2003 AMC 10B Problems/Problem 16

Revision as of 13:31, 10 August 2012 by Math padawan (talk | contribs) (Solution)

Problem

A restaurant offers three deserts, 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 $2003$?

$\textbf{(A) } 4 \qquad\textbf{(B) } 5 \qquad\textbf{(C) } 6 \qquad\textbf{(D) } 7 \qquad\textbf{(E) } 8$

Solution

Let $m$ be the number main courses the restaurant serves, and $2m$ be the number of appetizers. Then the number of combinations a dinner can have is $2m^2\timesm\times3=6m^2.$ (Error compiling LaTeX. Unknown error_msg) Since the customer wants to eat a different dinner in all $365$ days of $2003,$

\begin{align*} 6m^2 &\geq 365\\ m^2 &\geq 60.83\end{align*}

The smallest integer value that satisfies this is $\boxed{\textbf{(E)}\ 8}$.

See Also

2003 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 15
Followed by
Problem 17
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All AMC 10 Problems and Solutions