Difference between revisions of "2003 AMC 10B Problems/Problem 16"
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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{\ | + | 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 19:29, 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 ?
Solution
Let be the number main courses the restaurant serves, and 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 days of
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 |