AMC 10 · 2024 · #19

Grade 6 algebranumber-theory

Archetype Multiplicative Ratio with Integrality Constraints

sequences-geometricratio-proportionfactorsdigit-sum convert-to-algebrasystematic-enumerationoptimization-counting ↑ Prerequisites: sequences-geometricfraction-arithmeticfactors

📏 Medium solution 💡 3 insights

Problem

The first three terms of a geometric sequence are the integers $a,\,720,$ and $b,$ where $a<720<b.$ What is the sum of the digits of the least possible value of $b?$

$\textbf{(A) } 9 \qquad \textbf{(B) } 12 \qquad \textbf{(C) } 16 \qquad \textbf{(D) } 18 \qquad \textbf{(E) } 21$

Pick an answer.

(A)

(B)

(C)

(D)

(E)

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Toolkit + CCSS Solution

Understand

Restated: Three integers form a geometric sequence: $a$, then $720$, then $b$, with $a < 720 < b$. Among all valid choices, find the smallest possible $b$ and report the sum of its digits.

Givens: Middle term of the geometric sequence is $720$; Outer terms $a$ and $b$ are integers with $a < 720 < b$; Answer choices: (A) $9$, (B) $12$, (C) $16$, (D) $18$, (E) $21$

Unknowns: The least possible integer value of $b$; The sum of the digits of that least $b$

Understand

Restated: Three integers form a geometric sequence: $a$, then $720$, then $b$, with $a < 720 < b$. Among all valid choices, find the smallest possible $b$ and report the sum of its digits.

Givens: Middle term of the geometric sequence is $720$; Outer terms $a$ and $b$ are integers with $a < 720 < b$; Answer choices: (A) $9$, (B) $12$, (C) $16$, (D) $18$, (E) $21$

Plan

Primary tool: #13 Convert to Algebra

Secondary: #2 Systematic List

The words "geometric sequence" cry out for Tool #13 (Convert to Algebra): name the common ratio $r$, write $r = p/q$ in lowest terms, and the integer constraints on $a$ and $b$ collapse into a clean number-theory statement — both $p$ and $q$ must be divisors of $720$. To minimize $b = 720 \cdot p/q$ with $p > q$, we need $p/q$ as close to $1$ as possible, i.e. consecutive integers. Tool #2 (Systematic List) then sweeps through pairs of consecutive divisors of $720$ — small list, only need the largest pair. No need for full algebra of inequalities; the divisor structure does the work.

Execute — Answer: E

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

Set up the algebra.
Let $r$ be the common ratio, so $720 = a \cdot r$ and $b = 720 \cdot r$.
Write $r = p/q$ in lowest terms, with $p > q$ since $r > 1$.
Then $b = 720 \cdot p/q$ and $a = 720 \cdot q/p$.

$$a = \dfrac{720q}{p}, \quad b = \dfrac{720p}{q}, \quad \gcd(p,q) = 1, \quad p > q$$

💡 Naming the unknown ratio with a letter — and pinning it down as a reduced fraction — is the Grade 6 "letters stand for numbers" move that turns a vague "some sequence" into two clean formulas.

#13 Convert to Algebra 6.NS.B.4 Step 2

Translate the integer condition.
For $b = 720p/q$ to be an integer with $\gcd(p,q) = 1$, $q$ must divide $720$.
By the same logic on $a = 720q/p$, $p$ must divide $720$.
So $p$ and $q$ are coprime divisors of $720$ with $p > q$.

$$p, q \mid 720, \quad \gcd(p,q) = 1, \quad p > q$$

💡 Coprime plus divides-the-product is the GCF rule from Grade 6: if $q$ shares no factor with $p$, then $q$ has to take all its factors from $720$ alone.

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

Minimize $b$.
Since $b = 720 \cdot \dfrac{p}{q}$ and $720$ is fixed, minimizing $b$ (subject to $b > 720$) is the same as minimizing $p/q$ above $1$.
The ratio $(q+1)/q = 1 + 1/q$ shrinks as $q$ grows, so the smallest $p/q > 1$ comes from the largest pair of consecutive divisors of $720$.

$$\min b \iff \min \dfrac{p}{q} > 1 \iff \text{maximize } q \text{ with } q, q+1 \mid 720$$

💡 Consecutive integers are automatically coprime ($\gcd(n, n+1) = 1$), so the coprime condition comes for free — we just need both to divide $720$.

