Question

从前 $100$ 个正整数中随机选一个数，然后从该数的正整数因数中随机选一个。所选因数能被 $11$ 整除的概率是多少？

A number is chosen at random from among the first $100$ positive integers, and a positive integer divisor of that number is then chosen at random. What is the probability that the chosen divisor is divisible by $11?$

$\dfrac{4}{100}$

$\dfrac{9}{200}$

$\dfrac{1}{20}$

$\dfrac{11}{200}$

$\dfrac{3}{50}$

答案：B

知识点：因数个数基本概率

难度评级：1630

解答：

一个数 $n \le 100$ 只有在 $11 \mid n$ 时，才可能有能被 $11$ 整除的因数，所以 $n \in \{11, 22, \ldots, 99\}$ 。写成 $n = 11m$ ，其中 $m \le 9$ 。这里 $11 \nmid m$ ，所以 $d(11m) = 2\,d(m)$ ，而能被 $11$ 整除的因数恰好是 $11d$ 这种形式，共有 $d(m)$ 个。因此对于每个这样的 $n$ ，概率都是 $\frac{d(m)}{2\,d(m)} = \frac12$ 。对全部 $100$ 个起始数字取平均，概率为 $\frac{1}{100}\sum_{m=1}^{9}\frac12 = \frac{9}{200}$ 。因此，答案是 B。

A number $n \le 100$ can only have a divisor divisible by $11$ when $11 \mid n,$ so $n \in \{11, 22, \ldots, 99\}.$ Write $n = 11m$ with $m \le 9.$ Here $11 \nmid m,$ so $d(11m) = 2\,d(m),$ and the divisors that are multiples of $11$ are exactly the $d(m)$ numbers $11d.$ That makes the chance $\frac{d(m)}{2\,d(m)} = \frac12$ for each such $n.$ Averaging over all $100$ starting numbers, the probability is $\frac{1}{100}\sum_{m=1}^{9}\frac12 = \frac{9}{200}.$ Therefore, the answer is B.

2023 AMC 10A 第 14 题

答案：B

难度评级：1630

解答：

其他年份的第 14 题