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

Revision as of 11:10, 4 July 2013

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, so $2m$ is the number of appetizers. Then the number of dinner combinations is $2m\times m\times3=6m^2$ . Since the customer wants to eat a different dinner in all $365$ days of $2003$ , we must have

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

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

2003 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 15	Followed by Problem 17
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All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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 ==See Also==
 {{AMC10 box|year=2003|ab=B|num-b=15|num-a=17}}
+{{MAA Notice}}

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

Revision as of 11:10, 4 July 2013

Problem

Solution

See Also