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, | + | 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> | + | 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 
?
Solution
Let 
be the number main courses the restaurant serves, so 
is the number of appetizers. Then the number of dinner combinations is 
. Since the customer wants to eat a different dinner in all 
days of , we must have
The smallest integer value that satisfies this is 
.
See Also
| 2003 AMC 10B (Problems • Answer Key • Resources) | ||
| 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 | ||
