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Difference between revisions of "2004 AIME I Problems/Problem 12"
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== Problem ==
 
== Problem ==
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Let <math> S </math> be the set of ordered pairs <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 greatest integer that is less than or equal to <math> z. </math>
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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>
  
 
== Solution ==
 
== Solution ==
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{{solution}}
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<math>\left\lfloor\log_2\left(\frac{1}{x}\right)\right\rfloor</math> is even when
 +
 
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<cmath>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</cmath>
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Likewise:
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<math>\left\lfloor\log_5\left(\frac{1}{y}\right)\right\rfloor</math> is even when
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<cmath>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</cmath>
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Graphing this yields a series of [[rectangle]]s which become smaller as you move toward the [[origin]]. The <math>x</math> interval of each box is given by the [[geometric sequence]] <math>\frac{1}{2} , \frac{1}{8}, \frac{1}{32}, \cdots</math>, and the <math>y</math> interval is given by <math>\frac{4}{5} , \frac{4}{125}, \frac{4}{3125}, \cdots</math>
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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:
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 +
<cmath>\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},</cmath> and the answer is <math>m+n = 5 + 9 = \boxed{014}</math>.
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== See also ==
 
== See also ==
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* [[2004 AIME I Problems/Problem 11| Previous problem]]
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{{AIME box|year=2004|n=I|num-b=11|num-a=13}}
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* [[2004 AIME I Problems/Problem 13| Next problem]]
 
  
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* [[2004 AIME I Problems]]
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[[Category:Intermediate Algebra Problems]]
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{{MAA Notice}}

Latest revision as of 17:43, 16 October 2024

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

\[x \in \left(\frac{1}{2},1\right) \cup \left(\frac{1}{8},\frac{1}{4}\right) \cup \left(\frac{1}{32},\frac{1}{16}\right) \cup \cdots\]

Likewise: $\left\lfloor\log_5\left(\frac{1}{y}\right)\right\rfloor$ is even when

\[y \in \left(\frac{1}{5},1\right) \cup \left(\frac{1}{125},\frac{1}{25}\right) \cup \left(\frac{1}{3125},\frac{1}{625}\right) \cup \cdots\]

Graphing this yields a series of rectangles which become smaller as you move toward the origin. The $x$ interval of each box is given by the geometric sequence $\frac{1}{2} , \frac{1}{8}, \frac{1}{32}, \cdots$, and the $y$ interval is given by $\frac{4}{5} , \frac{4}{125}, \frac{4}{3125}, \cdots$

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:

\[\left(\frac{1}{2} + \frac{1}{8} + \frac{1}{32} \ldots \right)\left(\frac{4}{5} + \frac{4}{125} + \frac{4}{3125} \ldots \right)=\left(\frac{\frac{1}{2}}{1 - \frac{1}{4}}\right)\left(\frac{\frac{4}{5}}{1-\frac{1}{25}}\right)= \frac{2}{3} \cdot \frac{5}{6} = \frac{5}{9},\] and the answer is $m+n = 5 + 9 = \boxed{014}$.


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

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