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Difference between revisions of "2020 AMC 10A Problems/Problem 9"

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== Solution 2 ==
 
== Solution 2 ==
  
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This is similar to Solution 1, with the same basic idea, but a little different. Since both <math>7</math> and <math>11</math> are prime, their LCM must be their product. However, we don't want to find the LCM, but rather the number of seats required to fit the children and adults. So the answer would be <math>7 + 11 = \boxed{\text{(B)} 18}</math>.
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This is similar to Solution 1, with the same basic idea, but we don't need to calculate the LCM. Since both <math>7</math> and <math>11</math> are relatively prime, their LCM must be their product. So the answer would be <math>7 + 11 = \boxed{\text{(B) } 18}</math>.     ~Baolan
  
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==Video Solution==
  
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~Baolan
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Education, The Study of Everything
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 +
https://youtu.be/GKTQO99CKPM
  
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==Video Solution==
 
 
https://youtu.be/JEjib74EmiY
 
https://youtu.be/JEjib74EmiY
  
 
~IceMatrix
 
~IceMatrix
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 +
https://youtu.be/w2_H96-yzk8
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~savannahsolver
  
 
==See Also==
 
==See Also==

Revision as of 11:46, 7 November 2020

Problem

A single bench section at a school event can hold either $7$ adults or $11$ children. When $N$ bench sections are connected end to end, an equal number of adults and children seated together will occupy all the bench space. What is the least possible positive integer value of $N?$

$\textbf{(A) } 9 \qquad \textbf{(B) } 18 \qquad \textbf{(C) } 27 \qquad \textbf{(D) } 36 \qquad \textbf{(E) } 77$

Solution

The least common multiple of $7$ and $11$ is $77$. Therefore, there must be $77$ adults and $77$ children. The total number of benches is $\frac{77}{7}+\frac{77}{11}=11+7=\boxed{\text{(B) }18}$.


Solution 2

This is similar to Solution 1, with the same basic idea, but we don't need to calculate the LCM. Since both $7$ and $11$ are relatively prime, their LCM must be their product. So the answer would be $7 + 11 = \boxed{\text{(B) } 18}$. ~Baolan

Video Solution

Education, The Study of Everything

https://youtu.be/GKTQO99CKPM

https://youtu.be/JEjib74EmiY

~IceMatrix

https://youtu.be/w2_H96-yzk8

~savannahsolver

See Also

2020 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 8 		Followed by
Problem 10
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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