During AMC testing, the AoPS Wiki is in read-only mode. No edits can be made.

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

## 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. Since the number of adults is equal to the number of children we can write $N$ as $\frac{x}{7}+\frac{x}{11}=N$. Simplifying we get $\frac{18x}{77} = N$ Since both $n$ and $x$ have to be positive integers, $x$ has to equal $77$. Therefore, $N=\boxed{\text{(B) }18}$ is our final answer.

## Video Solution 1

Education, The Study of Everything

~IceMatrix

~savannahsolver

~ pi_is_3.14