2003 AMC 10B Problem 16

Problem 16 of 25IntermediateAlgebra Counting & Probability

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 the restaurant should offer so that a customer could have a different dinner each night in the year

2003?

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Solution

With

m

main courses, the number of dinners is

3 \cdot m \cdot 2m = 6m^2.

This must be at least

365.

m^2 \ge \dfrac{365}{6}\approx 60.8.

Since

7^2=49

is too small but

8^2=64

works,

m=8.

Thus, the correct answer is E.

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