AMC10 2003 10B, Problem 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 a restaurant should offer so that a customer could have a different dinner each night in the year $2003$?

Solution: Let $m$ be the number of main courses the restaurant serves, so $2m$ is the number of appetizers. Then the number of dinner combinations is $2m\times m\times3=6m^2$. Since the customer wants to eat a different dinner in all $365$ days of $2003$, we must have \begin{align} 6m^2 &\geq 365\\ m^2 &\geq 60.83\ldots.\end{align} 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 $\fbox{8}$.

Answer: 8