# 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 byProblem 17 Followed byProblem 19 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 10 Problems and Solutions

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