AMC12 2024 A

For a geometric sequence, we have $ab=720^2=2^8 3^4 5^2$, and we can test values for $b$. We find that $b=768$ and $a=675$ works, and we can test multiples of $5$ in between the two values. Finding that none of the multiples of 5 divide $720^2$ besides $720$ itself, we know that the answer is $7+6+8=\boxed{\textbf{(E) } 21}$. (Note: To find the value of $b$ without bashing, we can observe that $2^8=256$, and that multiplying it by $3$ gives us $768$, which is really close to $720$. ~ YTH) Note: The reason why $ab=720^2$ is because $b/720 = 720/a$. Rearranging this gives $ab = 720^2$ Note: Another reason that $ab=720^2$ is because the $\sqrt{ab}=720$ (as the middle term in a geometric series is always the geometric mean [the geometric mean is the square root of the product of the first and last terms of the series]) and squaring on both sides results in $ab=720^2$.