AMC10 2010 B

AMC10 2010 B · Q18

AMC10 2010 B · Q18. 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 $a,b,c$ 从集合 $\{1,2,3,\dots,2010\}$ 中独立随机有放回选取。$abc+ab+a$ 能被 3 整除的概率是多少？

(A) 1/3 1/3

(B) 29/81 29/81

(D) 11/27 11/27

(E) 13/27 13/27

Answer

Correct choice: (E)

正确答案：（E）

Solution

Answer (E): Let $N = abc + ab + a = a(bc + b + 1)$. If $a$ is divisible by 3, then $N$ is divisible by 3. Note that 2010 is divisible by 3, so the probability that $a$ is divisible by 3 is $\frac{1}{3}$. If $a$ is not divisible by 3 then $N$ is divisible by 3 if $bc + b + 1$ is divisible by 3. Define $b_0$ and $b_1$ so that $b = 3b_0 + b_1$ is an integer and $b_1$ is equal to 0, 1, or 2. Note that each possible value of $b_1$ is equally likely. Similarly define $c_0$ and $c_1$. Then \[ bc + b + 1 = (3b_0 + b_1)(3c_0 + c_1) + 3b_0 + b_1 + 1 = 3(3b_0c_0 + c_0b_1 + c_1b_0 + b_0) + b_1c_1 + b_1 + 1. \] Hence $bc + b + 1$ is divisible by 3 if and only if $b_1 = 1$ and $c_1 = 1$, or $b_1 = 2$ and $c_1 = 0$. The probability of this occurrence is $\frac{1}{3}\cdot\frac{1}{3} + \frac{1}{3}\cdot\frac{1}{3} = \frac{2}{9}$. Therefore the requested probability is $\frac{1}{3} + \frac{2}{3}\cdot\frac{2}{9} = \frac{13}{27}$.

答案（E）：令 $N = abc + ab + a = a(bc + b + 1)$。如果 $a$ 能被 3 整除，则 $N$ 能被 3 整除。注意 2010 能被 3 整除，因此 $a$ 能被 3 整除的概率是 $\frac{1}{3}$。如果 $a$ 不能被 3 整除，那么当且仅当 $bc + b + 1$ 能被 3 整除时，$N$ 才能被 3 整除。定义 $b_0$ 和 $b_1$，使得 $b = 3b_0 + b_1$ 为整数且 $b_1$ 取 0、1 或 2。注意 $b_1$ 的每个可能取值等可能。同理定义 $c_0$ 和 $c_1$。则 \[ bc + b + 1 = (3b_0 + b_1)(3c_0 + c_1) + 3b_0 + b_1 + 1 = 3(3b_0c_0 + c_0b_1 + c_1b_0 + b_0) + b_1c_1 + b_1 + 1. \] 因此，$bc + b + 1$ 能被 3 整除当且仅当 $b_1 = 1$ 且 $c_1 = 1$，或 $b_1 = 2$ 且 $c_1 = 0$。该事件的概率为 $\frac{1}{3}\cdot\frac{1}{3} + \frac{1}{3}\cdot\frac{1}{3} = \frac{2}{9}$。所以所求概率为 $\frac{1}{3} + \frac{2}{3}\cdot\frac{2}{9} = \frac{13}{27}$。

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