Difference between revisions of "2004 AIME I Problems/Problem 12"
Chickendude (talk | contribs) (Added Solution) |
I like pie (talk | contribs) (AIME box) |
||
| Line 29: | Line 29: | ||
<math>m+n = 5 + 9 = \boxed{014}</math> | <math>m+n = 5 + 9 = \boxed{014}</math> | ||
| − | |||
== See also == | == See also == | ||
| − | + | {{AIME box|year=2004|n=I|num-b=11|num-a=13}} | |
| − | |||
| − | |||
| − | |||
| − | |||
Revision as of 16:21, 27 April 2008
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
Likewise:
is even when
Graphing this creates a series of rectangles which become smaller as you move toward the origin.
The 
interval of each box is given by the sequence 
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:
Geometric sums are taken:
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 | ||
