2010 AMC 10B Problems/Problem 18

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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 $abc+bc+c$ into $a(b(c+1)+1)$. For $a(b(c+1)+1)$ to be divisible by three we can either have $a$ be a multiple of 3 or $b(c+1)+1$ be a multiple of three. Adding the probability of these two being divisible by 3 we get that the probability is $\boxed{\textbf{(E)}\ \frac{13}{27}}$

See Also

2010 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 17
Followed by
Problem 19
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All AMC 10 Problems and Solutions