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

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==Solution==
 
==Solution==
  
Let <math>m</math> be the number main courses the restaurant serves, and <math>2m</math> be the number of appetizers. Then the number of combinations a dinner can have is <math>2m^2\timesm\times3=6m^2.</math> Since the customer wants to eat a different dinner in all <math>365</math> days of <math>2003,</math>
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Let <math>m</math> be the number 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*}
 
<cmath>\begin{align*}
 
6m^2 &\geq 365\
 
6m^2 &\geq 365\
m^2 &\geq 60.83\end{align*}</cmath>
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m^2 &\geq 60.83\ldots.\end{align*}</cmath>
  
 
The smallest integer value that satisfies this is <math>\boxed{\textbf{(E)}\ 8}</math>.
 
The smallest integer value that satisfies this is <math>\boxed{\textbf{(E)}\ 8}</math>.

Revision as of 14:49, 10 August 2012

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 $\boxed{\textbf{(E)}\ 8}$.

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