2023 AMC 10A 第 14 题

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

小提示：

只有

100

以内的

11

的倍数才可能有能被

11

整除的因数

Only multiples of

11

up to

100

can have a divisor divisible by

11

大提示：

对于

n = 11m

且

m \le 9

，由于

m

不是

11

的倍数，

n

的因数中恰好一半是

11

的倍数。

For

n = 11m

with

m \le 9,

exactly half of

n

’s divisors are multiples of

11,

since

m

is not a multiple of

11

解答：

一个数

n \le 100

只有在

n

能被

11

整除时，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

m

不是

11

的倍数，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

n

is divisible by

11,

so

n \in \{11, 22, \ldots, 99\}.

Write

n = 11m

with

m \le 9.

Here

m

is not a multiple of

11,

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

小提示：

大提示：

解答：

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