2024 AMC 12A Problem 12

The middle term gives $720^2=ab,$ so $ab=518400$

To minimize $b,$ take the largest divisor $a\lt720$ of $518400$

Since the terms are geometric, $720^2=ab,$ so $ab=518400=2^8\cdot3^4\cdot5^2.$ Because $b=\frac{518400}{a},$ minimizing $b$ means finding the smallest divisor of $518400$ greater than $720.$ There is no divisor strictly between $720$ and $768:$ if its exponent of $5$ is $2,1,$ or $0,$ then after dividing by $25,5,$ or $1,$ respectively, a number of the form $2^i3^j$ would have to lie in $(28.8,30.72),\ (144,153.6),$ or $(720,768);$ the allowed powers $i\le8,j\le4$ give none. Since $768=2^8\cdot3$ is a divisor, it is the least possible $b.$ Its paired divisor is $a=\frac{518400}{768}=675,$ and the digit sum of $b$ is $7+6+8=21.$ Thus, the correct answer is E.