Competition · AMC preparation · step 4 of 4
AMC 8 · 2020 · #20
Grade 6 number-theorylogicPick an answer.
AMC 8 2020 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
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.
Fix the first three heights
Halving odd 11 gives 5.5, so both neighbors of Tree 2 must double: H₁ = H₃ = 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.
Both trees standing next to the 11-meter tree must be exactly 22 meters tall.
▸ Why?
A tree beside the 11-meter tree is either half of it (5.5 meters) or twice it (22 meters), and only 22 is a whole number of meters, so 22 is the only height the rule can leave.
▸ Why?
Half of 11 is 5.5 and not a whole number, because 11 is odd — it has no factor of 2, so it can never be split into two equal whole groups.
▸ Why?
Twice 11 is 22, which is a whole number, so the doubling choice gives a height that is actually allowed.
List the possible later pairs
From H₃ = 22, Tree 4 doubles or halves to H₄ = 11 or 44; then H₅ follows, keeping only integer values.
Doubling and halving small whole numbers like 11, 22, 44 is Grade 3 multiplication fluency.
3.OA.C.7Make A Systematic ListWrite each sequence and its sum
List all three valid sequences and total each: the sums are 88, 121, and 187.
Adding five two-digit numbers is the standard Grade 4 multi-digit addition skill.
4.NBT.B.4Make A Systematic ListEliminate the wrong averages
Divide each sum by 5: only 121 ÷ 5 = 24.2 ends in .2, so 88 and 187 (17.6 and 37.4) drop out.
Dividing a whole number by 5 and reading off the decimal result is Grade 5 decimal arithmetic.
5.NBT.B.7Eliminate PossibilitiesRead the surviving average
Only 22, 11, 22, 44, 22 survives, giving an average height of 24.2 meters — choice (B).
Computing the mean as sum ÷ count to summarize a data set is the Grade 6 measure-of-center idea.
6.SP.B.5Make A Systematic ListThis AMC 8 problem only needs Grade 6 averaging — sum divided by count — plus a careful list of doubling cases you already know!
- Fix the first three heights
- List the possible later pairs
- Write each sequence and its sum
- Eliminate the wrong averages
- Read the surviving average
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