, \frac{4}{3125}...</math>
 +
 
 +
Each box is the product of one term of each sequence. The sum of these boxes is simply the product of the sum of each sequence or:
 +
 
 +
<math>\left(\frac{1}{2} + \frac{1}{8} + \frac{1}{32} ...\right)\left(\frac{4}{5} + \frac{4}{125} + \frac{4}{3125}...\right)</math>
 +
 
 +
Geometric sums are taken:
 +
 
 +
<math>\left(\frac{\frac{1}{2}}{1 - \frac{1}{4}}\right)\left(\frac{\frac{4}{5}}{1-\frac{1}{25}}\right)</math>
 +
 
 +
<math>\frac{2}{3} \cdot \frac{5}{6} = \frac{5}{9}</math>
 +
 
 +
<math>m+n = 5 + 9 = \boxed{014}</math>
 +
 
  
 
== See also ==
 
== See also ==

Revision as of 23:20, 13 March 2008

Problem

Let $S$ be the set of ordered pairs $(x, y)$ such that $0 < x \le 1, 0<y\le 1,$ and $\left[\log_2{\left(\frac 1x\right)}\right]$ and $\left[\log_5{\left(\frac 1y\right)}\right]$ are both even. Given that the area of the graph of $S$ is $m/n,$ where $m$ and $n$ are relatively prime positive integers, find $m+n.$ The notation $[z]$ denotes the greatest integer that is less than or equal to $z.$

Solution

$\left\lfloor\log_2\left(\frac{1}{x}\right)\right\rfloor$ is even when

$x \in \left(\frac{1}{2},1\right) \cup \left(\frac{1}{8},\frac{1}{4}\right) \cup \left(\frac{1}{32},\frac{1}{16}\right) \cup...$

Likewise: $\left\lfloor\log_2\left(\frac{1}{y}\right)\right\rfloor$ is even when

$y \in \left(\frac{1}{5},1\right) \cup \left(\frac{1}{125},\frac{1}{25}\right) \cup \left(\frac{1}{3125},\frac{1}{625}\right) \cup...$

Graphing this creates a series of rectangles which become smaller as you move toward the origin.

The $x$ interval of each box is given by the sequence $\frac{1}{2} , \frac{1}{8}, \frac{1}{32} ...$

The $y$ interval is given by $\frac{4}{5} , \frac{4}{125}, \frac{4}{3125}...$

Each box is the product of one term of each sequence. The sum of these boxes is simply the product of the sum of each sequence or:

$\left(\frac{1}{2} + \frac{1}{8} + \frac{1}{32} ...\right)\left(\frac{4}{5} + \frac{4}{125} + \frac{4}{3125}...\right)$

Geometric sums are taken:

$\left(\frac{\frac{1}{2}}{1 - \frac{1}{4}}\right)\left(\frac{\frac{4}{5}}{1-\frac{1}{25}}\right)$

$\frac{2}{3} \cdot \frac{5}{6} = \frac{5}{9}$

$m+n = 5 + 9 = \boxed{014}$


See also

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2004_AIME_I_Problems/Problem_12&oldid=24015"