AMC 8 · 2020 · #20

Grade 6 number-theorylogic

modular-arithmeticparitymean-median-mode-rangesequences-geometric caseworktree-enumeration ↑ Prerequisites: modular-arithmeticparity

📏 Long solution 💡 4 insights

Problem

A scientist walking through a forest recorded as integers the heights of $5$ trees standing in a row. She observed that each tree was either twice as tall or half as tall as the one to its right. Unfortunately some of her data was lost when rain fell on her notebook. Her notes are shown below, with blanks indicating the missing numbers. Based on her observations, the scientist was able to reconstruct the lost data. What was the average height of the trees, in meters?

$\begingroup \setlength{\tabcolsep}{10pt} \renewcommand{\arraystretch}{1.5} \begin{tabular}{|c|c|} \hline Tree 1 & \rule{0.4cm}{0.15mm} meters \\ Tree 2 & 11 meters \\ Tree 3 & \rule{0.5cm}{0.15mm} meters \\ Tree 4 & \rule{0.5cm}{0.15mm} meters \\ Tree 5 & \rule{0.5cm}{0.15mm} meters \\ \hline Average height & \rule{0.5cm}{0.15mm}\text{ .}2 meters \\ \hline \end{tabular} \endgroup$
$\textbf{(A) }22.2 \qquad \textbf{(B) }24.2 \qquad \textbf{(C) }33.2 \qquad \textbf{(D) }35.2 \qquad \textbf{(E) }37.2$

Pick an answer.

(A)

22.2

(B)

24.2

(C)

33.2

(D)

35.2

(E)

37.2

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Toolkit + CCSS Solution

Understand

Restated: A scientist measured the integer heights (in meters) of $5$ trees in a row, where every tree's height is either double or half the height of the tree immediately to its right. Only Tree 2's height ($11$ meters) survived; the others got rained on. She also wrote down the average height of the $5$ trees, and the only thing readable from it is that the average ends in $.2$ (some integer followed by $.2$ meters). From these clues, find the average height.

Givens: Five trees in a row with integer heights $H_1, H_2, H_3, H_4, H_5$ (in meters); $H_2 = 11$; For each adjacent pair, one tree is exactly $2\times$ the other (equivalently, the right tree is $2\times$ or $\tfrac{1}{2}\times$ the left); The average $\tfrac{H_1+H_2+H_3+H_4+H_5}{5}$ has a decimal part of $.2$; Answer choices: (A) $22.2$, (B) $24.2$, (C) $33.2$, (D) $35.2$, (E) $37.2$

Unknowns: The average height (in meters) of the five trees

Understand

Plan

Primary tool: #2 Make a Systematic List

Secondary: #3 Eliminate Possibilities

Tool #2 (Systematic List) fits perfectly because the doubling rule plus the integer constraint leaves only a *small finite* set of possible $(H_1, H_2, H_3, H_4, H_5)$ sequences — we can enumerate them in order. The trick is that $H_2 = 11$ is odd, which forces its neighbors immediately, and then we branch from $H_3$ onward. After listing all valid integer sequences, Tool #3 (Eliminate Possibilities) lets us discard the ones whose average doesn't end in $.2$. We avoid Tool #13 (Algebra) because the listing is short and an elementary student can do it without setting up equations.

Execute — Answer: B

#3 Eliminate Possibilities 4.OA.B.4 Step 1

Pin down $H_1$ and $H_3$ from $H_2 = 11$.
Each neighbor of $H_2$ is either $2 \times 11 = 22$ or $\tfrac{1}{2} \times 11 = 5.5$.
Since $5.5$ isn't an integer, both $H_1$ and $H_3$ are forced to be $22$.

$$H_1 = 22, \; H_2 = 11, \; H_3 = 22$$

💡 Halving an odd number like $11$ gives a non-integer, so the only integer neighbor of $11$ in a doubling chain is $22$ — Grade 4 factor reasoning.

#2 Make a Systematic List 3.OA.C.7 Step 2

List the possible $(H_4, H_5)$ pairs systematically.
From $H_3 = 22$, $H_4$ is either $44$ or $11$.
From each $H_4$, $H_5$ is either $2 H_4$ or $\tfrac{1}{2} H_4$ — keeping only integers.

$H_4 \in \{11, 44\}$. \;If $H_4 = 11$: $H_5 = 22$ (since $5.5$ fails). \;If $H_4 = 44$: $H_5 \in \{22, 88\}$.

💡 Doubling and halving small whole numbers like $11, 22, 44$ is Grade 3 multiplication fluency.

