2024 AMC 10A Problem 19

1. Use the geometric-sequence rule. In $a, 720, b$, the ratio is the same each step: $\dfrac{720}{a} = \dfrac{b}{720}$. Cross-multiplying gives
$$a \cdot b = 720^2 = 518400.$$
2. Turn the question around. $b = \dfrac{518400}{a}$. To make $b$ as small as possible, make $a$ as large as possible, while keeping $a$ below $720$ and $b$ a whole number. So we want the largest divisor of $518400$ that is less than $720$.
3. What do divisors of $720^2$ look like? $720 = 2^4 \cdot 3^2 \cdot 5$, so $720^2 = 2^8 \cdot 3^4 \cdot 5^2$. A divisor can only use the primes $2$, $3$, $5$: at most eight $2$s, four $3$s and two $5$s.
4. Search just below $720$, sorted by how many $5$s are used.
- Two $5$s: $a = 25 \cdot m$, where $m$ uses only $2$s and $3$s. To be just below $720$, $m$ should be just below $28.8$. The numbers made of $2$s and $3$s near there are $24$, $27$, $32$. So $m = 27$ gives $a = 675$.
- One $5$: $a = 5 \cdot m$ with $m$ just below $144$. Numbers made of $2$s and $3$s near there: $108$, $128$, $144$. $144$ gives $720$ itself (not allowed), so the best is $5 \cdot 128 = 640$.
- No $5$s: numbers made of $2$s and $3$s near $720$ are $648$ and $729$. The best below $720$ is $648$.
The largest is $a = 675$.
5. Find $b$. $b = \dfrac{518400}{675} = 768$. Check: $\dfrac{720}{675} = \dfrac{16}{15}$ and $\dfrac{768}{720} = \dfrac{16}{15}$. The sequence $675, 720, 768$ works.
6. The digits of $768$ add to $7 + 6 + 8 = 21$.

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

In any geometric sequence of three terms, the middle term squared equals the product of the outer two. That turns "smallest $b$" into "largest $a$", and $a$ must be a divisor of $720^2$, so it can only be built from the primes $2$, $3$ and $5$. The ratio does not need to be a whole number; here it is $\tfrac{16}{15}$.

Think about the ratio instead. Write the ratio as a fraction $\dfrac{p}{q}$ in lowest terms. Then $b = 720 \cdot \dfrac pq$ is a whole number only if $q$ divides $720$, and $a = 720 \cdot \dfrac qp$ is a whole number only if $p$ divides $720$. So we want two divisors of $720$ whose ratio is as close to $1$ as possible (but bigger than $1$). The divisors of $720$ near the middle include $8, 9, 10, 12, 15, 16, 18$ and $40, 45, 48$. The ratios $\frac98$, $\frac{10}{9}$ and $\frac{16}{15} \approx 1.067$ are candidates, and $\frac{16}{15}$ is the smallest. So $b = 720 \cdot \frac{16}{15} = 768$.