2002 AMC 12A Problem 3

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

There are five ways to parenthesize a tower of four

2

’s

Every parenthesization evaluates to either

2^{16}

or

2^{8}

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.

Problem 3 in other years