Question

一个等比数列的前三项是整数 $a$ 、 $720$ 、 $b$ ，且 $a \lt 720 \lt b$ 。 $b$ 的最小可能值的各位数字之和是多少？

The first three terms of a geometric sequence are the integers $a,$ $720,$ and $b,$ where $a \lt 720 \lt b.$ What is the sum of the digits of the least possible value of $b?$

$9$

$12$

$16$

$18$

$21$

答案：E

知识点：等比数列整除性最优化

难度评级：1910

解答：

因为 $720^2 = ab$ ，公比 $r = \tfrac{720}{a} = \tfrac{b}{720}$ 是有理数。设 $r = \tfrac{p}{q}$ 为最简分数，且 $p \gt q$ 。则 $a = \tfrac{720q}{p}$ 和 $b = \tfrac{720p}{q}$ 都为整数，所以 $p \mid 720$ 、 $q \mid 720$ 。为了使 $b$ 最小，需要在 $p, q \mid 720$ 中选择大于一的最小比值 $\tfrac{p}{q} \gt 1$ ，即 $\tfrac{16}{15}$ 。因此 $b = 720 \cdot \tfrac{16}{15} = 768$ ， $a = 675$ ，各位数字之和为 $7 + 6 + 8 = 21$ ，正确答案是 E。

Since $720^2 = ab,$ the common ratio $r = \tfrac{720}{a} = \tfrac{b}{720}$ is rational. Write $r = \tfrac{p}{q}$ in lowest terms with $p \gt q.$ Then $a = \tfrac{720q}{p}$ and $b = \tfrac{720p}{q}$ are integers, which forces $p \mid 720$ and $q \mid 720.$ To make $b$ smallest, we want the smallest ratio $\tfrac{p}{q} \gt 1$ with both $p, q \mid 720,$ which is $\tfrac{16}{15}.$ That gives $b = 720 \cdot \tfrac{16}{15} = 768$ (and $a = 675$ ). The digit sum is $7 + 6 + 8 = 21.$ Thus, E is the correct answer.

2024 AMC 10A 第 19 题

答案：E

难度评级：1910

解答：

其他年份的第 19 题