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Difference between revisions of "2010 AMC 10B Problems/Problem 18"
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==Solution 1==
 
==Solution 1==
 
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>
 
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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==Solution 2==
 
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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.
 
  
 
==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 15:00, 21 August 2013

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 $\boxed{\textbf{(E)}\ \frac{13}{27}}$

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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