# 2003 AMC 10B Problems/Problem 16

## 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 $2003$? $\textbf{(A) } 4 \qquad\textbf{(B) } 5 \qquad\textbf{(C) } 6 \qquad\textbf{(D) } 7 \qquad\textbf{(E) } 8$

## Solution 1

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*}

The smallest integer value that satisfies this is $\boxed{\textbf{(E)}\ 8}$.

## Solution 2

Let $m$ denote the number of main courses needed to meet the requirement. Then the number of dinners available is $3\cdot m \cdot 2m = 6m^2$. Thus $m^2$ must be at least $365/6 \approx 61$. Since $7^2 = 49<61<64 = 8^2$, $\boxed{8}$ main courses is enough, but 7 is not. The smallest integer value that satisfies this is $\boxed{\textbf{(E)}\ 8}$.

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 