During AMC testing, the AoPS Wiki is in read-only mode. No edits can be made.

# Difference between revisions of "2002 AMC 12A Problems/Problem 3"

## Problem

According to the standard convention for exponentiation, $$2^{2^{2^{2}}} = 2^{(2^{(2^2)})} = 2^{16} = 65536.$$

If the order in which the exponentiations are performed is changed, how many other values are possible?

$\mathrm{(A) \ } 0\qquad \mathrm{(B) \ } 1\qquad \mathrm{(C) \ } 2\qquad \mathrm{(D) \ } 3\qquad \mathrm{(E) \ } 4$

## Solution

Note that $2^{2^2}$ has a unique value of $16$, because $2^4 = 4^2 = 16$

So $2^{2^{2^2}}$ can be perenthesized as either $2^{(2^{2^2})}=2^16$ or $({2^{2^2}})^{2}=16^2$

Therefore, there is one other possible value of $2^{2^{2^2}} \Rightarrow \mathrm {(B)}$

 2002 AMC 12A (Problems • Answer Key • Resources) Preceded byProblem 2 Followed byProblem 4 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 12 Problems and Solutions