2024 AMC 10A Problem 19

$720^2 = ab,$ so the common ratio is rational; write it as $\tfrac{p}{q}$ in lowest terms with $p \gt q$

Both $p$ and $q$ must divide $720;$ minimize $b = 720 \cdot \tfrac{p}{q}$ using the smallest ratio $\tfrac{p}{q} \gt 1$

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