Solution f4ff83fe-8d2f-4a9b-bac9-8ae938b6e317

Points: 1 pt

By the definition of a geometric sequence,
\[
720^2=a\cdot b.
\]
Rather than find the smallest factor of $720^2$ greater than $720$, we find its greatest factor less than $720$. We use the prime factorization
\[
720=2^4\cdot3^2\cdot5,
\qquad
720^2=2^8\cdot3^4\cdot5^2.
\]
Testing products of these prime factors, the largest divisors below $720$ containing zero, one, or two factors of $5$ are, respectively,
\[
2^3\cdot3^4=648,\qquad
2^7\cdot5=640,\qquad
3^3\cdot5^2=675.
\]
Hence the greatest possible value of $a$ is $675$, so the least possible value of $b$ is
\[
b=\frac{720^2}{675}=768.
\]
The sum of its digits is $7+6+8=21$.

Final answer:
21