Competition · AMC preparation · step 4 of 4
AMC 8 · 2023 · #22
Grade 6 arithmeticPick an answer.
AMC 8 2023 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
There are two unknown starting values (a₁ and a₂) tied together by a single end condition a₆ = 4000. That structure is exactly what Tool #13 (Convert to Algebra) is built for: name the unknowns a and b, propagate them through the recurrence, and read off one equation in a and b. Tool #5 (Look for a Pattern) supports this — when we multiply the terms one at a time, the exponents on a and on b each follow a Fibonacci-style pattern, and spotting that pattern saves us from a messier expansion. The final step compares prime factorizations of the two sides, which is a clean number-theory finish once the algebra has done its job.
Name the first two terms
Name the unknown starting terms: let a₁ = a and a₂ = b, both positive integers, so finding the first term means finding a.
Letting letters stand in for the unknown numbers is the Grade 6 move that lets us write one equation instead of guessing pairs.
6.EE.A.2Convert To AlgebraBuild terms up to the sixth
Apply the rule term by term; the exponents on a and b each grow in a Fibonacci-style pattern, so a₆ = a³ b⁵.
Generating each new term from a fixed rule and watching how the exponents grow is exactly the Grade 5 "generate patterns from rules" idea.
Running the rule out to the sixth term, the first term ends up used as a factor three times and the second term five times, so a₆ = a³ b⁵.
▸ Why?
Every term is the product of the two before it, so the number of first-term factors inside a term equals the sum of the first-term factor counts in the two previous terms, and the same holds for the second term; starting from counts (1,0) and (0,1) and adding the previous two each time reaches 3 and 5 by the sixth term.
▸ Why?
Multiplying one term by another simply pools all their factors into one product, so the count of first-term factors in that product is the two counts added together, and likewise for the second term.
▸ Why?
Inside a product you may regroup and reorder the factors freely, so every scattered copy of the first term can be gathered into a single running count and every copy of the second term into another.
▸ Why?
Re-bracketing which factors you multiply first never changes the product, so the loose copies can be regrouped to stand together.
▸ Why?
Swapping the order of two factors never changes the product, so like copies can be slid next to each other before they are counted.
Set the sixth term equal
Set the symbolic sixth term equal to the given value: a₆ = a³ b⁵ = 4000 gives one equation a³ b⁵ = 4000 in two positive-integer unknowns.
Reading and writing expressions with whole-number exponents like a³ and b⁵ is the Grade 6 entry point into algebraic expressions.
6.EE.A.1Convert To AlgebraFactor 4000 into primes
Break the target into primes so both sides share the same exponent shape: 4000 = 2⁵ · 5³.
Pulling out prime factors of a number is the same Grade 6 skill behind finding GCF and LCM.
6.NS.B.4Convert To AlgebraMatch the exponents
Match exponents on the unique factorization: a³ pairs with 5³ and b⁵ with 2⁵, so a = 5 and b = 2 — the first term is (D).
Matching exponents on a unique prime factorization is the Grade 6 "expressions with exponents" idea applied as a logic check.
6.EE.A.1Convert To AlgebraThis AMC 8 problem only needs Grade 6 exponents and prime factorization you already know!
- Name the first two terms
- Build terms up to the sixth
- Set the sixth term equal
- Factor 4000 into primes
- Match the exponents
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