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2024 AMC 10A Problem 19

Problem 19 of 25HarderAlgebraNumber Theory

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

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Solution

Since 7202=ab,720^2 = ab, the common ratio r=720a=b720r = \tfrac{720}{a} = \tfrac{b}{720} is rational. Write r=pqr = \tfrac{p}{q} in lowest terms with p>q.p \gt q. Then a=720qpa = \tfrac{720q}{p} and b=720pqb = \tfrac{720p}{q} are integers, which forces 720720 to be divisible by pp and by q.q. To make bb smallest, we want the smallest ratio pq>1\tfrac{p}{q} \gt 1 with 720720 divisible by both pp and q,q, which is 1615.\tfrac{16}{15}. That gives b=7201615=768b = 720 \cdot \tfrac{16}{15} = 768 (and a=675a = 675). The digit sum is 7+6+8=21.7 + 6 + 8 = 21. Thus, E is the correct answer.

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Concepts: geometric sequence · divisibility · optimization

Problem text and solution from the LIVE past-contest archive. See also the AoPS wiki page for community solutions.