Question

按照指数运算的标准约定， $2^{2^{2^2}} = 2^{\left(2^{\left(2^2\right)}\right)} = 2^{16} = 65{,}536\text{。}$ 如果改变指数运算的执行顺序，还可能得到多少个其他值？

According to the standard convention for exponentiation, $2^{2^{2^2}} = 2^{\left(2^{\left(2^2\right)}\right)} = 2^{16} = 65{,}536.$ If the order in which the exponentiations are performed is changed, how many other values are possible?

$0$

$1$

$2$

$3$

$4$

答案：B

知识点：运算顺序指数分类讨论

难度评级：1270

小提示：

四个 $2$ 组成的指数塔共有五种加括号方式。

There are five ways to parenthesize a tower of four $2$ ’s

大提示：

每种加括号方式的值都是 $2^{16}$ 或 $2^{8}$ 。

Every parenthesization evaluates to either $2^{16}$ or $2^{8}$

解答：

$2^{2^{2^2}}$ 的五种加括号方式给出 $\begin{aligned} &\left(\left(2^2\right)^2\right)^2 = 2^8, \\ &\left(2^{2^2}\right)^2 = 2^8, \\ &\left(2^2\right)^{2^2} = 2^8 \end{aligned}\text{，}$ $2^{\left(2^2\right)^2} = 2^{16},\quad 2^{2^{2^2}} = 2^{16}\text{。}$

因此只可能得到 $2^{16} = 65{,}536$ 和 $2^8 = 256$ 。除标准值外，恰好还有 $1$ 个其他值。

因此，正确答案是 B。

The five parenthesizations of $2^{2^{2^2}}$ give $\begin{aligned} &\left(\left(2^2\right)^2\right)^2 = 2^8, \\ &\left(2^{2^2}\right)^2 = 2^8, \\ &\left(2^2\right)^{2^2} = 2^8, \end{aligned}$ $2^{\left(2^2\right)^2} = 2^{16},\quad 2^{2^{2^2}} = 2^{16}.$

So the only values are $2^{16} = 65{,}536$ and $2^8 = 256.$ Besides the standard value there is exactly $1$ other.

Thus, the correct answer is B.

2002 AMC 12A 第 3 题

答案：B

难度评级：1270

小提示：

大提示：

解答：

其他年份的第 3 题