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

2002 AMC 12A Problems/Problem 3

Revision as of 21:53, 9 February 2009 by Castle (talk | contribs)

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 $2^{2^{2^2}} \Rightarrow \mathrm {(B)}$

See Also

2002 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 2
Followed by
Problem 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
Invalid username
Login to AoPS