Question

存在唯一一个由非负整数组成的有序三元组 $(a, k, m)$ ，使得

$\begin{aligned} &\small \frac{4^a + 4^{a+k} + 4^{a+2k} + \cdots + 4^{a+mk}}{2^a + 2^{a+k} + 2^{a+2k} + \cdots + 2^{a+mk}} \\ &= 964. \end{aligned}$

$a + k + m$ 是多少？

There is a unique ordered triple $(a, k, m)$ of nonnegative integers such that

$\begin{aligned} &\small \frac{4^a + 4^{a+k} + 4^{a+2k} + \cdots + 4^{a+mk}}{2^a + 2^{a+k} + 2^{a+2k} + \cdots + 2^{a+mk}} \\ &= 964. \end{aligned}$

What is $a + k + m?$

$8$

$9$

$10$

$11$

$12$

答案：A

知识点：等比数列指数因式分解

难度评级：2130

解答：

如果 $k=0$ 或 $m=0,$ 原比值就是 $2^a,$ 不可能等于 $964.$ 所以 $k,m\gt0.$

对等比级数求和，分子为 $4^a\dfrac{4^{k(m+1)} - 1}{4^k - 1}$ ，分母为 $2^a\dfrac{2^{k(m+1)} - 1}{2^k - 1}.$ 利用 $4^N - 1 = (2^N - 1)(2^N + 1),$ 比值化简为 $2^a \cdot \frac{2^{k(m+1)} + 1}{2^k + 1} = 964.$

左边的分数是奇数除以奇数，所以它的 $2$ 的指数为 $0.$ 因为 $964=2^2\cdot241,$ 必须有 $a=2,$ 并且 $2^{k(m+1)}+1=241(2^k+1).$ 将此式模 $2^k$ 化简，得到 $2^k\mid240,$ 所以 $k\le4.$ 当 $k=1,2,3,$ 时，所需的 $2$ 的幂分别为 $722,1204,2168,$ 都不是 $2.$ 的幂。然而当 $k=4,$ 时， $241(2^4+1)=4097=2^{12}+1,$ 所以 $k(m+1)=12$ 且 $m=2.$ 这也证明了解的唯一性。

于是 $a + k + m = 2 + 4 + 2 = 8.$

因此，正确答案是 A。

If $k=0$ or $m=0,$ the original ratio is just $2^a,$ which cannot equal $964.$ Hence $k,m\gt0.$

Summing the geometric series, the numerator is $4^a\dfrac{4^{k(m+1)} - 1}{4^k - 1}$ and the denominator is $2^a\dfrac{2^{k(m+1)} - 1}{2^k - 1}.$ Using $4^N - 1 = (2^N - 1)(2^N + 1),$ the ratio simplifies to $2^a \cdot \frac{2^{k(m+1)} + 1}{2^k + 1} = 964.$

The fraction on the left is odd over odd, so its power of $2$ is $0.$ Since $964=2^2\cdot241,$ we must have $a=2,$ and $2^{k(m+1)}+1=241(2^k+1).$ Reducing this equation modulo $2^k$ shows that $2^k\mid240,$ so $k\le4.$ For $k=1,2,3,$ the required powers of $2$ would be $722,1204,2168,$ none of which is a power of $2.$ For $k=4,$ however, $241(2^4+1)=4097=2^{12}+1,$ so $k(m+1)=12$ and $m=2.$ This also proves uniqueness.

Then $a + k + m = 2 + 4 + 2 = 8.$

Thus, the correct answer is A.

2025 AMC 12A 第 21 题

答案：A

难度评级：2130

解答：

其他年份的第 21 题