AMC12 2010 B

AMC12 2010 B · Q16

AMC12 2010 B · Q16. It mainly tests Probability (basic), Remainders & modular arithmetic.

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 $3$?

正整数 $a$、$b$ 和 $c$ 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？

(A) \frac{1}{3} \frac{1}{3}

(B) \frac{29}{81} \frac{29}{81}

(D) \frac{11}{27} \frac{11}{27}

(E) \frac{13}{27} \frac{13}{27}

Answer

Correct choice: (E)

正确答案：（E）

Solution

We group this into groups of $3$, because $3|2010$. This means that every residue class mod 3 has an equal probability. If $3|a$, we are done. There is a probability of $\frac{1}{3}$ that that happens. Otherwise, we have $3|bc+b+1$, which means that $b(c+1) \equiv 2\pmod{3}$. So either \[b \equiv 1 \pmod{3}, c \equiv 1 \pmod{3}\] or \[b \equiv 2 \pmod {3}, c \equiv 0 \pmod 3\] which will lead to the property being true. There is a $\frac{1}{3}\cdot\frac{1}{3}=\frac{1}{9}$ chance for each bundle of cases to be true. Thus, the total for the cases is $\frac{2}{9}$. But we have to multiply by $\frac{2}{3}$ because this only happens with a $\frac{2}{3}$ chance. So the total is actually $\frac{4}{27}$. The grand total is \[\frac{1}{3} + \frac{4}{27} = \boxed{\text{(E) }\frac{13}{27}.}\]

我们按模 $3$ 的剩余类来分类，因为 $3\mid 2010$，所以模 $3$ 的每个剩余类出现的概率相同。若 $3\mid a$，则必有 $3\mid (abc+ab+a)$。这种情况的概率为 $\frac{1}{3}$。否则，需有 $3\mid (bc+b+1)$，即 $b(c+1) \equiv 2\pmod{3}$。因此要么 \[b \equiv 1 \pmod{3},\ c \equiv 1 \pmod{3}\] 要么 \[b \equiv 2 \pmod {3},\ c \equiv 0 \pmod 3\] 这两种情形都会使条件成立。每一组情形发生的概率都是 $\frac{1}{3}\cdot\frac{1}{3}=\frac{1}{9}$，所以在 $a$ 不被 $3$ 整除的前提下，总概率为 $\frac{2}{9}$。再乘以 $a$ 不被 $3$ 整除的概率 $\frac{2}{3}$，得到该部分概率为 $\frac{4}{27}$。总概率为 \[\frac{1}{3} + \frac{4}{27} = \boxed{\text{(E) }\frac{13}{27}.}\]

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