Question

正整数 $a$ 、 $b$ 、 $c$ 从集合 $\{1, 2, 3,\dots, 2010\}$ 中随机且独立地有放回选取。

$abc + ab + a$ 能被 $3$ 整除的概率是多少？

Positive integers $a,$ $b,$ and $c$ are randomly and independently selected with replacement from the set $\{1, 2, 3,\dots, 2010\}.$

What is the probability that $abc + ab + a$ is divisible by $3?$

$\dfrac{1}{3}$

$\dfrac{29}{81}$

$\dfrac{31}{81}$

$\dfrac{11}{27}$

$\dfrac{13}{27}$

答案：E

知识点：模运算基本概率分类讨论

难度评级：1660

解答：

注意 $abc + ab + a = a(bc + b + 1).$ 因此若 $a$ 能被 $3,$ 整除，整个表达式也能被它整除。

因为 $2010$ 能被 $3,$ 整除， $a$ 能被 $3$ 整除的概率为 $\frac{1}{3}.$

现在考虑 $a$ 不能被 $3.$ 整除的情况。要使表达式能被 $3,$ 整除，必须有 $bc + b + 1$ 能被 $3.$ 整除。

这意味着 $bc + b = b(c + 1) \equiv 2 \pmod{3}.$

唯一可能是两个因子中一个模 $3$ 为 $2$ ，另一个模 $3$ 为 $1$ 。

两种次序中的每一种，两个因子各有 $\frac{1}{3}$ 的概率得到所需余数，因此概率为 $\dfrac{1}{3} \cdot \dfrac{1}{3} = \dfrac{1}{9}.$

共有两种次序，所以这种情况的概率为 $\dfrac{2}{9}$ 。

总概率为 $\dfrac{1}{3} \cdot 1 + \dfrac{2}{3} \cdot \dfrac{2}{9} = \dfrac{13}{27}.$

所以正确答案是 E。

Note that $abc + ab + a = a(bc + b + 1).$ This means that if $a$ is divisible by $3,$ the whole expression is as well.

Since $2010$ is divisible by $3,$ we have that $a$ is divisible by $3$ with probability $\frac{1}{3}.$

Now consider $a$ not divisible by $3.$ For the expression to be divisible by $3,$ we must have that $bc + b + 1$ is divisible by $3.$

This means that $bc + b = b(c + 1) \equiv 2 \pmod{3}.$

The only possibility for this is that one of the factors is $2$ mod $3$ and the other is $1$ mod $3.$

For each of the two cases, there is a $\frac{1}{3}$ chance that each of the factors is the desired modulus, for a probability of $\dfrac{1}{3} \cdot \dfrac{1}{3} = \dfrac{1}{9}.$

There are two cases, which means that this happens with a $\dfrac{2}{9}$ probability.

The total probability is then $\dfrac{1}{3} \cdot 1 + \dfrac{2}{3} \cdot \dfrac{2}{9} = \dfrac{13}{27}.$

Thus, E is the correct answer.

2010 AMC 10B 第 18 题

答案：E

难度评级：1660

解答：

其他年份的第 18 题