Question

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?

$0$

$1$

$2$

$3$

$4$

Answer: B

Concepts:order of operations exponent casework

Difficulty rating: 1270

Solution:

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 Problem 3

Answer: B

Difficulty rating: 1270

Solution:

Problem 3 in Other Years