AMC 8 · 2023 · #22

Grade 6 arithmetic

Archetype Multiplicative Ratio with Integrality Constraints

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

📏 Medium solution 💡 3 insights

Problem

In a sequence of positive integers, each term after the second is the product of the previous two terms. The sixth term is $4000$ . What is the first term?

Pick an answer.

(A)

(B)

(C)

(D)

(E)

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Toolkit + CCSS Solution

Understand

Restated: 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?

Givens: All terms are positive integers; Recurrence: for $n \ge 3$, $a_n = a_{n-1} \cdot a_{n-2}$; The sixth term satisfies $a_6 = 4000$; Answer choices: (A) $1$, (B) $2$, (C) $4$, (D) $5$, (E) $10$

Unknowns: The value of the first term $a_1$

Understand

Plan

Primary tool: #13 Convert to Algebra

Secondary: #5 Look for a Pattern

There are two unknown starting values ($a_1$ and $a_2$) tied together by a single end condition $a_6 = 4000$. 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.

Execute — Answer: D

#13 Convert to Algebra 6.EE.A.2 Step 1

Name the two unknown starting terms.
Let $a_1 = a$ and $a_2 = b$, where $a$ and $b$ are positive integers.
Every later term is determined by these two choices, so finding the first term means finding $a$.

$$a_1 = a, \quad a_2 = b$$

💡 Letting letters stand in for the unknown numbers is the Grade 6 move that lets us write one equation instead of guessing pairs.

#5 Look for a Pattern 5.OA.B.3 Step 2

Apply the recurrence $a_n = a_{n-1} \cdot a_{n-2}$ term by term and keep track of how many copies of $a$ and how many copies of $b$ appear in each product.
The exponent on $a$ and the exponent on $b$ each follow the Fibonacci-style pattern $0,1,1,2,3,5,\dots$ and $1,0,1,1,2,3,\dots$.

$$a_3 = ab, \;\; a_4 = a_3 \cdot a_2 = ab^2, \;\; a_5 = a_4 \cdot a_3 = a^2 b^3, \;\; a_6 = a_5 \cdot a_4 = a^3 b^5$$

💡 Generating each new term from a fixed rule and watching how the exponents grow is exactly the Grade 5 "generate patterns from rules" idea.

#13 Convert to Algebra 6.EE.A.1 Step 3

Set the symbolic sixth term equal to the given value.
Since $a_6 = a^3 b^5$ and we are told $a_6 = 4000$, we get one equation in two positive-integer unknowns.

$$a^3 b^5 = 4000$$

💡 Reading and writing expressions with whole-number exponents like $a^3$ and $b^5$ is the Grade 6 entry point into algebraic expressions.

#13 Convert to Algebra 6.NS.B.4 Step 4

Break $4000$ into prime factors so the right-hand side has the same "exponent shape" as the left.
Repeatedly divide by $2$, then by $5$: $4000 = 2 \cdot 2000 = 2^2 \cdot 1000 = 2^2 \cdot 10^3 = 2^2 \cdot (2 \cdot 5)^3 = 2^5 \cdot 5^3$.

$$4000 = 2^5 \cdot 5^3$$

💡 Pulling out prime factors of a number is the same Grade 6 skill behind finding GCF and LCM.

#13 Convert to Algebra 6.EE.A.1 Step 5

Match the equation $a^3 b^5 = 5^3 \cdot 2^5$ by exponent.
Because both sides must have the same prime factorization, the base raised to the $3$ on the left must equal the base raised to the $3$ on the right, and similarly for the $5$.
That forces $a = 5$ and $b = 2$.
The question asks for the first term, so the answer is $a = 5$, which is choice $(\textbf{D})$.

$$a^3 b^5 = 5^3 \cdot 2^5 \;\Rightarrow\; a = 5,\; b = 2 \;\Rightarrow\; a_1 = 5 \;\;(\textbf{D})$$

💡 Matching exponents on a unique prime factorization is the Grade 6 "expressions with exponents" idea applied as a logic check.

[1] #13 6.EE.A.2 Name the two unknown starting terms. Let $a_1 = a$ and $a_2 = b$, where $a$ and

[2] #5 5.OA.B.3 Apply the recurrence $a_n = a_{n-1} \cdot a_{n-2}$ term by term and keep track o

[3] #13 6.EE.A.1 Set the symbolic sixth term equal to the given value. Since $a_6 = a^3 b^5$ and

[4] #13 6.NS.B.4 Break $4000$ into prime factors so the right-hand side has the same "exponent sh

[5] #13 6.EE.A.1 Match the equation $a^3 b^5 = 5^3 \cdot 2^5$ by exponent. Because both sides mus

Review

Reasonableness: Plug $a_1 = 5$ and $a_2 = 2$ back into the sequence and march forward: $a_3 = 5 \cdot 2 = 10$, $a_4 = 2 \cdot 10 = 20$, $a_5 = 10 \cdot 20 = 200$, $a_6 = 20 \cdot 200 = 4000$. The sixth term comes out exactly $4000$, so the answer $a_1 = 5$ is consistent with every condition in the problem. The other answer choices fail this forward check: for example, $a_1 = 1$ would force $a^3 b^5 = b^5 = 4000$, but $4000$ is not a perfect fifth power.

Alternative: Tool #3 (Eliminate Possibilities) gives a quick multiple-choice attack: $a^3 b^5 = 4000 = 2^5 \cdot 5^3$ forces $a$ to be a cube root of $5^3 \cdot (\text{something})$ and $b$ to be a fifth root of $2^5 \cdot (\text{something})$. Testing each choice for $a$ against the prime factorization, only $a = 5$ leaves a clean $b^5 = 2^5$, eliminating (A), (B), (C), (E) and confirming (D).

CCSS standards used (min grade 6)

5.OA.B.3 Generate two numerical patterns using two given rules and identify relationships (Building the sequence term by term from the recurrence $a_n = a_{n-1} \cdot a_{n-2}$ and watching the exponents on $a$ and on $b$ grow in a Fibonacci-style pattern.)
6.EE.A.1 Write and evaluate numerical expressions involving whole-number exponents (Writing $a_6 = a^3 b^5$ with whole-number exponents and then matching exponents on $a^3 b^5 = 2^5 \cdot 5^3$ to read off $a$ and $b$.)
6.EE.A.2 Write, read, and evaluate expressions in which letters stand for numbers (Letting $a$ and $b$ stand for the unknown first two terms so the sixth term can be expressed as a single algebraic expression $a^3 b^5$.)
6.NS.B.4 Find greatest common factor and least common multiple of two numbers (Performing the prime factorization $4000 = 2^5 \cdot 5^3$ — the same prime-factor decomposition behind GCF and LCM work.)

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

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: D

Review

CCSS standards used (min grade 6)

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