2024 AMC 10A Problem 19

Let the common ratio be $r = \dfrac{720}{a} = \dfrac{b}{720} > 1$, and write $r = \dfrac pq$ in lowest terms. Then $a = \dfrac{720q}{p}$ is an integer only if $p \mid 720$ (since $\gcd(p,q) = 1$), and $b = \dfrac{720p}{q}$ is an integer only if $q \mid 720$. Conversely any coprime divisors $p > q$ of $720$ give a valid sequence.

Since $b = 720\cdot\dfrac pq$, we want the smallest ratio $p/q > 1$ between two divisors of $720 = 2^4\cdot 3^2\cdot 5$. The divisors in order are
$$1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 16, 18, 20, 24, 30, 36, 40, 45, 48, 60, 72, 80, 90, 120, 144, 180, 240, 360, 720.$$
The smallest ratio of consecutive divisors is $\dfrac{16}{15}$ (compare $\dfrac98 = 1.125$, $\dfrac{10}{9} \approx 1.111$, $\dfrac{48}{45} = \dfrac{16}{15}$ again; every other consecutive ratio is larger). So $r = \dfrac{16}{15}$, and
$$b = 720\cdot\frac{16}{15} = 768, \qquad a = 720\cdot\frac{15}{16} = 675.$$
Check: $675, 720, 768$ has ratio $\tfrac{16}{15}$ throughout. The digit sum of $768$ is $7 + 6 + 8 = 21$.

The answer is $\boxed{\textbf{(E)}\ 21}$.

Integer geometric sequences have a rational ratio whose numerator and denominator are controlled by the middle term: $p$ must divide it (so the earlier term is an integer) and $q$ must divide it (so the later term is). Minimizing $b$ is then minimizing a ratio of divisors, which a sorted divisor list answers by inspection. Equivalently, $ab = 720^2$, so $b$ is smallest when $a$ is the largest divisor of $720^2$ below $720$, namely $675$.

• Assuming the ratio must be an integer (b = 1440, digit sum 9) or grabbing 9/8 (b = 810) without checking that 15 and 16 both divide 720.
• Reporting $b = 768$ or $a = 675$ instead of the digit sum $21$.