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 - \displaystyle\frac{1}{2} + \displaystyle\frac{1}{4} - \displaystyle\frac{1}{8} \cdots)(1 - \displaystyle\frac{1}{5} + \displaystyle\frac{1}{25} - \displaystyle\frac{1}{125} \cdots) = \displaystyle\frac{3}{2} \cdot \displaystyle\frac{5}{6} = \displaystyle\frac{5}{9} = \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 14:40, 5 September 2020

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 $m+n = 5 + 9 = \boxed{014}$.


Alternate way of computing the end result

Note that actually you can just multiply $(1 -\frac{1}{2} +\frac{1}{4} -\frac{1}{8} \cdots)(1 -\frac{1}{5} +\frac{1}{25} -\frac{1}{125} \cdots) =\frac{3}{2} \cdot\frac{5}{6} =\frac{5}{9} \implies m+n \boxed{014}.$

See also

2004 AIME I (ProblemsAnswer KeyResources)
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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