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

Revision as of 21:15, 17 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

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 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

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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