#2 Make a Systematic List 4.NBT.B.4 Step 3

Write out the complete list of valid integer sequences and compute each sum.
There are exactly three candidate sequences after step 2.

Seq A: $22, 11, 22, 11, 22 \Rightarrow S = 88$. \;Seq B: $22, 11, 22, 44, 22 \Rightarrow S = 121$. \;Seq C: $22, 11, 22, 44, 88 \Rightarrow S = 187$.

💡 Adding five two-digit numbers is the standard Grade 4 multi-digit addition skill.

#3 Eliminate Possibilities 5.NBT.B.7 Step 4

Eliminate sequences whose average does NOT end in $.2$.
Divide each sum by $5$ and check the decimal part.

Seq A: $88 \div 5 = 17.6$ (ends in $.6$ — eliminated). \;Seq B: $121 \div 5 = 24.2$ (ends in $.2$ — KEEP). \;Seq C: $187 \div 5 = 37.4$ (ends in $.4$ — eliminated).

💡 Dividing a whole number by $5$ and reading off the decimal result is Grade 5 decimal arithmetic.

#2 Make a Systematic List 6.SP.B.5 Step 5

Only Sequence B survives, so the trees are $22, 11, 22, 44, 22$ and the average is $24.2$ meters, matching choice (B).

$$\text{Average} = \dfrac{121}{5} = 24.2 \text{ m} \;\Rightarrow\; \textbf{(B)}$$

💡 Computing the mean as $\text{sum} \div \text{count}$ to summarize a data set is the Grade 6 measure-of-center idea.

[1] #3 4.OA.B.4 Pin down $H_1$ and $H_3$ from $H_2 = 11$. Each neighbor of $H_2$ is either $2 \t

[2] #2 3.OA.C.7 List the possible $(H_4, H_5)$ pairs systematically. From $H_3 = 22$, $H_4$ is e

[3] #2 4.NBT.B.4 Write out the complete list of valid integer sequences and compute each sum. The

[4] #3 5.NBT.B.7 Eliminate sequences whose average does NOT end in $.2$. Divide each sum by $5$ a

[5] #2 6.SP.B.5 Only Sequence B survives, so the trees are $22, 11, 22, 44, 22$ and the average

Review

Reasonableness: The surviving sequence $22, 11, 22, 44, 22$ obeys every doubling rule: $22 \leftrightarrow 11$ (half), $11 \leftrightarrow 22$ (double), $22 \leftrightarrow 44$ (double), $44 \leftrightarrow 22$ (half). All heights are positive integers. The average $24.2$ ends in $.2$ as the problem requires. Among the answer choices, $24.2$ is the only one we can actually realize with this rule — sequences giving $22.2$, $33.2$, $35.2$, or $37.2$ would require non-integer heights or break the doubling chain, so the multiple-choice format also corroborates (B).

Alternative: Use Tool #16 (Change Focus / Count the Complement) on the answer choices themselves: an average of $N.2$ means the sum is $5N + 1$. So the sums implied by (A)–(E) are $111, 121, 166, 176, 187$. Only $121$ and $187$ are realizable by some valid doubling chain starting with $H_1, 11, H_3 = 22, 22$ — and $187$ requires $H_5 = 88$, which forces an unmatched halving chain that fails the integer test for some neighbor — leaving $121 \Rightarrow$ average $24.2 = $ (B). This back-solving is faster on contest day.

CCSS standards used (min grade 6)

4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite (Recognizing that $11$ is odd (not divisible by $2$), which forces both $H_1$ and $H_3$ to be $22$ rather than $5.5$.)
3.OA.C.7 Fluently multiply and divide within 100 (Doubling and halving small whole numbers ($11 \times 2 = 22$, $22 \times 2 = 44$, $44 \times 2 = 88$) to generate the candidate sequences.)
4.NBT.B.4 Fluently add and subtract multi-digit whole numbers (Adding the five tree heights in each candidate sequence to obtain the sums $88$, $121$, and $187$.)
5.NBT.B.7 Add, subtract, multiply, and divide decimals to hundredths (Dividing each candidate sum by $5$ to get $17.6$, $24.2$, $37.4$ and reading off the decimal part to test the $.2$ condition.)
6.SP.B.5 Summarize numerical data sets by reporting number of observations and measures (Interpreting and computing the mean (average height) of the five tree heights as a single summary statistic.)

⭐ This AMC 8 problem only needs Grade 6 averaging — sum divided by count — plus a careful list of doubling cases you already know!

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: B

Review

CCSS standards used (min grade 6)

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