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

Latest revision as of 07:06, 6 September 2022

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

$\begin{align*} 6m^2 &\geq 365\\ m^2 &\geq 60.83\ldots.\end{align*}$ 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

@@ Line 1: / Line 1: @@
-{{delete|What's the point of this page?}}
+==Problem==
+<!-- don't remove the following tag, for PoTW on the Wiki front page--><onlyinclude>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 <math>2003</math>?<!-- don't remove the following tag, for PoTW on the Wiki front page--></onlyinclude>
+<math>\textbf{(A) } 4 \qquad\textbf{(B) } 5 \qquad\textbf{(C) } 6 \qquad\textbf{(D) } 7 \qquad\textbf{(E) } 8</math>
+==Solution 1==
+Let <math>m</math> be the number of main courses the restaurant serves, so <math>2m</math> is the number of appetizers. Then the number of dinner combinations is <math>2m\times m\times3=6m^2</math>. Since the customer wants to eat a different dinner in all <math>365</math> days of <math>2003</math>, we must have
+<cmath>\begin{align*}
+m^2 &\geq 365\\
+m^2 &\geq 60.83\ldots.\end{align*}</cmath>
+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 <math>\boxed{\textbf{(E)}\ 8}</math>.
+==Solution 2==
+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 number 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*}
+x^2 \ge 365
+\end{align*}</cmath>
+<cmath>\begin{align*}
+x^2 \ge 61
+\end{align*}</cmath>
+<cmath>\begin{align*}
+x \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
+==Video Solution by WhyMath==
+https://youtu.be/-Yv2R46IO4E
+~savannahsolver
+==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
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

Page

Toolbox

Search