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Difference between revisions of "2010 AMC 12B Problems/Problem 16"
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== Solution ==
 
== Solution ==
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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.
 
  
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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
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We group this into groups of <math>3</math>, because <math>3|2010</math>.  
  
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<math>bc+b+1\equiv0\pmod3</math>, which is the same as
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If <math>3|a</math>, we are done. There is a probability of <math>\frac{1}{3}</math> that that happens.
  
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<math>b(c+1)\equiv2\pmod3</math>.
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Otherwise, we have <math>3|bc+b+1</math>, which means that <math>b(c+1) \equiv 2\pmod{3}</math>. So either <cmath>b \equiv 1 \pmod{3}, c \equiv 1 \pmod{3}</cmath> or <cmath>b \equiv 2 \pmod {3}, c \equiv 0 \pmod 3</cmath> which will lead to the property being true. There are a <math>\frac{1}{3}\cdot\frac{1}{3}=\frac{1}{9}</math> chance for each bundle of cases to be true. Thus, the total for the cases is <math>\frac{2}{9}</math>. But we have to multiply by <math>\frac{2}{3}</math> because this only happens with a <math>\frac{2}{3}</math> chance. So the total is actually <math>\frac{4}{27}</math>.
  
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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.
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The grand total is <cmath>\frac{1}{3} + \frac{4}{27} = \boxed{\text{(E) }\frac{13}{27}.}</cmath>
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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 ==

Revision as of 17:44, 9 October 2017

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

We group this into groups of $3$, because $3|2010$.

If $3|a$, we are done. There is a probability of $\frac{1}{3}$ that that happens.

Otherwise, we have $3|bc+b+1$, which means that $b(c+1) \equiv 2\pmod{3}$. So either \[b \equiv 1 \pmod{3}, c \equiv 1 \pmod{3}\] or \[b \equiv 2 \pmod {3}, c \equiv 0 \pmod 3\] which will lead to the property being true. There are a $\frac{1}{3}\cdot\frac{1}{3}=\frac{1}{9}$ chance for each bundle of cases to be true. Thus, the total for the cases is $\frac{2}{9}$. But we have to multiply by $\frac{2}{3}$ because this only happens with a $\frac{2}{3}$ chance. So the total is actually $\frac{4}{27}$.

The grand total is \[\frac{1}{3} + \frac{4}{27} = \boxed{\text{(E) }\frac{13}{27}.}\]

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

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