Difference between revisions of "2004 AIME I Problems/Problem 12"
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== Problem == | == Problem == | ||
| − | Let <math> S </math> be the set of [[ordered pair]]s <math> (x, y) </math> such that <math> 0 < x \le 1, 0<y\le 1, </math> and <math> \left[\log_2{\left(\frac 1x\right)}\right] </math> and <math> \left[\log_5{\left(\frac 1y\right)}\right] </math> are both even. Given that the area of the graph of <math> S </math> is <math> m/n, </math> where <math> m </math> and <math> n </math> are relatively prime positive integers, find <math> m+n. </math> The notation <math> [z] </math> denotes the [[floor function|greatest integer]] that is less than or equal to <math> z. </math> | + | Let <math> S </math> be the set of [[ordered pair]]s <math> (x, y) </math> such that <math> 0 < x \le 1, 0<y\le 1, </math> and <math> \left\floor[\log_2{\left(\frac 1x\right)}\floor\right] </math> and <math> \left[\log_5{\left(\frac 1y\right)}\right] </math> are both even. Given that the area of the graph of <math> S </math> is <math> m/n, </math> where <math> m </math> and <math> n </math> are relatively prime positive integers, find <math> m+n. </math> The notation <math> [z] </math> denotes the [[floor function|greatest integer]] that is less than or equal to <math> z. </math> |
== Solution == | == Solution == | ||
Revision as of 19:31, 14 February 2016
Problem
Let 
be the set of ordered pairs 
such that 
and $\left\floor[\log_2{\left(\frac 1x\right)}\floor\right]$ (Error compiling LaTeX. Unknown error_msg) and ![$\left[\log_5{\left(\frac 1y\right)}\right]$](http://latex.artofproblemsolving.com/c/3/3/c33380736cf1c503a9bc24013f134f19b4cae85e.png)
are both even. Given that the area of the graph of is 
where 
and 
are relatively prime positive integers, find 
The notation ![$[z]$](http://latex.artofproblemsolving.com/d/4/d/d4d87c4df7a861f6bc1a1be5600dcf4111c49841.png)
denotes the greatest integer that is less than or equal to 
Solution
is even when
![\[x \in \left(\frac{1}{2},1\right) \cup \left(\frac{1}{8},\frac{1}{4}\right) \cup \left(\frac{1}{32},\frac{1}{16}\right) \cup \cdots\]](http://latex.artofproblemsolving.com/0/6/9/069dbb13543f5e6f7ff10f5b8f6b4f1da25eacba.png)
Likewise:
is even when
![\[y \in \left(\frac{1}{5},1\right) \cup \left(\frac{1}{125},\frac{1}{25}\right) \cup \left(\frac{1}{3125},\frac{1}{625}\right) \cup \cdots\]](http://latex.artofproblemsolving.com/1/d/6/1d64ebd466362a66efef98d7c5d974f9854bd18e.png)
Graphing this yields a series of rectangles which become smaller as you move toward the origin. The 
interval of each box is given by the geometric sequence 
, and the 
interval is given by 
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:
![\[\left(\frac{1}{2} + \frac{1}{8} + \frac{1}{32} \ldots \right)\left(\frac{4}{5} + \frac{4}{125} + \frac{4}{3125} \ldots \right)=\left(\frac{\frac{1}{2}}{1 - \frac{1}{4}}\right)\left(\frac{\frac{4}{5}}{1-\frac{1}{25}}\right)= \frac{2}{3} \cdot \frac{5}{6} = \frac{5}{9},\]](http://latex.artofproblemsolving.com/6/3/0/63087ddd18ec20f71058f90f4746b0b2993fc34f.png)
and the answer is 
.
See also
| 2004 AIME I (Problems • Answer Key • Resources) | ||
| Preceded by Problem 11 |
Followed by Problem 13 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
