AMC 10 · 2024 · #19

Grade 6 algebranumber-theory

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

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

📏 Medium solution 💡 3 insights

Problem

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.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 10 2024 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Convert to Algebra

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 · 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.

1STEP 1

Name the common ratio r = p/q in lowest terms with p ＞ q (since r ＞ 1), giving b = 720·p/q and a = 720·q/p.

a = 720q/p, b = 720p/q, gcd(p,q) = 1, p ＞ q

2STEP 2

Since gcd(p,q) = 1, b and a being integers forces both p and q to divide 720.

p, q ∣ 720, gcd(p,q) = 1, p ＞ q

3STEP 3

Minimizing b = 720·p/q means making p/q as close to 1 as possible — the largest consecutive divisor pair of 720.

min b ⇔ min p/q ＞ 1 ⇔ maximize q with q, q+1 ∣ 720

4STEP 4

Scanning divisors of 720 = 2⁴·3²·5, the consecutive pairs top out at (15, 16).

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

5STEP 5

With r = $\frac{16}{15}$ , b = 720· $\frac{16}{15}$ = 768, whose digits sum to 21 (sequence 675, 720, 768 checks out).

b = 768, 7 + 6 + 8 = 21 → (E)

Answer

Verify the sequence directly: 675 ·

\frac{16}{15}

= 720 and 720 ·

\frac{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 ·

\frac{10}{9}

= 800 ＞ 768, and (8, 9) gives b = 720 ·

\frac{9}{8}

= 810 ＞ 768. Both are worse, confirming (15, 16) wins. Also 7 + 6 + 8 = 21 matches choice (E).

💡Key takeaway

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 · $\frac{16}{15}$ = 768.

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