AMC 8 · 2020 · #20

Grade 6 number-theorylogic

Archetype Branching State Enumeration └ Sub-archetype Numeric Operation Tree

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

📏 Long solution 💡 4 insights

Problem

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.

Pick an answer.

(A)

22.2

(B)

24.2

(C)

33.2

(D)

35.2

(E)

37.2

AMC 8 2020 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Make a Systematic List

Tool #2 (Systematic List) fits perfectly because the doubling rule plus the integer constraint leaves only a small finite set of possible (H₁, H₂, H₃, H₄, H₅) sequences — we can enumerate them in order. The trick is that H₂ = 11 is odd, which forces its neighbors immediately, and then we branch from H₃ 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.

1STEP 1

Halving odd 11 gives 5.5, so both neighbors of Tree 2 must double: H₁ = H₃ = 22.

H₁ = 22, H₂ = 11, H₃ = 22

2STEP 2

From H₃ = 22, Tree 4 doubles or halves to H₄ = 11 or 44; then H₅ follows, keeping only integer values.

H₄ ∈ {11, 44}. If H₄ = 11: H₅ = 22 (since 5.5 fails). If H₄ = 44: H₅ ∈ {22, 88}.

3STEP 3

List all three valid sequences and total each: the sums are 88, 121, and 187.

Seq A: 22, 11, 22, 11, 22 → S = 88. Seq B: 22, 11, 22, 44, 22 → S = 121. Seq C: 22, 11, 22, 44, 88 → S = 187.

4STEP 4

Divide each sum by 5: only 121 ÷ 5 = 24.2 ends in .2, so 88 and 187 (17.6 and 37.4) drop out.

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

5STEP 5

Only 22, 11, 22, 44, 22 survives, giving an average height of 24.2 meters — choice (B).

Average =

\frac{121}{5}

= 24.2 m → (B)

Answer

24.2

The surviving sequence 22, 11, 22, 44, 22 obeys every doubling rule: 22 ⇔ 11 (half), 11 ⇔ 22 (double), 22 ⇔ 44 (double), 44 ⇔ 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).

💡Key takeaway

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

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