2003 AMC 10B Problem 16

Question

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?

4

5

6

7

8

Answer: E

Concepts:multiplication principle inequality

Difficulty rating: 1370

Small hint:

With

m

main courses there are

2m

appetizers, giving

3 \cdot m \cdot 2m

dinners

Big hint:

The year

2003

is not a leap year, so require

6m^2 \ge 365

Solution:

With

m

main courses, the number of dinners is

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

This must be at least

365.

So

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.