Question

Let $\mathcal{S}$ be the set of ordered pairs $(x, y)$ such that $0 \lt x \le 1,$ $0 \lt y \le 1,$ and $\left\lfloor \log_2\left(\frac{1}{x}\right) \right\rfloor$ and $\left\lfloor \log_5\left(\frac{1}{y}\right) \right\rfloor$ are both even. Given that the area of the graph of $\mathcal{S}$ is $m/n,$ where $m$ and $n$ are relatively prime positive integers, find $m + n.$ The notation $\lfloor z \rfloor$ denotes the greatest integer that is less than or equal to $z.$

Answer

Answer: 14

Concepts:floor and ceiling functions logarithm geometric sequence

Difficulty rating: 2840

Solution:

For $0 \lt x \le 1,$ the condition $\lfloor \log_2(1/x) \rfloor = 2k$ (for an integer $k \ge 0$ ) means $2k \le \log_2(1/x) \lt 2k + 1,$ i.e. $x \in \left(2^{-2k-1}, 2^{-2k}\right].$ These intervals have total length $\sum_{k \ge 0} 2^{-2k-1} = \frac{1/2}{1 - 1/4} = \frac{2}{3}.$ Similarly, $\lfloor \log_5(1/y) \rfloor$ is even for $y \in \left(5^{-2k-1}, 5^{-2k}\right],$ intervals of total length $\sum_{k \ge 0} \frac{4}{5} \cdot 25^{-k} = \frac{4/5}{1 - 1/25} = \frac{5}{6}.$

The graph of $\mathcal{S}$ is the product of these two sets, so its area is $\frac{2}{3} \cdot \frac{5}{6} = \frac{5}{9},$ and $m + n = 5 + 9 = 14.$

2004 AIME I Problem 12

Answer: 14

Difficulty rating: 2840

Solution:

Problem 12 in Other Years