Difference between revisions of "2003 AMC 10B Problems/Problem 16"
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==Problem== | ==Problem== | ||
− | A restaurant offers three | + | <!-- 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> | <math>\textbf{(A) } 4 \qquad\textbf{(B) } 5 \qquad\textbf{(C) } 6 \qquad\textbf{(D) } 7 \qquad\textbf{(E) } 8</math> | ||
− | ==Solution== | + | ==Solution 1== |
− | Let <math>m</math> be the number main courses the restaurant serves, | + | 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*} | <cmath>\begin{align*} | ||
6m^2 &\geq 365\\ | 6m^2 &\geq 365\\ | ||
− | m^2 &\geq 60.83\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== | ||
+ | |||
+ | 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*} | ||
+ | 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== | ==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}} |
Latest revision as of 08:06, 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 ?
Solution 1
Let be the number of main courses the restaurant serves, so
is the number of appetizers. Then the number of dinner combinations is
. Since the customer wants to eat a different dinner in all
days of
, we must have
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
.
Solution 2
Let denote the number of main courses needed to meet the requirement. Then the number of dinners available is
. Thus
must be at least
. Since
,
main courses is enough, but 7 is not. The smallest integer value that satisfies this is
.
Solution 3
Let be the number of main courses and let
be the number of appetizers.
Since there are 3 desserts, the number of possible dinner choices would be
for any number
. Since a year has
days, we can assume that:
The least option that is greater than
is
, so the answer is
.
~ Alfi06
Video Solution by WhyMath
~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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.