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| − | ==Problem==
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| − | <!-- don't remove the following tag, for PoTW on the Wiki front page--><onlyinclude>A restaurant offers three desserts, and exactly twice as many appetizers as main courses. A dinner consists of an appetizer, a main course, and a dessert. What is the least number of main courses that a restaurant should offer so that a customer could have a different dinner each night in the year <math>2003</math>?<!-- don't remove the following tag, for PoTW on the Wiki front page--></onlyinclude>
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| − | <math>\textbf{(A) } 4 \qquad\textbf{(B) } 5 \qquad\textbf{(C) } 6 \qquad\textbf{(D) } 7 \qquad\textbf{(E) } 8</math>
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| − | ==Solution==
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| − | Let <math>m</math> be the number of main courses the restaurant serves, so <math>2m</math> is the number of appetizers. Then the number of dinner combinations is <math>2m\times m\times3=6m^2</math>. Since the customer wants to eat a different dinner in all <math>365</math> days of <math>2003</math>, we must have
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| − | <cmath>\begin{align*}
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| − | 6m^2 &\geq 365\\
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| − | m^2 &\geq 60.83\ldots.\end{align*}</cmath>
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| − | The smallest integer value that satisfies this is <math>\boxed{\textbf{(E)}\ 8}</math>.
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| − | ==See Also==
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| − | {{AMC10 box|year=2003|ab=B|num-b=15|num-a=17}}
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| − | {{MAA Notice}}
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