2002 AMC 12A 第 3 题

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

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

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}

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.

其他年份的第 3 题