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2004 AIME I Problems/Problem 12

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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 \...

Likewise: \left\lfloor\log_2\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) ...

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... and the answer is m+n = 5 + 9 = \boxed{014}.

See also

2004 AIME I (ProblemsResources)
Preceded by
Problem 11
Followed by
Problem 13
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
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