Difference between revisions of "2003 AMC 10B Problems/Problem 16"

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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>
 
 
 
<math>\textbf{(A) } 4 \qquad\textbf{(B) } 5 \qquad\textbf{(C) } 6 \qquad\textbf{(D) } 7 \qquad\textbf{(E) } 8</math>
 
 
 
==Solution==
 
 
 
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
 
 
 
<cmath>\begin{align*}
 
6m^2 &\geq 365\\
 
m^2 &\geq 60.83\ldots.\end{align*}</cmath>
 
 
 
The smallest integer value that satisfies this is <math>\boxed{\textbf{(E)}\ 8}</math>.
 
 
 
==See Also==
 
{{AMC10 box|year=2003|ab=B|num-b=15|num-a=17}}
 
{{MAA Notice}}
 

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