# Difference between revisions of "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 $abc + ab + a$ 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 $\boxed{E}$.

## Solution 2

We see that since $2010$ is divisible by $3$, the probability that any one of $a$, $b$, or $c$ being divisible by $3$ is $\frac{1}{3}$. Because of this, we can shrink the set of possibilities for $a$, $b$, and $c$ to the set $\{1,2,3\}$ without affecting the probability in question.

Listing out all possible combinations for $a$, $b$, and $c$, we see that the answer is $\bold{\boxed{\left( E\right) \frac{13}{27}}}$.