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 1

First we factor $abc+ab+a$ 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 $\boxed{\textbf{(E)}\ \frac{13}{27}}$

Solution 2

We look at the probability of each term being 3. 1/3 for the first term, 1/9 for the second term, and 1/27 for the third term. So the solution is 13/27 or 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

Contents

Problem

Solution 1

Solution 2

See Also