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

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

$\begin{align*} 6m^2 &\geq 365\\ m^2 &\geq 60.83\ldots.\end{align*}$ 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 $\boxed{\textbf{(E)}\ 8}$ .

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 $\boxed{\textbf{(E)}\ 8}$ .

Solution 3

Let $x$ be the number of main courses and let $2x$ be the number of appetizers. Since there are 3 desserts, the number of possible dinner choices would be $2x \cdot x \cdot 3 = 6x^2$ for any number $x$ . Since a year has $365$ days, we can assume that: $\begin{align*} 6x^2 \ge 365 \end{align*}$ $\begin{align*} x^2 \ge 61 \end{align*}$ $\begin{align*} x \ge 7.8 \end{align*}$ The least option that is greater than $7.8$ is $8$ , so the answer is $\boxed{\textbf{(E)}\ 8}$ .

~ Alfi06

Video Solution by WhyMath

https://youtu.be/-Yv2R46IO4E

~savannahsolver

2003 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 15	Followed by Problem 17
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