Question

$2^{24}-1$ 有多少个正的两位数因数？

How many positive two-digit integers are factors of $2^{24}-1?$

$4$

$8$

$10$

$12$

$14$

答案：D

知识点：平方差质因数分解因数

难度评级：1740

解答：

分解得即 $63\cdot65\cdot17\cdot241$ $=3^2\cdot5\cdot7\cdot13\cdot17\cdot241$ 。 $\begin{aligned} 2^{24}-1 &=(2^{12}-1)(2^{12}+1) \\ &=(2^6-1)(2^6+1) \\ &\quad {}\cdot(2^4+1)(2^8-2^4+1), \end{aligned}$

因为 $241$ 是三位质数，两位数因数只来自 $3^2\cdot5\cdot7\cdot13\cdot17$ 。它们是共 $12$ 个。 $\begin{gathered} 13,15,17,21,35,39,45,51, \\ 63,65,85,91, \end{gathered}$

所以正确答案是 D。

Factoring, $\begin{aligned} 2^{24}-1 &=(2^{12}-1)(2^{12}+1) \\ &=(2^6-1)(2^6+1) \\ &\quad {}\cdot(2^4+1)(2^8-2^4+1), \end{aligned}$ which equals $63\cdot65\cdot17\cdot241$ $=3^2\cdot5\cdot7\cdot13\cdot17\cdot241.$

Since $241$ is a three-digit prime, the two-digit factors come from $3^2\cdot5\cdot7\cdot13\cdot17.$ They are $\begin{gathered} 13,15,17,21,35,39,45,51, \\ 63,65,85,91, \end{gathered}$ for a total of $12.$

Thus, the correct answer is D.

2011 AMC 12B 第 15 题

答案：D

难度评级：1740

解答：

其他年份的第 15 题