2003 AMC 10B Problem 16

Below is the professionally curated solution for Problem 16 of the 2003 AMC 10B, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2003 AMC 10B solutions, or check the answer key.

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Concepts:multiplication principle inequality

Difficulty rating: 1370

16.

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 $2003?$

$4$

$5$

$6$

$7$

$8$

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.

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