2010 AMC 10B Problems/Problem 18

Problem

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 $3$ ?

$\textbf{(A)}\ \dfrac{1}{3} \qquad \textbf{(B)}\ \dfrac{29}{81} \qquad \textbf{(C)}\ \dfrac{31}{81} \qquad \textbf{(D)}\ \dfrac{11}{27} \qquad \textbf{(E)}\ \dfrac{13}{27}$

Solution

First we factor $abc + ab + a$ as $a(bc + b + 1)$ , so in order for the number to be divisible by 3, either $a$ is divisible by $3$ , or $bc + b + 1$ is divisible by $3$ .

We see that $a$ is divisible by $3$ with probability $\frac{1}{3}$ . We only need to calculate the probability that $bc + b + 1$ is divisible by $3$ .

We need $bc + b + 1 \equiv 0\pmod 3$ or $b(c + 1) \equiv 2\pmod 3$ . Using some modular arithmetic, $b \equiv 2\pmod 3$ and $c \equiv 0\pmod 3$ or $b \equiv 1\pmod 3$ and $c \equiv 1\pmod 3$ . The both cases happen with probability $\frac{1}{3} * \frac{1}{3} = \frac{1}{9}$ so the total probability is $\frac{2}{9}$ .

Then the answer is $\frac{1}{3} + \frac{2}{3}\cdot\frac{2}{9} = \frac{13}{27}$ or $\boxed{E}$ .

2010 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 17	Followed by Problem 19
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2010 AMC 10B Problems/Problem 18

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