Difference between revisions of "2003 AMC 10B Problems/Problem 16"
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==Solution 2== | ==Solution 2== | ||
− | + | Let <math>m</math> denote the number of main courses needed to meet the requirement. Then the number of dinners available is <math>3\cdot m \cdot 2m = 6m^2</math>. Thus <math>m^2</math> must be at least <math>365/6 \approx 61</math>. Since <math>7^2 = 49<61<64 = 8^2</math>, <math>\boxed{8}</math> main courses is enough, but 7 is not. 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}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 13:31, 31 March 2018
Contents
Problem
A restaurant offers three desserts, 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 1
Let be the number of 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 .
Solution 2
Let denote the number of main courses needed to meet the requirement. Then the number of dinners available is . Thus must be at least . Since , main courses is enough, but 7 is not. 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 |
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