Difference between revisions of "2004 AIME I Problems/Problem 12"
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=== Alternate way of computing the end result === | === Alternate way of computing the end result === | ||
| − | Note that actually you can just multiply <math>(1 -\frac{1}{2} +\frac{1}{4} -\frac{1}{8} \cdots)(1 -\frac{1}{5} +\frac{1}{25} -\frac{1}{125} \cdots) =\frac{3}{2} \cdot\frac{5}{6} =\frac{5}{9} \implies m+n \boxed{014}.</math> | + | Note that actually you can just multiply <math>(1 -\frac{1}{2} +\frac{1}{4} -\frac{1}{8} \cdots)(1 -\frac{1}{5} +\frac{1}{25} -\frac{1}{125} \cdots) =\frac{3}{2} \cdot\frac{5}{6} =\frac{5}{9} \implies m+n = \boxed{014}.</math> |
== See also == | == See also == | ||
Revision as of 13:41, 5 September 2020
Problem
Let 
be the set of ordered pairs 
such that 
and ![$\left[\log_2{\left(\frac 1x\right)}\right]$](http://latex.artofproblemsolving.com/1/0/b/10b4ebe0c91d961ffd0a61a8a4bee5f7090f1916.png)
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 
.
Alternate way of computing the end result
Note that actually you can just multiply 
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. 
