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

(Video Solution 3)
(Solution 2)
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== Solution 2 ==
 
== Solution 2 ==
  
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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Let <math>x</math> denote how many adults there are. Then we can write <math>N</math> as <math>x/7</math> + <math>x/11</math> = <math>N</math>. Simplifying we get <math>18x/77 = N</math>
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Since both <math>n</math> and <math>x</math> have to be integers, <math>x</math> has to equal 77. Therefore, <math>N=18.</math>
  
 
==Video Solution 1==
 
==Video Solution 1==

Revision as of 17:27, 24 January 2021

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 1

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 $\frac{77}{7}+\frac{77}{11}=11+7=\boxed{\text{(B) }18}$.~taarunganesh

Solution 2

Let $x$ denote how many adults there are. Then we can write $N$ as $x/7$ + $x/11$ = $N$. Simplifying we get $18x/77 = N$ Since both $n$ and $x$ have to be integers, $x$ has to equal 77. Therefore, $N=18.$

Video Solution 1

Education, The Study of Everything

https://youtu.be/GKTQO99CKPM

Video Solution 2

https://youtu.be/JEjib74EmiY

~IceMatrix

Video Solution 3

https://youtu.be/w2_H96-yzk8

~savannahsolver

Video Solution 4

https://youtu.be/ZhAZ1oPe5Ds?t=1616

~ pi_is_3.14

See Also

2020 AMC 10A (ProblemsAnswer KeyResources)
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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