AMC12 2005 A

AMC12 2005 A · Q23

AMC12 2005 A · Q23. It mainly tests Basic counting (rules of product/sum), Remainders & modular arithmetic.

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?

从集合 $\{2, 2^2, 2^3, \dots , 2^{25}\}$ 中随机选择两个不同的数 $a$ 和 $b$。$\log_a b$ 是整数的概率是多少？

(A) \frac{2}{25} \frac{2}{25}

(B) \frac{31}{300} \frac{31}{300}

(D) \frac{7}{50} \frac{7}{50}

(E) \frac{1}{2} \frac{1}{2}

Answer

Correct choice: (B)

正确答案：（B）

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 $\frac{62}{600} = \boxed{\textbf{(B) }\frac{31}{300}}$.

设 $\log_{a}b = z$，则 $a^z = b$。令 $a = 2^x$, $b = 2^y$；则 $\left(2^x\right)^z = 2^{xz}= 2^y$，所以 $x \mid y$。这里可以列一个表，按每个 $x$ 统计对应的 $y$ 的个数。$x$ 的最大可能值是 $12$，因此得到 $\sum_{x=1}^{12} \left\lfloor\frac{25}{x}-1\right\rfloor = 24 + 11 + 7 + 5 + 4 + 3 + 2 + 2 + 1 + 1 + 1 + 1 = 62$。选取两个不同的数（考虑顺序）的总方法数为 $\frac{25!}{(25-2)!}= 25 \cdot 24 = 600$，所以概率为 $\frac{62}{600} = \boxed{\textbf{(B) }\frac{31}{300}}$。

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