Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2010 AMC 12B Problems/Problem 16"
(→‎Solution)
Line 5: Line 5:
  
 
== Solution ==
 
== Solution ==
 +
The value of <math>2010</math> is arbitrary other than it is divisible by <math>3</math>, so the set <math>\{1,2,3,...,2010\}</math> can be grouped into threes.
 +
 +
Obviously, if <math>a</math> is divisible by <math>3</math> (which has probability <math>\frac{1}{3}</math>) then the sum is divisible by <math>3</math>. In the event that <math>a</math> is not divisible by <math>3</math> (which has probability <math>\frac{2}{3})</math>, then the sum is divisible by <math>3</math> if
 +
 +
<math>bc+b+1\equiv0\pmod3</math>, which is the same as
 +
 +
<math>b(c+1)\equiv2\pmod3</math>.
 +
 +
This only occurs when one of the factors <math>b</math> or <math>c+1</math> is equivalent to <math>2\pmod3</math> and the other is equivalent to <math>1\pmod3</math>. All four events <math>b\equiv1\pmod3</math>, <math>c+1\equiv2\pmod3</math>, <math>b\equiv2\pmod3</math>, and <math>c+1\equiv1\pmod3</math> have a probability of <math>\frac{1}{3}</math> because the set is grouped in threes.
 +
 +
In total the probability is <math>\frac{1}{3}+\frac{2}{3}(2(\frac{1}{3}\times\frac{1}{3}))=\frac{13}{27}\Rightarrow\boxed{E}</math>
  
 
== See also ==
 
== See also ==
 
{{AMC12 box|year=2010|num-b=15|num-a=17|ab=B}}
 
{{AMC12 box|year=2010|num-b=15|num-a=17|ab=B}}

Revision as of 15:18, 30 January 2011

Problem 16

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

The value of $2010$ is arbitrary other than it is divisible by $3$, so the set $\{1,2,3,...,2010\}$ can be grouped into threes.

Obviously, if $a$ is divisible by $3$ (which has probability $\frac{1}{3}$) then the sum is divisible by $3$. In the event that $a$ is not divisible by $3$ (which has probability $\frac{2}{3})$, then the sum is divisible by $3$ if

$bc+b+1\equiv0\pmod3$, which is the same as

$b(c+1)\equiv2\pmod3$.

This only occurs when one of the factors $b$ or $c+1$ is equivalent to $2\pmod3$ and the other is equivalent to $1\pmod3$. All four events $b\equiv1\pmod3$, $c+1\equiv2\pmod3$, $b\equiv2\pmod3$, and $c+1\equiv1\pmod3$ have a probability of $\frac{1}{3}$ because the set is grouped in threes.

In total the probability is $\frac{1}{3}+\frac{2}{3}(2(\frac{1}{3}\times\frac{1}{3}))=\frac{13}{27}\Rightarrow\boxed{E}$

See also

2010 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 15 		Followed by
Problem 17
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 12 Problems and Solutions
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_12B_Problems/Problem_16&oldid=36514"