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

(Undo revision 92691 by Tuktuk24 (talk))
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6m^2 &\geq 365\\
 
6m^2 &\geq 365\\
 
m^2 &\geq 60.83\ldots.\end{align*}</cmath>
 
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==
  
−
The smallest integer value that satisfies this is <math>\boxed{\textbf{(E)}\ 8}</math>.
+
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*}
 +
6x^2 \ge 365
 +
\end{align*}</cmath>
 +
<cmath>\begin{align*}
 +
x^2 \ge 61
 +
\end{align*}</cmath>
 +
<cmath>\begin{align*}
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x \ge 7.8
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\end{align*}</cmath>
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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
  
−
==Solution 2==
+
==Video Solution by WhyMath==
 +
https://youtu.be/-Yv2R46IO4E
  
−
et <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>.
+
~savannahsolver
  
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2003|ab=B|num-b=15|num-a=17}}
 
{{AMC10 box|year=2003|ab=B|num-b=15|num-a=17}}
 
{{MAA Notice}}
 
{{MAA Notice}}

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

See Also

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

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