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| − | ==Problem==
| + | #redirect [[2010 AMC 12B Problems/Problem 16]] |
| − | Positive integers <math>a</math>, <math>b</math>, and <math>c</math> are randomly and independently selected with replacement from the set <math>\{1, 2, 3,\dots, 2010\}</math>. What is the probability that <math>abc + ab + a</math> is divisible by <math>3</math>?
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| − | <math>\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}</math>
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| − | ==Solution==
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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\pmod 3</math> or <math>b(c + 1) \equiv 2\pmod 3</math>. Using some modular arithmetic, <math>b \equiv 2\pmod 3</math> and <math>c \equiv 0\pmod 3</math> or <math>b \equiv 1\pmod 3</math> and <math>c \equiv 1\pmod 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}\cdot\frac{2}{9} = \frac{13}{27}</math> or <math>\boxed{E}</math>.
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| − | ==See Also==
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| − | {{AMC10 box|year=2010|ab=B|num-b=17|num-a=19}}
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| − | {{MAA Notice}}
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