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

Revision as of 21:55, 9 February 2009

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
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

@@ Line 12: / Line 12: @@
 Note that <math>2^{2^2}</math> has a unique value of <math>16</math>, because <math>2^4 = 4^2 = 16</math>
-So <math>2^{2^{2^2}}</math> can be perenthesized as either <math>2^{(2^{2^2})}=2^16</math> or <math>({2^{2^2}})^{2}=16^2</math>
+So <math>2^{2^{2^2}}</math> can be perenthesized as either <math>2^{(2^{2^2})}=2^{16}</math> or <math>({2^{2^2}})^{2}=16^2</math>
 Therefore, there is one other possible value of <math>2^{2^{2^2}} \Rightarrow \mathrm {(B)}</math>