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

2002 AMC 12A Problem 3 — Solution

2002 AMC 12A Problem 3

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Concepts:order of operations exponent casework

Difficulty rating: 1270

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?

$0$

$1$

$2$

$3$

$4$

Solution:

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

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.

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