#2 Systematic List 4.OA.B.4 Step 4

Systematic list of consecutive divisor pairs of $720 = 2^4 \cdot 3^2 \cdot 5$.
Scan divisors and check whether the next integer is also a divisor.

$$(q, q+1) \in \{(1,2), (2,3), (3,4), (4,5), (5,6), (8,9), (9,10), (15,16)\}$$

💡 After $(15, 16)$ the next candidates would need $q \ge 16$ with $q+1$ also dividing $720$. The divisors of $720$ jump from $16$ to $18$ to $20$ to $24$ — no more consecutive pairs. So $(15, 16)$ is the winner.

#13 Convert to Algebra 6.NS.B.2 Step 5

Compute the answer.
The largest consecutive divisor pair is $(15, 16)$, so the optimal ratio is $r = 16/15$.
Then $b = 720 \cdot 16/15 = 48 \cdot 16 = 768$.
The corresponding $a = 720 \cdot 15/16 = 45 \cdot 15 = 675$, and the sequence $675, 720, 768$ checks out.

$$b = 768, \quad 7 + 6 + 8 = 21 \;\Rightarrow\; \textbf{(E)}$$

💡 Plugging the optimal $(p, q) = (16, 15)$ back into $b = 720p/q$ is just division and multiplication — divide $720$ by the small factor $15$ first to keep the arithmetic light.

[1] #13 6.EE.A.2 Set up the algebra. Let $r$ be the common ratio, so $720 = a \cdot r$ and $b = 7

[2] #13 6.NS.B.4 Translate the integer condition. For $b = 720p/q$ to be an integer with $\gcd(p,

[3] #13 6.RP.A.1 Minimize $b$. Since $b = 720 \cdot \dfrac{p}{q}$ and $720$ is fixed, minimizing

[4] #2 4.OA.B.4 Systematic list of consecutive divisor pairs of $720 = 2^4 \cdot 3^2 \cdot 5$. S

[5] #13 6.NS.B.2 Compute the answer. The largest consecutive divisor pair is $(15, 16)$, so the o

Review

Reasonableness: Verify the sequence directly: $675 \cdot \tfrac{16}{15} = 720$ and $720 \cdot \tfrac{16}{15} = 768$. All three are integers, and $675 < 720 < 768$, so the constraints hold. Sanity-check minimality: the next-best consecutive pair is $(9, 10)$, giving $b = 720 \cdot 10/9 = 800 > 768$, and $(8, 9)$ gives $b = 720 \cdot 9/8 = 810 > 768$. Both are worse, confirming $(15, 16)$ wins. Also $7 + 6 + 8 = 21$ matches choice (E).

Alternative: Tool #6 (Guess and Check) on the answer choices: the digit sum being $21$ for a $3$-digit $b$ in the range $720 < b < 1000$ forces $b \in \{768, 786, 858, 876, 939, 948, 984\}$ etc. Check $768$ first since it is closest to $720$: $720^2 / 768 = 518400 / 768 = 675$, an integer — so $(a, b) = (675, 768)$ works, and we are done since nothing in $(720, 768)$ has digit sum matching any other listed choice with a valid integer $a$. Same answer (E).

CCSS standards used (min grade 6)

6.EE.A.2 Write, read, and evaluate expressions in which letters stand for numbers (Naming the common ratio $r = p/q$ in lowest terms so that $b = 720p/q$ and $a = 720q/p$ become clean algebraic expressions.)
6.NS.B.4 Find greatest common factor and least common multiple of two numbers (Using $\gcd(p, q) = 1$ to conclude that both $p$ and $q$ must individually divide $720$ for $a$ and $b$ to be integers.)
6.RP.A.1 Understand the concept of a ratio and use ratio language (Recognizing that minimizing $b$ above $720$ is the same as minimizing the ratio $p/q$ above $1$, which points to consecutive-integer ratios.)
4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite (Listing divisors of $720 = 2^4 \cdot 3^2 \cdot 5$ and scanning for the largest pair of consecutive divisors, $(15, 16)$.)
6.NS.B.2 Fluently divide multi-digit numbers using the standard algorithm (Computing $b = 720 \cdot 16 / 15 = 48 \cdot 16 = 768$ and confirming $a = 720 \cdot 15 / 16 = 675$.)

⭐ When a geometric sequence has to land on integers, the common ratio is a fraction $p/q$ in lowest terms where both $p$ and $q$ divide the middle term. To squeeze $b$ as close to $720$ as possible, find the largest pair of consecutive divisors of $720$ — that pair is $(15, 16)$, so $b = 720 \cdot 16/15 = 768$.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: E

Review

CCSS standards used (min grade 6)

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