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 | + | 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\lfloor\log_2{\left(\frac 1x\right)}\right\rfloor </math> and <math>\left\lfloor\log_5{\left(\frac 1y\right)}\right\rfloor </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 17:41, 16 October 2024
Problem
Let be the set of ordered pairs such that and and are both even. Given that the area of the graph of is where and are relatively prime positive integers, find The notation denotes the greatest integer that is less than or equal to
Solution
is even when
Likewise: is even when
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:
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.