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?

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

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 by Problem 2	Followed by Problem 4
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