2010 AMC 12B Problems/Problem 16

Problem 16

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

The value of $2010$ is arbitrary other than it is divisible by $3$ , so the set $\{1,2,3,...,2010\}$ can be grouped into threes.

Obviously, if $a$ is divisible by $3$ (which has probability $\frac{1}{3}$ ) then the sum is divisible by $3$ . In the event that $a$ is not divisible by $3$ (which has probability $\frac{2}{3})$ , then the sum is divisible by $3$ if

$bc+b+1\equiv0\pmod3$ , which is the same as

$b(c+1)\equiv2\pmod3$ .

This only occurs when one of the factors $b$ or $c+1$ is equivalent to $2\pmod3$ and the other is equivalent to $1\pmod3$ . All four events $b\equiv1\pmod3$ , $c+1\equiv2\pmod3$ , $b\equiv2\pmod3$ , and $c+1\equiv1\pmod3$ have a probability of $\frac{1}{3}$ because the set is grouped in threes.

In total the probability is $\frac{1}{3}+\frac{2}{3}(2(\frac{1}{3}\times\frac{1}{3}))=\frac{13}{27}\Rightarrow\boxed{E}$

2010 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 15	Followed by Problem 17
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2010 AMC 12B Problems/Problem 16

Problem 16

Solution

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