AMC 8 · 2023 · #22

Grade 6 arithmetic

Archetype Multiplicative Ratio with Integrality Constraints └ Sub-archetype Integrality / Divisibility Screen

prime-factorizationexponentsrecursive-sequence pattern-recognitioncaseworkprime-factorization ↑ Prerequisites: prime-factorizationexponents

📏 Medium solution 💡 3 insights

Problem

A sequence of positive integers is built so that, from the third term on, each term equals the product of the two terms immediately before it. The sixth term of this sequence is 4000. Working backward from that piece of information, what is the very first term?

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 8 2023 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Convert to Algebra

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.

1STEP 1

Name the unknown starting terms: let a₁ = a and a₂ = b, both positive integers, so finding the first term means finding a.

a₁ = a, a₂ = b

2STEP 2

Apply the rule term by term; the exponents on a and b each grow in a Fibonacci-style pattern, so a₆ = a³ b⁵.

a₃ = ab, a₄ = a₃ · a₂ = ab², a₅ = a₄ · a₃ = a² b³, a₆ = a₅ · a₄ = a³ b⁵

3STEP 3

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.

a³ b⁵ = 4000

4STEP 4

Break the target into primes so both sides share the same exponent shape: 4000 = 2⁵ · 5³.

4000 = 2⁵ · 5³

5STEP 5

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).

a³ b⁵ = 5³ · 2⁵ → a = 5, b = 2 → a₁ = 5 (D)

Answer

Plug a₁ = 5 and a₂ = 2 back into the sequence and march forward: a₃ = 5 · 2 = 10, a₄ = 2 · 10 = 20, a₅ = 10 · 20 = 200, a₆ = 20 · 200 = 4000. The sixth term comes out exactly 4000, so the answer a₁ = 5 is consistent with every condition in the problem. The other answer choices fail this forward check: for example, a₁ = 1 would force a³ b⁵ = b⁵ = 4000, but 4000 is not a perfect fifth power.

💡Key takeaway

This AMC 8 problem only needs Grade 6 exponents and prime factorization you already know!

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