Difference between revisions of "2003 AMC 10B Problems/Problem 16"

Revision as of 12:09, 10 January 2021

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 umber 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: $\begin{align*} 6n^2 \ge 365 \end{align*}$ $\begin{align*} n^2 \ge 61 \end{align*}$ $\begin{align*} n \ge 7.8 \end{align*}$ The least option that is greater than $7.8$ is $8$ , so the answer is $\boxed{\textbf{(E)}\ 8}$ .

~ Alfi06

@@ Line 19: / Line 19: @@
 Let <math>m</math> denote the number of main courses needed to meet the requirement. Then the number of dinners available is <math>3\cdot m \cdot 2m = 6m^2</math>. Thus <math>m^2</math> must be at least <math>365/6 \approx 61</math>. Since <math>7^2 = 49<61<64 = 8^2</math>, <math>\boxed{8}</math> main courses is enough, but 7 is not. The smallest integer value that satisfies this is <math>\boxed{\textbf{(E)}\ 8}</math>.
+===Solution 3===
+Let <math>x</math> be the number of main courses and let <math>2x</math> be the umber of appetizers.
+Since there are 3 desserts, the number of possible dinner choices would be <math>2x \cdot x \cdot 3 = 6x^2</math> for any number <math>x</math>. Since a year has <math>365</math> days, we can assume that:
+<cmath>\begin{align*}
+n^2 \ge 365
+\end{align*}</cmath>
+<cmath>\begin{align*}
+n^2 \ge 61
+\end{align*}</cmath>
+<cmath>\begin{align*}
+n \ge 7.8
+\end{align*}</cmath>
+The least option that is greater than <math>7.8</math> is <math>8</math>, so the answer is <math>\boxed{\textbf{(E)}\ 8}</math>.
+~ Alfi06
 ==See Also==
 {{AMC10 box|year=2003|ab=B|num-b=15|num-a=17}}
 {{MAA Notice}}

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