2003 AMC 10B Problems/Problem 16
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
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
Also, year 2003 is not a leap year, because 2003 divided by 4 does not equal an integer. 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 .
Solution 3
Let be the number of main courses and let be the number of appetizers. Since there are 3 desserts, the number of possible dinner choices would be for any number . Since a year has days, we can assume that: The least option that is greater than is , so the answer is .
~ Alfi06
Video Solution by WhyMath
~savannahsolver
See Also
2003 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 15 |
Followed by Problem 17 | |
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All AMC 10 Problems and Solutions |
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