2020 AMC 8 Problems/Problem 20

Problem

Let $a_{1}, a_{2}, a_{3}, a_{4},$ and $a_{5}$ be five positive integers such that $a_{2}=11$ and $\frac{a_{i}}{a_{i+1}}\in\{\frac{1}{2}, 2\}$ for $1\leq n\leq 4$ . If $\{\frac{a_{1}+a_{2}+a_{3}+a_{4}+a_{5}}{5}\}=0.2$ , compute $\frac{a_{1}+a_{2}+a_{3}+a_{4}+a_{5}}{5}$ .

Note: $\{x\}$ denotes the fractional part of $x$ .

$\textbf{(A) }22.2 \qquad \textbf{(B) }24.2 \qquad \textbf{(C) }33.2 \qquad \textbf{(D) }35.2 \qquad \textbf{(E) }37.2$

Solution 1

We will show that $22$ , $11$ , $22$ , $44$ , and $22$ meters are the heights of the trees from left to right. We are given that all tree heights are integers, so since Tree 2 has height $11$ meters, we can deduce that Trees 1 and 3 both have a height of $22$ meters. There are now three possible cases for the heights of Trees 4 and 5 (in order for them to be integers), namely heights of $11$ and $22$ , $44$ and $88$ , or $44$ and $22$ . Checking each of these, in the first case, the average is $17.6$ meters, which doesn't end in $.2$ as the problem requires. Therefore, we consider the other cases. With $44$ and $88$ , the average is $37.4$ meters, which again does not end in $.2$ , but with $44$ and $22$ , the average is $24.2$ meters, which does. Consequently, the answer is $\boxed{\textbf{(B) }24.2}$ .

Solution 2

Notice the average height of the trees ends with $0.2$ therefore the sum of all five heights of the trees must end with $1$ . ( $0.2$ * $5$ = $1$ ) We already know Tree $2$ is $11$ meters tall. Both Tree $1$ and Tree $3$ must $22$ meters tall - since neither can be $5.5$ . Once again, apply our observation for solving for the Tree $4$ 's height. Tree $4$ can't be $11$ meters for the sum of the five tree heights to still end with $1$ . Therefore, the Tree $4$ is $44$ meters tall. Now the Tree $5$ can either be $22$ or $88$ . Find the average height for both cases of Tree $5$ . Doing this, we realize the Tree $5$ must be $22$ for the average height to end with $0.2$ and that the average height is $\boxed{\textbf{(B) }24.2}$ .

Solution 3

As in Solution 1, we shall show that the heights of the trees are $22$ , $11$ , $22$ , $44$ , and $22$ meters. Let $S$ be the sum of the heights, so that the average height will be $\frac{S}{5}$ meters. We note that $0.2 = \frac{1}{5}$ , so in order for $\frac{S}{5}$ to end in $.2$ , $S$ must be one more than a multiple of $5$ . Moreover, as all the heights are integers, the heights of Tree 1 and Tree 3 are both $22$ meters. At this point, our table looks as follows: $\begingroup \setlength{\tabcolsep}{10pt} \renewcommand{\arraystretch}{1.5} \begin{tabular}{|c|c|} \hline Tree 1 & 22 meters \\ Tree 2 & 11 meters \\ Tree 3 & 22 meters \\ Tree 4 & \rule{0.5cm}{0.15mm} meters \\ Tree 5 & \rule{0.5cm}{0.15mm} meters \\ \hline Average height & \rule{0.5cm}{0.15mm}\text{ .}2 meters \\ \hline \end{tabular} \endgroup$

If Tree 4 now has a height of $11$ , then Tree 5 would need to have height $22$ , but in that case $S$ would equal $88$ , which is not $1$ more than a multiple of $5$ . So we instead take Tree 4 to have height $44$ . Then the sum of the heights of the first 4 trees is $22+11+22+44 = 99$ , so using a height of $22$ for Tree 5 gives $S=121$ , which is $1$ more than a multiple of $5$ (whereas $88$ gives $S = 187$ , which is not). Thus the average height of the trees is $\frac{121}{5} = \boxed{\textbf{(B) }24.2}$ meters.

Video Solution

https://youtu.be/VnOecUiP-SA

2020 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 19	Followed by Problem 21
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 AJHSME/AMC 8 Problems and Solutions

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