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Difference between revisions of "2010 AMC 12B Problems/Problem 16"

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Revision as of 09:59, 4 July 2013

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

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