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Difference between revisions of "2010 AMC 10B Problems/Problem 18"

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==Solution==
 
==Solution==
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First we factor <math>abc+ab+a</math> into <math>a(b(c+1)+1)</math>. For <math>a(b(c+1)+1)</math> to be divisible by three we  can either have <math>a</math> be a multiple of 3 or <math>b(c+1)+1</math> be a multiple of three. Adding the probability of these two being divisible by 3 we get that the probability is <math>\boxed{\textbf{(E)}\ \frac{13}{27}}</math>
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First we factor <math>abc + ab + a</math> as <math>a(bc + b + 1)</math>, so in order for the number to be divisible by 3, either <math>a</math> is divisible by <math>3</math>, or <math>bc + b + 1</math> is divisible by <math>3</math>.  
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We see that <math>a</math> is divisible by <math>3</math> with probability <math>\frac{1}{3}</math>. We only need to calculate the probability that <math>bc + b + 1</math> is divisible by <math>3</math>.
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We need <math>bc + b + 1 \equiv 0</math>(mod <math>3</math>) or <math>b(c + 1) \equiv 2</math>(mod <math>3</math>). Using some modular arithmetic, <math>b \equiv 2</math> (mod <math>3</math>) and <math>c \equiv 0</math> (mod <math>3</math>) or <math>b \equiv 1</math> (mod <math>3</math>) and <math>c \equiv 1</math>(mod <math>3</math>). The both cases happen with probability <math>\frac{1}{3} * \frac{1}{3} = \frac{1}{9}</math> so the total probability is <math>\frac{2}{9}</math>.
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Then the answer is <math>\frac{1}{3} + \frac{2}{3} * \frac{2}{9} = \frac{13}{27}</math> or <math>\boxed{E}</math>.
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==See Also==
 
==See Also==
 
{{AMC10 box|year=2010|ab=B|num-b=17|num-a=19}}
 
{{AMC10 box|year=2010|ab=B|num-b=17|num-a=19}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 21:03, 18 January 2015

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 $bc + b + 1 \equiv 0$(mod $3$) or $b(c + 1) \equiv 2$(mod $3$). Using some modular arithmetic, $b \equiv 2$ (mod $3$) and $c \equiv 0$ (mod $3$) or $b \equiv 1$ (mod $3$) and $c \equiv 1$(mod $3$). 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 $\frac{1}{3} + \frac{2}{3} * \frac{2}{9} = \frac{13}{27}$ or $\boxed{E}$.


See Also

2010 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 17 		Followed by
Problem 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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