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2002 AMC 12A Problems/Problem 3

Revision as of 18:26, 6 October 2008 by Qntty (talk | contribs) (New page: ==Problem== According to the standard convention for exponentiation, <cmath> 2^{2^{2^{2}}} = 2^{(2^{(2^2)})} = 2^{16} = 65536. </cmath> If the order in which the exponentiations are perf...)
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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?

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

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