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

(Created page with "==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...")
 
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m^2 &\geq 60.83\end{align*}</cmath>
 
m^2 &\geq 60.83\end{align*}</cmath>
  
The smallest integer value that satisfies this is <math>\boxed{\mathrm{(E) \ } 8}</math>.
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The smallest integer value that satisfies this is <math>\boxed{\textbf{(E)}\ 8}</math>.
  
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2003|ab=B|num-b=15|num-a=17}}
 
{{AMC10 box|year=2003|ab=B|num-b=15|num-a=17}}

Revision as of 18:29, 26 November 2011

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 diner can have is $2m\timesm\times3=6m^2.$ (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
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions