2005 AMC 12A Problems/Problem 23

Problem

Two distinct numbers $a$ and $b$ are chosen randomly from the set $\{2, 2^2, 2^3, \ldots, 2^{25}\}$ . What is the probability that $\log_{a}b$ is an integer?

$\mathrm{(A)}\ \frac{2}{25}\qquad \mathrm{(B)}\ \frac{31}{300}\qquad \mathrm{(C)}\ \frac{13}{100}\qquad \mathrm{(D)}\ \frac{7}{50}\qquad \mathrm{(E)}\ \frac{1}{2}$

Solution

Let $\log_{a}b = z$ , so $a^z = b$ . Define $a = 2^x$ , $b = 2^y$ ; then $\left(2^x\right)^z = 2^{xz}= 2^y$ , so $x \mid y$ . Here we can just make a table and count the number of values of $y$ per value of $x$ . The largest possible value of $x$ is $12$ , and so we get $\sum_{x=1}^{12} \left\lfloor\frac{25}{x}-1\right\rfloor = 24 + 11 + 7 + 5 + 4 + 3 + 2 + 2 + 1 + 1 + 1 + 1 = 62$ .

The total number of ways to pick two distinct numbers (where the order matters) is $\frac{25!}{(25-2)!}= 25 \cdot 24 = 600$ , so we get a probability of $\frac{62}{600} = \boxed{\textbf{(B) }\frac{31}{300}}$ .

2005 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 22	Followed by Problem 24
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All AMC 12 Problems and Solutions

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2005 AMC 12A Problems/Problem 23

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